Statistical Inference

Problem 101

Rachel, a city employee, would like to make the claim that the average amount that residents spend per month on public transit fare is less than $\$ 140$ . Rachel samples 25 city residents and obtains a sample mean of $\$ 125.80$ spent per month on public transit.
At the $5 \%$ significance level, should Rachel reject or fail to reject the null hypothesis given the sample data below? - $H_{0}: \mu=\$ 140$ per month; $H_{a}: \mu<\$ 140$ per month - $\alpha=0.05$ (significance level) - test statistic $=-0.52$
Use the graph below to select the type of test (left-, right-, or two-tailed). Then set the $\alpha$ and the test statistic to determine the $p$ -value. Use the results to determine whether to reject or fail to reject the null hypothesis.
Alternatively, find the $p$ -value using the table below: \begin{tabular}{|l|l|l|l|l|l|l|l|l|l|l|} \hline $Z$ & 0.00 & 0.01 & 0.02 & 0.03 & 0.04 & 0.05 & 0.06 & 0.07 & 0.08 & 0.09 \\ \hline-0.9 & 0.184 & 0.181 & 0.179 & 0.176 & 0.174 & 0.171 & 0.169 & 0.166 & 0.164 & 0.161 \\ \hline-0.6 & 0.274 & 0.271 & 0.268 & 0.264 & 0.261 & 0.258 & 0.255 & 0.251 & 0.248 & 0.245 \\ \hline-0.5 & 0.309 & 0.305 & 0.302 & 0.298 & 0.295 & 0.291 & 0.288 & 0.284 & 0.281 & 0.278 \\ \hline-0.4 & 0.345 & 0.341 & 0.337 & 0.334 & 0.330 & 0.326 & 0.323 & 0.319 & 0.316 & 0.312 \\ \hline-0.3 & 0.382 & 0.378 & 0.374 & 0.371 & 0.367 & 0.363 & 0.359 & 0.356 & 0.352 & 0.348 \\ \hline \end{tabular}
Select the correct answer below: Reject the null hypothesis because the value of $z$ is negative. Do not reject the null hypothesis because $|-0.52|>0.05$ . Reject the null hypothesis because the $p$ -value 0.3015 is greater than the significance level $\alpha=0.05$ . Do not reject the null hypothesis because the $p$ -value 0.3015 is greater than the significance level $\alpha=0.05$ . Reject the null hypothesis because $|-0.52|>0.05$ .

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Problem 102

The recommended daily amount of vitamin C for an adult is 80 milligrams. A pharmacist would like to test the claim that the average amount of daily vitamin C intake is different than the generally accepted amount of 80 milligrams. To test this claim, at the $1 \%$ significance level, the pharmacist collects the following data on a sample of 40 adults and records the daily intake of vitamin C. The following is the data from this study:
Sample size $=40$ adults Sample mean $=85$ milligrams From past data, it is known that the population standard deviation is 14 milligrams.
Identify the null and alternative hypothesis for this study by filling in the blanks with the correct symbol ( $=, \neq,<$ , or $>$ to represent the correct hypothesis.)
Provide your answer below: null hypothesis : $\mu \square 80$ alternative hypothesis : $\mu \square 80$

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Problem 103

Polyanne, a hospital administrator, noticed a significant change in the weight of babies born at the hospital during the last few months. She randomly selected 20 babies born in the last month and found the weights. Historically, the weight of babies born at the hospital is normally distributed with a population mean weight of 7.63 pounds and a population standard deviation of 1.02 pounds. To test the weights, Polyanne decided to use the historical standard deviation as the population standard deviation. The hospital administrator conducts a one-mean hypothesis at the $5 \%$ significance level, to test if there was a change in the weight of babies being born at the hospital in the last few months from 7.63 pounds. (a) Which answer choice shows the correct null and alternative hypotheses for this test?
Select the correct answer below: $H_{0}: \mu=7.63 ; H_{a}: \mu>7.63$ , which is a right-tailed test. $H_{0}: \mu=1.02 ; H_{a}: \mu<1.02$ , which is a left-tailed test. $H_{0}: \mu=1.02 ; H_{a}: \mu \neq 1.02$ , which is a two-tailed test. $H_{0}: \mu=7.63 ; H_{a}: \mu \neq 7.63$ , which is a two-tailed test.

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Problem 104

Determine whether the statement makes sense or does not make sense, and explain your reasoning. The death rate from this new strain of flu is catastrophic because $25 \%$ of the people hospitalized with the disease have died.
Choose the correct answer below. A. This does not make sense because it is not clear whether the facts are derived from a sample of patients which is representative of the entire population of the patients suffering from the disease. B. This does not make sense because the percentage of the hospitalized people is less than $50 \%$ . C. This makes sense because the percentage of the hospitalized people is less than $50 \%$ . D. This makes sense because the sample of patients is representative of the entire population of patients suffering from the disease.

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Problem 105

Students in a large lecture class want to know who has, on average, more Facebook friends, male or female students. The data for the students are displayed in the provided dotplots and summary statistics are available in the provided table. \begin{tabular}{|l|c|c|c|} \hline Gender & $n$ & $\bar{x}$ & $s$ \\ \hline Male & 68 & 678.9 & 317.6 \\ \hline Female & 91 & 616.1 & 330.6 \\ \hline \end{tabular}
Construct a 95\% confidence interval for the difference in mean number of Facebook friends for male and female s,tudents at this university. Use two decimal places in your margin of error. -40.62 to 166.22 -38.75 to 164.35 -33.95 to 159.55 -22.43 to 148.03

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Problem 106

In a random sample of 16 residents of the state of Washington, the mean waste recycled per person per day was 2.8 pounds with a standard deviation of 0.24 pounds. Determine the $80 \%$ confidence interval for the mean waste recycled per person per day for the population of Washington. Assume the population is approximately normal.
Step 2 of 2: Construct the $80 \%$ confidence interval. Round your answer to one decimal place.
Answer How to enter your answer (opens in new window) Tables Keypad Keyboard Shortcuts Previous step answers
Lower endpoint: $\square$ Upper endpoint: $\square$

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Problem 107

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A group of students estimated the length of one minute without reference to a watch or clock, and the times (seconds) are listed below. Use a 0.01 significance level to test the claim that these times are from a population with a mean equal to 60 seconds. Does it appear that students are reasonably good at estimating one minute? $\begin{array}{lllllllllllllll} 76 & 93 & 47 & 76 & 48 & 35 & 72 & 70 & 77 & 59 & 74 & 80 & 102 & 95 & 71 \end{array}$
Assuming all conditions for conducting a hypothesis test are met, what are the null and alternative hypotheses? A. $\mathrm{H}_{0}: \mu=60$ seconds B. $H_{0}: \mu=60$ seconds $H_{1}: \mu>60$ seconds $H_{1}: \mu \neq 60$ seconds C. $\mathrm{H}_{0}: \mu=60$ seconds D. $H_{0}: \mu \neq 60$ seconds $\mathrm{H}_{1}: \mu<60$ seconds $H_{1}: \mu=60$ seconds
Determine the test statistic. (Round to two decimal places as needed.) Determine the P -value. (Round to three decimal places as needed.) State the final conclusion that addresses the original claim.

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Problem 108

For a hypothesis test of the claim that the mean amount of sleep for adults is less than 6 hours, technology output shows that the hypothesis test has power of 0.4945 of supporting the claim that $\mu<6$ hours of sleep when the actual population mean is 4.0 hours of sleep. Interpret this value of the power, then identify the value of $\beta$ and interpret that value.
Interpret this value of the power. A. The chance of failing to recognize that $\mu=4.0$ hours is not very high when in reality $\mu=4.0$ hours. B. The chance of failing to recognize that $\mu<6$ hours is not very high when in reality $\mu=4.0$ hours. C. The chance of recognizing that $\mu<6$ hours is very high when in reality $\mu=4.0$ hours. D. The chance of recognizing that $\mu<6$ hours is not very high when in reality $\mu=4.0$ hours.

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Problem 109

Which source of bias is most relevant to the following situation? A research study funded by a shoe polish company found that scuffed shoes were the number one reason to not hire a job applicant. self-interest study voluntary response bias nonresponse bias or missing data perceived lack of anonymity loaded or leading question

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Problem 110

QUESTION 11 A researcher wants to identify the variables that affect how likely college students are to watch their college's sports teams. The researcher surveys students to measure how often students watch their school's games. They plan to analyze these watch frequency data while considering several independent variables: IV1: Seasons during which the school has televised sports (3 levels): Fall vs. Summer vs. Spring) IV2: How often school teams win ( 5 levels): Mostly Win vs. More Wins than Losses vs. Equal Wins and Losses vs. More Losses than Wins vs. Mostly Lose) IV3: Whether the student identifies generally as a sports fan (2 levels): Sports Fan vs. Not a Sports Fan How many groups (or cells) are required for this factorial design? $30$

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Problem 111

Find the regression equation, letting the first variable be the predictor $(x)$ variable. Using the listed lemon/crash data, where lemon imports are in metric tons and the fatality rates are per 100,000 people, find the best predicted crash fatality rate for a year in which there are 425 metric tons of lemon imports. Is the prediction worthwhile? Use a significance level of 0.05 . \begin{tabular}{lccccc} \hline Lemon Imports & 227 & 264 & 358 & 460 & 535 \\ Crash Fatality Rate & 16 & 15.9 & 15.6 & 15.4 & 15.1 \\ \hline \end{tabular}
Find the equation of the regression line. $\hat{y}=\square+(-0.0028) x$ (Round the $y$ -intercept to three decimal places as needed. Round the slope to four decimal places as needed.)

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Problem 112

Elvin collected the scores of a random sample 41 students on the first exam in a certain class and their corresponding scores on the second exam in that class. Here is computer output on the sample data:
Summary statistics \begin{tabular}{lrrrr} Variable & n & Mean & StDev & SE Mean \\ \hline $x=$ first exam score & 41 & 59.5 & 19.7 & 3.1 \\ $y=$ second exam score & 41 & 59.4 & 21.7 & 3.4 \end{tabular}
Regression: second exam score vs. first exam score \begin{tabular}{lrr} Predictor & Coef & SE Coef \\ \hline Constant & 6.985 & 6.65 \\ First exam score & 0.881 & 0.11 \\ S $=13.2$ & R-sq $=62.89 \%$ & \end{tabular}
Assume that all conditions for inference have been met. Which of these is an appropriate test statistic for testing the null hypothesis that the population slope in this setting is 0 ?
Choose 1 answer: (A) $t=\frac{6.985}{6.65}$ (B) $t=\frac{0.881}{\frac{0.11}{\sqrt{41}}}$ (C) $t=\frac{0.881}{0.11}$ (D) $t=\frac{59.4}{3.4}$ (ㄷ) $t=\frac{6.985}{\frac{6.65}{\sqrt{41}}}$

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Problem 113

A Gallup survey of 1,012 randomly selected U.S. adults (age 18 and over), $53 \%$ said that they were dissatisfied with the quality of education students receive in kindergarten through grade 12. The bootstrap distribution (based on 5,000 samples) is provided.
Would it be appropriate to use the normal distribution to construct the confidence interval in this situation? Yes No

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Problem 114

Jason and Yolanda both drew a simple random sample from a non-normally distributed population of 25,000. Jason's sample consisted of $0.24 \%$ of the population, while Yolanda's sample consisted of $0.32 \%$ of the population. Whose sample can be used to make inferences about the population? neither Jason's sample nor Yolanda's sample only Jason's sample only Yolanda's sample both Jason's sample and Yolanda's sample

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Problem 115

Exercise A survey found the $21 \%$ of adults never exercise at all. A researcher selected a random sample of 120 adults and found that 21 adults said that they do not exercise at all. At $\alpha=0.05$ , is there sufficient evidence that less than $21 \%$ of adults do not exercise at all? Use the critical value method. Do not round intermediate steps.
Part 1 of 5 (a) State the hypotheses and identify the claim. $\begin{array}{l} H_{0}: p=0.21 \\ H_{1}: p<0.21 \end{array}$ not claim claim
This hypothesis test is a $\square$ one-tailed test.
Part: $1 / 5$ $\square$
Part 2 of 5 (b) Find the critical value. Round the answer to at least two decimal places.
The critical value is $\square$ .

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Problem 116

Test the claim that the mean body temperature of normal and healthy adults is equal to $37^{\circ} \mathrm{C}$ . Sample data consist of 20 randomly selected healthy adults who have body temperatures with a mean of $36.78^{\circ}$ and a standard deviation of $0.35^{\circ}$ . Use a 0.05 level of significance,, $\cup=37$ $\begin{array}{ll} \text { ificance } J=37 & t=36.78-37 \\ H_{0}: J \neq 37 & 0.351 \sqrt{20} \\ H_{a}: 0.81 & t=-\frac{0.22}{0.0783} \end{array}$
Jsing the same sample data given in problem 1, test the claim that the standard leviation of body temperatures for normal healthy adults is less than $1.00^{\circ} \mathrm{C}$ . Use 05 level of significance.

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Problem 117

For the given confidence level and values of $x$ and $n$ , find the following. $x=46, n=98, \text { confidence level } 80 \%$
Part: $0 / 3$
Part 1 of 3 (a) Find the point estimate. Round the answers to at least four decimal places, if necessary
The point estimate for the given data is $\square$

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Problem 118

Sleep apnea: Sleep apnea is a disorder in which there are pauses in breathing during sleep. People with this condition must wake up frequently to breathe. In a sample of 434 people aged 65 and over, 106 of them had sleep apnea.
Part: 0 / 3
Part 1 of 3 (a) Find a point estimate for the population proportion of those aged 65 and over who have sleep apnea. Round the answer to three decimal places.
The point estimate for the population proportion of those aged 65 and over who have sleep apnea is $\square$

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Problem 119

Conndence interval tor the popilation standard deviation
The standard deviation of the daily demand for a product is an important factor for inventory control for the product. Suppose that a pharmacy wants to estimate the standard deviation of the daily demand for a certain antibiotic. It is known that the daily demand for this antibiotic follows an approximately normal distribution. A random sample of 30 days has a sample mean of 124 orders for this antibiotic with a standard deviation of 10.1 orders. Find a $99 \%$ confidence interval for the population standard deviation of the daily demand for this antibiotic. Then give its lower limit and upper limit.
Carry your intermediate computations to at least three decimal places, Round your answers to at least two decimal places. (If necessary, consult a list of formulas.)
Lower limit: $\square$ Upper limits $\square$

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Problem 120

Module Knowledge Cheals Question 9
Pilots who cannot maintain regular sleep hours due to their work schedule often suffer from insomnia. A random sample of 26 commercial airline pilots was taken, and the pilots in the sample reported the time at which they went to sleep on their most recent working day. These times measured in hours after midnight. (Thus, if the pilot reported going to sleep at $11 \mathrm{p} . \mathrm{m}$ ., the measurement was -1 .) The sample mean was 0.5 hours, and the standard deviation was 1.9 hours. Assume that the sample is drawn from a normally distributed population. Find a $90 \%$ confidence interval for the population standard deviation, that is, the standard deviation of the time (hours after midnight) at which pilots go to sleep on their work days. Then give its lower limit and upper limit. Carry your intermediate computations to at least three decimal places. Round your answers to at least two decimal places. (If necessary, consult a list. of formulas.)
Lower limit: $\square$ Upper limit: $\square$

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Problem 121

Module Knowlsdge Chsck Question 9 Sophi
Pilots who cannot maintain regular sleep hours due to their work schedule often suffer from insominia. A random sample of 26 commercial airline pilots was taken, and the pilots in the sample reported the time at which they went to sleep on their most recent working day. These times measured in hours after midnight. (Thus, if the pilot reported going to sleep at 11 p.m., the measurement was -1 .) The sample mean was 0.5 hours, and the standard deviation was 19 hours. Assume that the sample is drawn from a normally distributed population. Find a $90 \%$ confidence interval for the population standard deviation, that is, the standard deviation of the time (hours after midnight) at which pilots go to sleep on their work days. Then give its lower limit and upper limit. Carry your intermediate computations to at least three decimal places. Round your answers to at least two decimal places. (If necessary, consult a list of formulas.) $\square$

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Problem 122

Confidence in banks: A poll conducted in 2012 asked a random sample of 1259 adults in the United States how much confidence they had in banks and other financial institutions. A total of 154 adults said that they had a great deal of confidence. An economist claims that greater than $13 \%$ of U.S. adults have a great deal of confidence in banks. Can you conclude that the economist's claim is true? Use both $\alpha=0.01$ and $\alpha=0.05$ levels of significance and the $P$ -value method with the TI-84 Plus calculator.
Part: $0 / 4$
Part 1 of 4 (a) State the appropriate null and alternate hypotheses. $H_{0}$ : $\square$ $H_{1}$ : $\square$
This hypothesis test is a (Choose one) $\nabla$ test. $\square$ - $\square$ $\square$ $\square$ $\square$ $\square$ $1 \neq$ $\square$ $p$
3

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Problem 123

You work for a consumer watchdog publication and are testing the advertising claims of a tire manufacturer. The manufacturer claims th normally distributed, with a mean of 40,000 miles and a standard deviation of 7500 miles. You test 16 tires and get the following life spar \begin{tabular}{llllllll} 48,363 & 39,091 & 24,794 & 36,299 & 30,068 & 40,606 & 40,514 & 35,085 \\ 24,654 & 33,554 & 36,128 & 36,278 & 30,630 & 38,308 & 35,665 & 47,237 \end{tabular}
Life spans of tires Life spans of tires
Distance (in miles) Dystance (it miles) Distance
Is it reasonable to assume that the life spans are normally distributed? Why? Choose the correct answer below. A. No, because the histogram is neither symmetric nor bell-shaped. B. Yes, because the histogram is symmetric and bell-shaped. C. No, because the histogram is symmetric and bell-shaped. D. Yes, because the histogram is neither symmetric nor bell-shaped

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Problem 124

Big babies: The National Health Statistics Reports described a study in which a sample of 97 one-year-old baby boys were weighed. Their mean weight was 25.1 pounds with standard deviation 5.3 pounds. A pediatrician claims that the mean weight of one-year-old boys is greater than 25 pounds, Do the data provide convincing evidence that the pediatrician's claim is true? Use the $\alpha=0.05$ level of significance and the $P$ -value method and Excel.
Part: $0 / 5$
Part 1 of 5 (a) State the appropriate null and alternate hypotheses. $\begin{array}{l} H_{0}: \square \\ H_{1}: \square \end{array}$
This hypothesis test is a (Choose one) $\boldsymbol{\nabla}$ test. $\square$ Nexhrart Submit Assignment 1020erM 1 14eres4

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Problem 125

College tuition: The mean annual tuition and fees in the 2013-2014 academic year for a sample of 13 private colleges in California was $\$ 33,000$ with a standard deviation of $\$ 7300$ . A dotplot shows that it is reasonable to assume that the population is approximately normal. Can you conclude that the mean tuition and fees for private institutions in California is less than $\$ 35,000$ ? Use the $\alpha=0.01$ level of significance and the $P$ -value method and Excel.
Part 1 of 5 (a) State the appropriate null and alternate hypotheses. $\begin{array}{l} H_{0}: \mu=35,000 \\ H_{1}: \mu<35,000 \end{array}$
This hypothesis test is a left-tailed $\quad \mathbf{}$ test. Part 2 of 5 Skip Part Check Save For Later Submit Assi Q 2024 McGraw Hill LLC. All Rights Reserved. Terms of Use I Privacy Center

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Problem 126

A student was asked to find a 98\% confidence interval for widget width using data from a random sample of size $n=16$ . Which of the following is a correct interpretation of the interval $12.9<\mu<$ 29.5?
Check all that are correct. There is a 98\% chance that the mean of a sample of 16 widgets will be between 12.9 and 29.5 . The mean width of all widgets is between 12.9 and $29.5,98 \%$ of the time. We know this is true because the mean of our sample is between 12.9 and 29.5. With $98 \%$ confidence, the mean width of all widgets is between 12.9 and 29.5. With $98 \%$ confidence, the mean width of a randomly selected widget will be between 12.9 and 29.5. There is a 98\% chance that the mean of the population is between 12.9 and 29.5.

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Problem 127

You want to obtain a sample to estimate a population mean. Based on previous evidence, you believe the population standard deviation is approximately $\sigma=48.5$ . You would like to be $99 \%$ confident that your esimate is within 2 of the true population mean. How large of a sample size is required? $n=$ $\square$
Do not round mid-calculation. However, use a critical value accurate to three decimal places - this is important for the system to be able to give hints for incorrect answers.

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Problem 128

If $n=140$ and $\widehat{p}=0.55$ , construct a $99 \%$ confidence interval. Give your answers to three decimals $\square$ $<p<$ $\square$

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Problem 129

If $n=520$ and $\hat{p}=0.62$ , construct a $99 \%$ confidence interval.
Give your answers to three decimals. $\square$ $<p<$ $\square$

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Problem 130

What is P-value? Give a formal definition.
Problem 7) State the Central Limit Theorem.

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Problem 131

The formula used to compute a large-sample confidence interval for $p$ is $\hat{p} \pm(z \text { critical value }) \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$
What is the appropriate $z$ critical value for each of the following confidence levels? (Round your answers to two decimal places.) (a) $95 \%$ $\square$ (b) $90 \%$ $\square$ (c) $99 \%$ $\square$ (d) $80 \%$ $\square$ (e) $81 \%$ $\square$

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Problem 132

$\mu=23$ $\mu>32$ $\mu<352$ $\mu \neq 10$ a) Which statement could be used for the alternative hypothesis for a left-tail test? b) Which statement could be used for the null hypothesis? c) Which statement could be used for the alternative hypothesis for a two-tail test? d) Which statement could be used for the alternative hypothesis for a right-tail test?

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Problem 133

question 1, 10.3.1 Part 1 of 3 HW Scor
Complete parts (a) through (c) below. Point: (a) Determine the critical value(s) for a right-tailed test of a population mean at the $\alpha=0.05$ level of significance with 20 degrees of freedom. (b) Determine the critical value(s) for a left-tailed test of a population mean at the $\alpha=0.01$ level of significance based on a sample size of $n=15$ (c) Determine the critical value(s) for a two-tailed test of a population mean at the $\alpha=0.10$ level of significance based on a sample size of $n=16$ .
Click here to view the $t$ -Distribution Area in Right Tail (a) $t_{\text {crit }}=$ $\square$ $\square$ (Round to three decimal places as needed.)

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Problem 134

To test $\mathrm{H}_{0} \cdot \mu=40$ versus $\mathrm{H}_{1} \cdot \mu<40$ , a random sample of size $n=25$ is obtained from a population that is known to be normally distributed. Complete parts (a) through (d) below. Click here to view the $t$ -Distribution Area in Right Tail (a) If $\bar{x}=37.5$ and $s=11.9$ , compute the test statistic, $\mathrm{t}_{0}=$ $\square$ (Round to three decimal places as needed.)

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Problem 135

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Problem 136

2. Find the required sample size for $95 \%$ confidence level with a margin of error of $4 \%$ and standard deviation $=0.35$ .

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Problem 137

Out of 100 people sampled, 71 preferred Candidate A. Based on this, estimate what proportion of the voting population (p) prefers Candidate A. To construct a confidence interval with a $99 \%$ confidence level, you need to use which one of the following calculators? Confidence Interval for a Population Mean Given Statistics Hypothesis Test for a Population Mean Given Statistics Hypothesis Test for a Population Proportion Chi-Square Test for Independence Confidence Interval for a Population Proportion Confidence Interval for a Population Mean Given Data Two Independent Sample Means Comparison Given Statistics Two Independent Sample Means Comparison Given Data Chi-Square Test for Goodness of Fit Hypothesis Test for a Population Mean Given Data One-Way ANOVA Two Independent Proportions Comparison Two Dependent Sample Means Comparison Given Data a. Enter your answer as an open-interval (i.e., parentheses) using decimals (not percents) accurate to three decimal places.
Confidence interval = $\square$ b. Express the same answer as a tri-linear inequality using decimals (not percents) accurate to three decimal places. $\square$ $<p<$ $\square$ c. Express the same answer using the point estimate and margin of error. Give your answers as decimals, to three places. $p=\square \pm \square$

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Problem 138

Suppose that an accounting firm does a study to determine the time needed to complete one person's tax forms. It randomly surveys 175 people. The sample mean is standard deviation of 6.0 hours. The population distribution is assumed to be normal. NOTE: If you are using a Student's $t$ -distribution, you may assume that the underlying population is normally distributed. (In general, you must first prove that assumption, though.) Part (a) Part (b) In words, define the random variables $X$ and $\bar{X}$ $\bar{X}$ is the time needed to complete one person's tax forms, and $X$ is the mean time needed to complete tax forms from a sample of 175 customers. $\bar{X}$ is the number of tax forms that an accounting firm completes, and $X$ is the mean number of tax forms that an accounting firm completes $X$ is the time needed to complete one person's tax forms, and $\bar{X}$ is the mean time needed to complete tax forms from a sample of 175 customers. $X$ is the number of tax forms that an accounting firm completes, and $\bar{X}$ is the mean number of tax forms that an accounting firm completes.

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Problem 139

2.2 The following are measurements of the total dissolved salts (TDS) and hardness index of 10 samples of water: \begin{tabular}{cc|cccccccccc} TDS & $\boldsymbol{X}$ & 202 & 327 & 112 & 463 & 582 & 926 & 681 & 290 & 776 & 378 \\ \hline \begin{tabular}{c} Hardness \\ index \end{tabular} & $\boldsymbol{Y}$ & 14 & 23 & 9 & 35 & 44 & 68 & 50 & 21 & 57 & 28 \end{tabular}
Given that $S_{X}=262.435$ and $S_{Y}=19.393$ , compute the correlation coefficient.

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Problem 140

Islam Spre. Copy of Pattems of - Unblocked Games -- Unblocked Garnes -- Vandalizing My Ow... Ne \#2b ur math teacher recorded the grades students received based the number of hours they studied for a test. The results are wn below and can be represented by a linear function. the table below to answer the questions. \begin{tabular}{|c|c|} \hline Hours Studied & Grade on Test \\ \hline 5 & 85 \\ \hline 7 & 92 \\ \hline 9 & 93 \\ \hline 9 & 95 \\ \hline 12 & 108 \\ \hline \end{tabular}
What is the correlation coefficie 0.15 $-0.15$ 0.96 $-0.96$

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Problem 141

2. You are interested in examining whether criminal justice majors are more likely than non-criminal justice majors to support the death penalty. You take a sample of 300 college students and ask their major and if they support use of the death penalty for murder. What statistic would you calculate to test your hypothesis? Set up Step 1 for your hypothesis testing

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Problem 142

The Cadet is a popular model of sport utility vehicle. The bivariate data given below were taken from a sample of fifteen Cadets, each houghe "ney" tyo yeaf ago and each sold "used" within the past month. For each Cadet in the sample, we have ligted heth the mileage, x (in thaysanda), that the Cader had at the time it was sold used, and the price, y (in thousands of dollars), at which the Cadet was sold used, These data are shown graphically in the geatrer plat in Figure 1. \begin{tabular}{|c|c|} \hline \begin{tabular}{l} Mileage, x \\ (in thousands) \end{tabular} & \begin{tabular}{l} Used selling price, y \\ (in thousands of dollars) \end{tabular} \\ \hline 25.8 & 26.7 \\ \hline 15.5 & 34,1 \\ \hline 23.9 & 20.1 \\ \hline 21.1 & 30.6 \\ \hline 22.0 & 31.3 \\ \hline 20.9 & 31.6 \\ \hline 28.1 & 25.6 \\ \hline 23.2 & 33.1 \\ \hline 24.3 & 30.5 \\ \hline 29.2 & 20.0 \\ \hline 24.5 & 28.0 \\ \hline 27.0 & 30.7 \\ \hline 37.8 & 22.0 \\ \hline 28.0 & 29.7 \\ \hline 34.3 & 26.0 \\ \hline \end{tabular} Send data to calculator Send data to Excel
The value of the sample correlation coefficient $r$ for these data is approximately -0.856 . Answer the following. Carry your intermediate computations to at least four decimal places. (If necessary, consult a bat of formular.) (a) What is the value of the slope of the least-squares regression line for these data? Round your answer to at least two decimal places. (b) What is the value of the $y$ -intercept of the least-squares regression line for these data? Round your answer to at least two decimal places. $\square$

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Problem 143

3. (t Test (in Excel) for Two Paired Samples) The owner of a chain of mini-markets wants to compare the sales perfomance of two of her stores, Store 1 and Store 2. Though the two stores have been comparable in the past, the owner has made several improvements to Store 1 and wishes to see if the improvements have made Store 1 more popular than Store 2. Sales can vary considerably depending on the day of the week and the season of the year, so she decides to eliminate such effects by making sure to record each store's sales on the same sample of days. After choosing a random sample of days, she records the sales (in dollars) for each store on these days, as shown in the table. \begin{tabular}{|r|r|r|c|} \hline Day & Store 1 & Store 2 & \begin{tabular}{l} Difference \\ (Store 1 - Store 2) \end{tabular} \\ \hline 1 & 397 & 297 & 100 \\ \hline 2 & 345 & 375 & -30 \\ \hline 3 & 365 & 168 & 197 \\ \hline 4 & 290 & 270 & 20 \\ \hline 5 & 812 & 731 & 81 \\ \hline 6 & 869 & 823 & 46 \\ \hline 7 & 975 & 668 & 307 \\ \hline 8 & 848 & 469 & 379 \\ \hline 9 & 980 & 605 & 375 \\ \hline 10 & 220 & 14 & 206 \\ \hline 11 & 692 & 670 & 22 \\ \hline 12 & 498 & 370 & 128 \\ \hline \end{tabular}
Based on these data, can the owner conclude, at the 0.10 level of significance, that the mean daily sales of Store 1 exceeds that of Store 2? Answer this question by performing a hypothesis test regarding $\mu_{d}$ , the population mean daily sales difference between the two stores. Assume that this population of differences is normally distributed. Use Excel to perform the hypothesis test. a. State the null and alternate hypotheses that would be appropriate for this hypothesis test. Use proper notation and formatting.

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Problem 144

In an attempt to increase business on Monday nights, a restaurant offers a free dessert with every dinner order. Before the offer, the mean number of dinner customers on Monday was 150. Following are the numbers of diners on a random sample of 12 days while the offer was in effect. $\begin{array}{llllllllllll} 206 & 169 & 191 & 152 & 212 & 139 & 142 & 151 & 174 & 220 & 192 & 153 \end{array}$
A hypothesis test will be performed to determine if the mean number of diners increased while the free dessert offer was in effect. A significance level of $\alpha=0.05$ will be used. Answer the questions below. a. Will this be a left-tailed test, a right-tailed test, or a two-tailed test? $\square$ b. Give the value of the test statistic. Round all values to three places. $\square$ c. Identify any critical value(s). Use the 0.05 level of significance. Keep three places. $\square$ d. Should $H_{0}$ be rejected at the 0.05 level of significance? Answer Yes or No. $\square$

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Problem 145

The three confidence intervals below were constructed from the same sample. One of them was computed at a confidence level of $90 \%$ , another at a confidence level of $95 \%$ , and another at a confidence level of $98 \%$ . Which one is the $98 \%$ confidence interval? a. $36.5<\mu<49.5$ b. $38.4<\mu<47.6$ c. $37.6<\mu<48.4$ d. cannot be determined a b c d

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Problem 146

Estimating a Question 6, 9.2.27-T Part 2 of 4 HW Score: $79.17 \%, 4.75$ of 6 points Points: 0 of 1 Save
A researcher wishes to estimate the average blood alcohol concentration (BAC) for drivers involved in fatal accidents who are found to have positive BAC values. He randomly selects records from 60 such drivers in 2021 and determines the sample mean BAC to be 0.169 gram per deciliter ( $\mathrm{g} / \mathrm{dL}$ ) with a standard deviation of $0.010 \mathrm{~g} / \mathrm{dL}$ . Complete parts (a) through (d). (a) A histogram of blood alcohol concentrations in fatal accidents shows that BACs are highly skewed right. Explain why a large sample size is needed to construct a confidence interval for the mean BAC of fatal crashes with a positive BAC. Choose the correct answer. A. Since the distribution of blood alcohol concentrations is highly skewed right, a large sample size is needed to maximize the margin of error to ensure that both tails are accounted for in the confidence interval. B. Since the distribution of blood alcohol concentrations is highly skewed right, a large sample size is needed to ensure that it contains at least $5 \%$ of the population. C. Since the distribution of blood alcohol concentrations is highly skewed right, a large sample size is necessary to ensure that the distribution of the sample mean is approximately normal. D. Since the distribution of blood alcohol concentrations is highly skewed right, a large sample size is needed to minimize the margin of error to ensure only the peak of the sampling distribution is captured in the confidence interval. (b) In 2021, there were approximately 25,000 fatal crashes in which the driver had a positive BAC. Explain why this, along with the fact that the data were obtained using a simple random sample, satisfies the requirements for constructing a confidence interval. Choose the correct answer. A. The sample size is likely greater than $10 \%$ of the population. B. The sample size is likely less than $5 \%$ of the population. C. The sample size is likely greater than $5 \%$ of the population. D. The sample size is likely less than $10 \%$ of the population.

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Problem 147

A new vaccination is being used in a laboratory experiment to investigate whether it is effective. There are 227 subjects in the study. Is there sufficient evidence to determine if vaccination and disease status are related? \begin{tabular}{|c|c|c|c|} \hline Vaccination Status & Diseased & Not Diseased & Total \\ \hline Vaccinated & 41 & 103 & 144 \\ \hline Not Vaccinated & 59 & 24 & 83 \\ \hline Total & 100 & 127 & 227 \\ \hline \end{tabular} Copy Data
Step 1 of 8: State the null and alternative hypothesis.
Answer Tables Keypad Keyboard Shortcuts $H_{0}$ : vaccination and disease status are independent $H_{a}$ : vaccination and disease status are dependent $H_{0}$ : vaccination and disease status are dependent $H_{a}:$ vaccination and disease status are independent Submit Answer

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Problem 148

Find the regression equation, letting the first variable be the predictor ( $x$ ) variable. Using the listed actress/actor ages in various years, find the best predicted age of the Best Actor winner given that the age of the Best Actress winner that year is 30 years. Is the result within 5 years of the actual Best Actor winner, whose age was 39 years? Use a significance level of 0.05 . \begin{tabular}{cllllllllllll} \hline Best Actress & 28 & 30 & 30 & 63 & 30 & 34 & 43 & 30 & 64 & 23 & 42 & 53 \\ Best Actor & 44 & 39 & 37 & 47 & 52 & 48 & 58 & 49 & 37 & 58 & 43 & 31 \\ \hline \end{tabular}
Find the equation of the regression line. $\hat{\mathrm{y}}=\square+(\square) \mathrm{x}$ (Round the $y$ -intercept to one decimal place as needed. Round the slope fo three decimal places as needed.)

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Problem 149

Two of the hottest smartphones on the market are the newly released iPhone6 and the Samsung Galaxy S6. CNet.com offers online reviews of all major cell phones, including battery life tests. In a review of the iPhone6, the talk-time battery life of 35 iPhones was measured. Similarly, the talk-time battery life of 30 Galaxy S6s was measured.
Two outputs are given below. Which is appropriate for analyzing the data collected? ``` Output 1 Hhi Mean of iPhone6 \muz: Mean of Galaxy S6 ``` \begin{tabular}{|l|c|c|} \hline Difference & Sample Diff. & Std. Err. \\ \hline $\mu_{1}-\mu_{2}$ & -0.71759861 & 0.189403 \\ \hline \end{tabular}
Possible p-values: $0.0001,0.0002,0.9999$ Output 2 $H_{R}=\mathrm{m}$ ean of the paired difference between iPhone6 and Galaxy 56 \begin{tabular}{|c|c|c|} \hline Difference & Sample Diff. & Std. Err. \\ \hline iPhone6 - Galaxy S6 & -0.754246 & 0.192151 \\ \hline \end{tabular}
Possible p-values: $0.0002,0.0004,0.9998$ Output 1 Output 2
Using the StatCrunch output chosen above, determine if there is a difference in the mean battery life for the two phones. Use a significance level of 0.01 when conducting the test. - Select the appropriate hypotheses. Make sure the notation used in the hypotheses agrees with the type of samples selected in the output. $\begin{array}{llllll} H_{o}: \mu_{d}=0 & H_{o}: \mu_{d}=0 & H_{o}: \mu_{d}=0 & H_{o}: \mu_{1}=\mu_{2} & H_{o}: \mu_{1}=\mu_{2} \quad H_{o}: \mu_{1}=\mu_{2} \\ H_{a}: \mu_{d}>0 & H_{a}: \mu_{d}<0 & H_{a}: \mu_{d} \neq 0 & H_{a}: \mu_{1}<\mu_{2} & H_{a}: \mu_{1}>\mu_{2} & H_{a}: \mu_{1} \neq \mu_{2} \end{array}$ - $\alpha=$ $\square$ reject $H_{o}$ if probability $\square$ $\alpha$ - $\mathrm{TS}: \mathrm{t}=$ $\square$ (make sure you reference the probabilities in the output you selected in the - probability = ) first question) - decision: Select an answer (6) - At the 0.01 level, there Select an answer significant evidence to conclude the mean battery life for an iPhone 6 is Select an answer (0) than the mean for a Galaxy S6.

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Problem 150

Use the results from a survey of a simple random sample of 1055 adults. Among the 1055 respondents, $65 \%$ rated themselves as above average drivers. We want to test the claim that $\frac{11}{20}$ of adults rate themselves as above average drivers. Complete parts (a) through (c). a. Identify the actual number of respondents who rated themselves as above average drivers.
686 (Round to the nearest whole number as needed.) b. Identify the sample proportion and use the symbol that represents it. $\square$ (Type an integer or a decimal rounded to two decimal places as needed.) c. For the hypothesis test, identify the value used for the population proportion and use the symbol that represents it. $\square$ $\square$ (Type an integer or a decimal rounded to two decimal places as needed.)

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Problem 151

A certain drug is used to treat asthma. In a clinical trial of the drug, 25 of 284 treated subjects experienced headaches (based on data from the manufacturer). The accompanying calculator display shows results from a test of the claim that less than $10 \%$ of treated subjects experienced headaches. Use the normal distribution as an approximation to the binomial distribution and assume a 0.01 significance level to ``` 1-PropzTest prop<0.1 ``` ``` p=0.2506296634 \hat{p}=0.0880281690 n=284 ``` complete parts (a) through (e) below. a. Is the test two-tailed, left-tailed, or right-tailed? Left-tailed test Right tailed test Two-tailed test b. What is the test statistic? $z=-0.67$ (Round to two decimal places as needed.) c. What is the P -value? $P$ -value $=0.2506$ (Round to four decimal places as needed.) d. What is the null hypothesis, and what do you conclude about it?
Identify the null hypothesis. A. $H_{0}: p=0.1$ B. $H_{0}: p<0.1$ c. $H_{0}: p \neq 0.1$ D. $H_{0}: p>0,1$

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Problem 152

A television show conducted an experiment to study what happens when buttered toast is dropped on the floor. When 46 buttered slices of toast were dropped, 25 of them landed with the buttered side up and 21 landed with the buttered side down. Use a 0.10 significance level to test the claim that toast will land with the buttered side down $50 \%$ of the time. Use the P-value method. Use the normal distribution as an approximation to the binomial distribution. After that, supposing the intent of the experiment was to assess the claim that toast will land with the buttered side down more than $50 \%$ of the time, write a conclusion that addresses the intent of the experiment.
Let p denote the population proportion of all buttered toast that will land with the buttered side down when dropped. Identify the null and alternative hypotheses to test the claim that buttered toast will land with the buttered side down $50 \%$ of the time. $\mathrm{H}_{0}$ : p $\square$ $\square$ $\mathrm{H}_{1} \mathrm{p}$ $\square$ $\square$ (Type integers or decimals. Do not round.)

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Problem 153

Is there a doctor in the house? A market research firm reported the mean annual earnings of all family practitioners in the United States was $\$ 178,258$ . A random sample of 43 family practitioners in Los Angeles had mean earnings of $\bar{x}=\$ 193,010$ with a standard deviation of $\$ 42,777$ . Do the data provide sufficient evidence to conclude that the mean salary for family practitioners in Los Angeles is greater than the national average? Use the $\alpha=0.05$ level of significance and the $P$ -value method with the T1-84 Plus calculator.
Part 1 of 5 (a) State the appropriate null and alternate hypotheses. $\begin{array}{l} H_{0}: \mu=178,258 \\ H_{1}: \mu>178,258 \end{array}$
This hypothesis test is a $\square$ right-tailed test.
Part 2 of 5 (b) Compute the value of the test statistic. Round the answer to two decimal places. $t=2.26$
Part 3 of 5 (c) Compute the $P$ -value. Round the answer to at least four decimal pla $P \text {-value }=0.014$ $\square$
Part: $3 / 5$
Part 4 of 5 (d) Determine whether to reject $H_{0}$ . (Choose one) $\nabla$ the null hypothesis $H_{0}$ .

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Problem 154

College tuition: The mean annual tuition and fees for a sample of 15 private colleges in California was $\$ 37,500$ with a standard deviation of $\$ 7850$ . A dotplot shows that it is reasonable to assume that the population is approximately normal. Can you conclude that the mean tuition and fees for private institutions in California is greater than $\$ 35,000$ ? Use the $\alpha=0.05$ level of significance and the $P$ -value method with the TI-84 Plus calculator.
Part: $0 / 5$
Part 1 of 5 (a) State the appropriate null and alternate hypotheses. $\begin{array}{l} H_{0}: \square \\ H_{1}: \mu>35,000 \end{array}$
This hypothesis test is a right-tailed $\square$ test.

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Problem 155

Big babies: The National Health Statistics Reports described a study in which a sample of 332 one-year-old baby boys were weighed. Their mean weight was 25.5 pounds with standard deviation 5.3 pounds. A pediatrician claims that the mean weight of one-year-old boys differs from 25 pounds. Do the data provide convincing evidence that the pediatrician's claim is true? Use the $\alpha=0.01$ level of significançe and the $P$ -value method with the TI-84 Plus calculator.
Part 1 of 5 (a) State the appropriate null and alternate hypotheses. $\begin{array}{l} H_{0}: \mu=25 \\ H_{1}: \mu \neq 25 \end{array}$
This hypothesis test is a two-tailed $\quad \boldsymbol{t}$ test.
Part 2 of 5 (b) Compute the value of the test statistic. Round the answer to two decimal places. $t=1.72$
Part: $2 / 5$
Part 3 of 5 (c) Compute the $P$ -value. Round the answer to at least four decimal places. $P \text {-value }=\square$

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Problem 156

Commuting to work: A community survey sampled 1923 people in Colorado and asked them how long it took them to commute to work each day. The sample mean one-way commute time was 25.2 minutes with a standard deviation of 13 minutes. A transportation engineer claims that the mean commute time is greater than 25 minutes. Do the data provide convincing evidence that the engineer's claim is true? Use the $\alpha=0.10$ level of significance and the $P$ -value method with the TI-84 Plus calculator.
Part: $0 / 5$
Part 1 of 5 (a) State the appropriate null and alternate hypotheses. $\begin{array}{l} H_{0}: \square \\ H_{1}: \square \end{array}$
This hypothesis test is a $\square$ (Choose one) test.

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Problem 157

Commuting to work: A community survey sampled 1923 people in Colorado and asked them how long it took them to commute to work each day. The sample mean one-way-commute time was 25.2 minutes with a standard deviation of 13 minutes. A transportation engineer claims that the mean commute time is greater than 25 minutes. Do the data provide convincing evidence that the engineer's claim is true? Use the $\alpha=0.10$ level of significance and the $P$ -value method with th TI-84 Plus calculator.
Part 1 of 5 (a) State the appropriate null and alternate hypotheses. $\begin{array}{l} H_{0}: \mu=25 \\ H_{1}: \mu>25 \end{array}$
This hypothesis test is a right-tailed $\square$ test.
Part: $1 / 5$
Part 2 of 5 (b) Compute the value of the test statistic. Round the answer to two decimal places. $t=\square$

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Problem 158

In a simple random sample of size 58 , there were 36 individuals in the category of interest.
Part 1 of 4 (a) Compute the sample proportion $\hat{p}$ . Round the answer to at least three decimal places.
The sample proportion is 0.621
Part 2 of 4 (b) Are the assumptions for a hypothesis test satisfied? Explain.
Yes $\square$ , the number of individuals in each category is greater than 10.
Part 3 of 4 (c) It is desired to test $H_{0}: p=0.6$ versus $H_{1}: p>0.6$ . Compute the test statistic $z$ . Round the answer to at least two decimal places.
The test statistic is 0.33 .
Part: $3 / 4$
Part 4 of 4 (d) Compute the $P$ -value. Round the answer to at least four decimal places. $P$ -value: $\square$ Do you reject $H_{0}$ at the 0,1 level? Yes No

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Problem 159

Determine whether the $z$ -test or the $t$ -test should be performed, or whether a statistician should be consulted. Part 1 of 3
A simple random sample of size 33 has mean $\bar{x}=14.3$ and the standard deviation is $s=1.9$ . Can you conclude that the population mean is less than 10 ?
The population standard deviation (Choose one) $\boldsymbol{\nabla}$ known. The sample size $n$ (Choose one) $\boldsymbol{\nabla}$ greater than 30. The correct decision is to (Choose one)
Part 2 of 3
A simple random sample of size 33 has mean $\bar{x}=41.8$ . The population standard deviation is $\sigma=3.72$ . The population is normally distributed. Can you conclude that the population mean differs from 40?
The population standard deviation $\square$ (Choose one) known.
The sample size $n$ $\square$ (Choose one) greater than 30.
The correct decision is to (Choose one) $\square$ .
Part 3 of 3
A simple random sample of size 15 has mean $\bar{x}=7.26$ . The population standard deviation is $\sigma=3.72$ . The population is not approximately normal. Can you conclude that the population mean differs from 9 ?
The population standard deviation (Choose one) known. The sample size $n$ (Choose one) $\nabla$ greater than 30. $\square$ The population (Choose one) $\boldsymbol{\nabla}$ approximately normal. The correct decision is to (Choose one) $\square$

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Problem 160

Good credit: The Fair Isaac Corporation (FICO) credit score is used by banks and other lenders to determine whether someone is a good credit risk. Scores range from 300 to 850 , with a score of 720 or more indicating that a person is a very good credit risk. An economist wants to determine whether the mean FICO score is lower than the cutoff of 720 . She finds that a random sample of 50 people had a mean FICO score of 707 with a standard deviation of 79 . Can the economist conclude that the mean FICO score is less than 720? Use the $\alpha=0.10$ level of significance and the $P$ -value method with the TI-84 Plus calculator.
Part: $0 / 5$ $\square$
Part 1 of 5 (a) State the appropriate null and alternate hypotheses. $\begin{array}{l} H_{0}: \square \\ H_{1}: \square \end{array}$
This hypothesis test is a (Choose one) $\boldsymbol{\nabla}$ test. $\square$

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Problem 161

Good credit: The Fair Isaac Corporation (FICO) credit score is used by banks and other lenders to determine whether someone is a good credit risk. Scores range from 300 to 850 , with a score of 720 or more indicating that a person is a very good credit risk. An economist wants to determine whether the mean FICO score is lower than the cutoff of 720 . She finds that a random sample of 50 people had a mean FICO score of 707 with a standard deviation of 79 . Can the economist conclude that the mean FICO score is less than 720 ? Use the $\alpha=0.10$ level of significance and the $P$ -value method with the TI-84 Plus calculator.
Part 1 of 5 (a) State the appropriate null and alternate hypotheses. $\begin{array}{l} H_{0}: \mu=720 \\ H_{1}: \mu<720 \end{array}$
This hypothesis test is a left-tailed $\nabla$ test.
Part 2 of 5 (b) Compute the value of the test statistic. Round the answer to two decimal places. $t=-1.16$
Part: $2 / 5$
Part 3 of 5 (c) Compute the $P$ -value. Round the answer to at least four decimal places. $P \text {-value }=\square$

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Problem 162

College tuition: The mean annual tuition and fees for a sample of 13 private colleges in California was $\$ 33,000$ with a standard deviation of $\$ 7300$ . A dotplot shows that it is reasonable to assume that the population is approximately normal. Can you conclude that the mean tuition and fees for private institutions in California is less than $\$ 35,000$ ? Use the $\alpha=0.01$ level of significance and the $P$ -value method with the TI-84 Plus calculator.
Part: $0 / 5$ $\square$
Part 1 of 5 (a) State the appropriate null and alternate hypotheses. $\begin{array}{l} H_{0}: \square \\ H_{1}: \square \end{array}$
This hypothesis test is a (Choose one) $\nabla$ test. $\square$

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Problem 163

College tuition: The mean annual tuition and fees for a sample of 13 private colleges in California was $\$ 33,000$ with a standard deviation of $\$ 7300$ . A dotplot shows that it is reasonable to assume that the population is approximately normal. Can you condude that the mean tuition and fees for private institutions in California is less than $\$ 35,000$ ? Use the $\alpha=0.01$ tevel of significance and the $P$ -value method with the $\pi$ - 84 Plus calculator.
Part 1 of 5 (a) State the appropriate null and alternate hypotheses. $\begin{array}{l} H_{0}=\mu=35,000 \\ H_{1}=\mu<35,000 \end{array}$
This hypothesis test is a left-tailed $\quad$ test.
Part 2 of 5 (b) Compute the value of the test statistic. Round the answer to two decimal places. $t=-0.99$
Part: $2 / 5$
Part 3 of 5 (c) Compute the $P$ -value. Round the $P$ -value to at least four decimal places. $P$ -vatue $=$ $\square$

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Problem 164

Keep cool: Following are prices, in dollars, of a random sample of ten 7.5-cubic-foot refrigerators. A consumer organization reports that the mean price of 7.5 -cubic-foot refrigerators is less than $\$ 370$ . Do the data provide convincing evidence of this claim? Use the $\alpha=0.01$ level of significance and the $P$ -method with the (3) Critical Values for the Student's $t$ Distribution Table. \begin{tabular}{lllll} \hline 350 & 414 & 360 & 313 & 353 \\ 318 & 369 & 383 & 329 & 339 \\ \hline \end{tabular} Send data to Excel
Part: $0 / 6$ $\square$
Part 1 of 6
Following is a dotplot for these data. Is it reasonable to assume that the conditions for performing a hypothesis test are satisfied? Explain.
The dotplot shows that there (Choose one) $\mathbf{\nabla}$ outliers. The dotplot shows that there (Choose one) $\boldsymbol{\nabla}$ evidence of strong skewness. We (Choose one) $\boldsymbol{\nabla}$ assume that the population is approximately normal. It (Choose one) $\boldsymbol{\nabla}$ reasonable to assume that the conditions are satisfied.

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Problem 165

7.36 Trading skills of institutional investors. Managers of stock portfolios make decisions as to what stocks to buy and sell in a given quarter. The trading skills of these institutional investors were quantified and analyzed in The Journal of Finance (April 2011). The study focused on "round-trip" trades, i.e., trades in which the same stock was both bought and sold in the same quarter. Consider a random sample of 200 round-trip trades made by institutional investors. Suppose the sample mean rate of return is $2.95 \%$ and the sample standard deviation is $8.82 \%$ . If the true mean rate of return of round-trip trades is positive, then the population of institutional investors is considered to have performed successfully. a. Specify the null and alternative hypotheses for determining whether the population of institutional investors performed successfully. b. Find the rejection region for the test using $\alpha=.05$ . c. Interpret the value of $\alpha$ in the words of the problem. d. A Minitab printout of the analysis is shown below. Locate the test statistic and $p$ -value on the printout. [Note: For large samples, $z \approx t$ .] e. Give the appropriate conclusion in the words of the problem.

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Problem 166

The test statistic of $z=2.08$ is obtained when testing the claim that $p \neq 0.621$ . a. Identify the hypothesis test as being two-tailed, left-tailed, or right-tailed. b. Find the P-value. c. Using a significance level of $\alpha=0.05$ , should we reject $H_{0}$ or should we fail to reject $H_{0}$ ?
Click here to view page 1 of the standard normal distribution table. Click here to view page 2 of the standard normal distribution table. a. This is a $\square$ two-tailed test. b. $P$ -value $=$ $\square$ 0.038 (Round to three decimal places as needed.) c. Choose the correct conclusion below. A. Fail to reject $H_{0}$ . There is not sufficient evidence to support the claim that $p \neq 0.621$ . B. Fail to reject $H_{0}$ . There is sufficient evidence to support the claim that $p \neq 0.621$ . C. Reject $H_{0}$ . There is sufficient evidence to support the claim that $p \neq 0.621$ . D. Reject $\mathrm{H}_{0}$ . There is not sufficient evidence to support the claim that $\mathrm{p} \neq 0.621$ .

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Problem 167

Determine the area under the standard normal curve that lies to the right of (a) $Z=1.72$ , (b) $Z=-0.09$ , (c) $Z=-1.11$ , and $(\mathrm{d}) \mathrm{Z}=1.37$ .
Click here to view the standard normal distribution table (page 1). Click here to view the standard normal distribution table (page 2). (a) The area to the right of $Z=1.72$ is $\square$ (Round to four decimal places as needed.)

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Problem 168

Crime rates: A governmental agency computed the proportion of violent crimes in the United States in a particular year falling into each of four categories. A simple random sample of 500 violent crimes committed in California during that year were categorized in the same way. The following table presents the results. \begin{tabular}{lcc} \hline \multicolumn{1}{c}{ Category } & \begin{tabular}{c} U.S. \\ Proportion \end{tabular} & \begin{tabular}{c} California \\ Frequency \end{tabular} \\ \hline Murder & 0.018 & 4 \\ Forcible Rape & 0.07 & 35 \\ Robbery & 0.390 & 218 \\ Aggravated Assault & 0.522 & 243 \\ \hline \end{tabular} Send data to Excel
Can you conclude that the proportions of crimes in the various categories in California differ from those in the United States as a whole? Use the 0.05 level of significance and the $P$ -value method with the TI-84 Plus calculator.
Part: 0 / 4 $\square$
Part 1 of 4 (a) State the null and alternate hypotheses. $H_{0}$ : The proportions of crimes in the various categories in California (Choose one) $\boldsymbol{\nabla}$ the same as those in the United States as a whole. $\square$ $H_{1}$ : The proportions of crimes in the various categories in California (Choose one) $\boldsymbol{\nabla}$ the same as those in the United States as a whole.

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Problem 169

Destiny performs a hypothesis test on a population with $\mu=115$ and $\sigma=30$ . Her sample size is 25 with a mean of 120.4. Which of the following correctly depicts the $z$ -statistic for this data? 0.04 0.90 3.53 3.93

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Problem 170

Confidence interval for the population mean: Use of the $t$ distribution
A corporation that maintains a large fleet of company cars for the use of its sales staff is interested in the mean distance driven monthly per salesperson. The Español following list gives the monthly distances in miles driven by a random sample of 15 salespeople. $2346,2026,2561,2437,2244,2180,2108,2271,2382,1953,2356,2309,2391,1958,2605$ Send data to calculator Send data to Excel
Based on this sample, find a $99 \%$ confidence interval for the mean number of miles driven monthly by members of the sales staff, assuming that monthly driving distances are normally distributed. Give the lower limit and upper limit of the $99 \%$ confidence interval.
Carry your intermediate computations to at least three decimal places. Round your answers to one decimal place. (If necessary, consult a list of formulas.)
Lower limit: $\square$ Upper limit: $\square$

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Problem 171

College student spends studying each week. They take a simple random sample of 91 students and compute a sample mean of 5.4 hours per week with a standard deviation of 1.2 hours. Find the $95 \%$ confidence interval for the population mean. Follow the PANIC acronym and answer each part.

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Problem 172

In a survey of 2084 adults in a recent year, 750 made a New Year's resolution to eat healthier. Construct $90 \%$ and $95 \%$ confidence intervals for the population proportion. Interpret the results and compare the widths of the confidence intervals.
The $90 \%$ confidence interval for the population proportion $p$ is ( $0.343,0.377$ ). (Round to three decimal places as needed.) The 95\% confidence interval for the population proportion p is ( $0.339,0.381$ ). (Round to three decimal places as needed.) With the given confidence, it can be said that the $\square$ of adults who say they have made a New Year's resolution to eat healthier is $\square$ Iof the given confidence interval.

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Problem 173

InIs question: 1 point(S) possible
Rhino viruses typically cause common colds. In a test of the effectiveness of echinacea, 39 of the 45 subjects treated with echinacea developed rhinovirus infections. In a placebo group, 89 of the 105 subjects developed rhinovirus infections. Use a 0.01 significance level to test the claim that echinacea has an effect on rinovirus infections. Complete parts (a) through (c) below.
Identify the P -value. P -value $=0.764$ (Round to three decimal places as needed.) What is the conclusion based on the hypothesis test? The P -value is $\square$ greater than the significance level of $\alpha=0.01$ , so reject $\square$ the null hypothesis. There is not sufficient evidence to support the claim that echinacea treatment has an effect. b. Test the claim by constructing an appropriate confidence interval.
The $99 \%$ confidence interval is $-0.143<\left(p_{1}-p_{2}\right)<0.181$ . (Round to three decimal places as needed.) What is the conclusion based on the confidence interval? Because the confidence interval limits $\square$ 0 , there $\square$ appear to be a significant difference between the two proportions. There $\square$ evidence to support the claim that echinacea treatment has an effect. c. Based on the results, does echinacea appear to have any effect on the infection rate? A. Echinacea does appear to have a significant effect on the infection rate. There is evidence that it increases the infection rate. B. Echinacea does not appear to have a significant effect on the infection rate. C. Echinacea does appear to have a significant effect on the infection rate. There is evidence that it lowers the infection rate. D. The results are inconclusive. Submit quiz

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Problem 174

A company that makes cola drinks states that the mean caffeine content per 12 -ounce bottle of cola is 40 milligrams. You want to test this claim. During your tests, you find that a random sample of thirty 12-ounce bottles of cola has a mean caffeine content of 37.8 milligrams. Assume the population is normally distributed and the population standard deviation is 6.6 milligrams. At $\alpha=0.05$ , can you reject the company's claim? Complete parts (a) through (e). (d) Decide whether to reject or fail to reject the null hypothesis. A. Since $z$ is in the rejection region, fail to reject the null hypothesis. B. Since $z$ is not in the rejection region, reject the null hypothesis. C. Since $z$ is in the rejection region, reject the null hypothesis. D. Since $z$ is not in the rejection region, fail to reject the null hypothesis. (e) Interpret the decision in the context of the original claim.
At the $5 \%$ significance level, there $\square$ enough evidence to $\square$ the company's claim that the mean caffeine content per 12-ounce bottle of cola $\square$ $\square$ milligrams.

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Problem 175

Researchers conducted a study to determine whether magnets are effective in treating back pain. Pain was measured using the visual analog scale, and the results shown below are among the results obtained in the study. Higher scores correspond to greater pain levels. Assume that the two samples are independent simple random samples selected from normally distributed populations, and do not assume that the population standard deviations are equal. Complete parts (a) to (c) below.
Reduction in Pain Level After Magnet Treatment $\left(\mu_{1}\right): n=22, \bar{x}=0.52, s=0.95$ Reduction in Pain Level After Sham Treatment $\left(\mu_{2}\right): n=22, \bar{x}=0.48, s=1.24$ $\pi_{1} \cdot \mu_{1}=\mu_{2} \quad \Pi_{1} \cdot \mu_{1}+\mu_{2}$
The test statistic, t , is 0.12 . (Round to two decimal places as needed.) The P -value is 0.453 . (Round to three decimal places as needed.) State the conclusion for the test. Fail to reject the null hypothesis. There is not sufficient evidence to support the claim that those treated with magnets have a greater mean reduction in pain than those given a sham treatment. b. Construct a confidence interval appropriate for the hypothesis test in part (a). $\square<\mu_{1}-\mu_{2}<\square$ (Round to two decimal places as needed.)

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Problem 176

The weights of four randomly and independently selected bags of potatoes labeled 20.0 pounds were found to be 20.8, $21.4,20.9$ , and 21.2 pounds. $\square$ Assume Normality. Answer parts (a) and (b) below. a. Find a $95 \%$ confidence interval for the mean weight of all bags of potatoes. $(20.64$ , 21.51) (Type integers or decimals rounded to the nearest hundredth as needed. Use ascending order.) b. Does the interval capture 20.0 pounds? Is there enough evidence to reject a mean weight of 20.0 pounds? A. The interval captures 20.0 pounds, su there is not enough evidence to reject a mean weight of 20.0 pounds. It is plausible the population mean weight is 20.0 pounds. B. The interval does not capture 20.0 pounds, so there not is enough evidence to reject a mean weight of 20.0 pounds. It is plausible the population mean weight is 20.0 pounds. C. The interval captures 20.0 pounds, so there is enough evidence to reject a mean weight of 20.0 pounds. It is not plausible the population mean weight is 20.0 pounds. D. The interval does not capture 20.0 pounds, so there is enough evidence to reject a mean weight of 20.0 pounds. It is not plausible the population mean weight is 20.0 pounds. E. There is insufficient information to make a decision regarding the rejection of 20.0 pounds. The sample size of 4 bags is less than the required 25.

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Problem 177

In a previous poil, $42 \%$ of adults with children under the age of 18 reported that their family ate dinner together seven nights a week. Suppose that, in a more recent poll, 1198 adults with children under the age of 18 were selected at random, and 481 of those 1198 adults reported that their family ate dinner together seven nights a wook. Is there sufficient ovidence that the proportion of families with children under the age of 18 who eat dinner together seven nights a week has decreased? Use the $\alpha=0.05$ significance level.
Because $n p_{0}\left(1-p_{0}\right)=$ $\square$ $\square$ 10 and the sample size is $\square$ $5 \%$ of the population size, and the adults in the sample $\square$ selected at random, all of the requirements for testing the hypothesis (Round to one decimal place as needed.)

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Problem 178

In a previous poll, $42 \%$ of adults with children under the age of 18 reported that their family ate dinner together seven nights a week. Suppose that, in a more recent poll, 1198 adults with children under the age of 18 were selected at random, and 481 of those 1198 adults reported that their family ate dinner together seven nights a week. Is there sufficient evidence that the proportion of families with children under the age of 18 who eat dinner together seven nights a week has decreased? Use the $\alpha=0.05$ significance level.
Because $\mathrm{np}_{0}\left(1-\mathrm{p}_{0}\right)=291.8>10$ and the sample size is less than $5 \%$ of the population size, and the adults in the sample were selected at random, all of the requirements for testing the hypothesis are satisfied. (Round to one decimal place as needed.) What are the null and alternative hypotheses? $\mathrm{H}_{0}$ : $\square$ $\square$ versus $\mathrm{H}_{1}$ : $\square$ $\square$ $\square$ (Type integers or decimals. Do not round.)

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Problem 179

Points: 0 of 1 Save
The brand manager for a brand of toothpaste must plan a campaign designed to increase brand recognition. He wants to first determine the percentage of adults who have heard of the brand. How many adults must he survey in order to be $95 \%$ confident that his estimate is within seven percentage points of the true population percentage? Complete parts (a) through (c) below. a) Assume that nothing is known about the percentage of adults who have heard of the brand. $n=196$ (Round up to the nearest integer.) b) Assume that a recent survey suggests that about $81 \%$ of adults have heard of the brand. $n=121$ (Round up to the nearest integer.) c) Given that the required sample size is relatively small, could he simply survey the adults at the nearest college? A. No, a sample of students at the nearest college is a convenience sample, not a simple random sample, so it is very possible that the results would not be representative of the population of adults. B. No, a sample of students at the nearest college is a cluster sample, not a simple random sample, so it is very possible that the results would not be representative of the population of adults. C. No, a sample of students at the nearest college is a stratified sample, not a simple random sample, so it is very possible that the results would not be representative of the population of adults. D. Yes, a sample of studerts at the nearest college is a simple random sample, so the results should be representative of the population of adults. Clear all Final check

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Problem 180

Part 2 of 4 0 Points: 0 of 1 Save
Here are summary statistics for the weights of Pepsi in randomly selected cans: $n=36, \bar{x}=0.82409 \mathrm{lb}, s=0.00567 \mathrm{lb}$ . Use a confidence level of $90 \%$ to complete parts (a) through (d) below. a. Identify the critical value $t_{\alpha / 2}$ used for finding the margin of error. $t_{\alpha / 2}=1.69$ (Round to two decimal places as needed.) b. Find the margin of error. $\mathrm{E}=\square \mathrm{lb}$ (Round to five decimal places as needed.) an example Get more help - Clear all Check answer

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Problem 181

Part 3 of 4 HW Score: $76.59 \%, 16.08$ of 21 points Points: 0 of 1
Use the sample data and confidence level given below to complete parts (a) through (d). Save
A research institute poll asked respondents if they felt vulnerable to identity theft. In the poll, $n=1068$ and $x=517$ who said "yes." Use a $99 \%$ confidence level. Click the icon to view a table of z scores. a) Find the best point estimate of the population proportion $p$ . 0.484 (Round to three decimal places as needed.) b) Identify the value of the margin of error $E$ . $E=0.039$ (Round to three decimal places as needed.) c) Construct the confidence interval. $\square$ $\square$ < p < (Round to three decimal places as needed.)

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Problem 182

Part 4 of 4 st
Use the sample data and confidence level given below to complete parts (a) through (d). HW Score: $76.59 \%, 16.08$ of 21 points Points: 0 of 1 Save
A research institute poll asked respondents if they felt vulnerable to identity theft. In the poll, $n=1068$ and $x=517$ who said "yes." Use a $99 \%$ confidence level. Click the icon to view a table of $z$ scores. (Round to three decimal places as needed.) b) Identify the value of the margin of error E . $E=0.039$ (Round to three decimal places as needed.) c) Construct the confidence interval. $0.445<p<0.523$ (Round to three decimal places as needed.) d) Write a statement that correctly interprets the confidence interval. Choose the correct answer below. A. One has $99 \%$ confidence that the sample proportion is equal to the population proportion. B. There is a $99 \%$ chance that the true value of the population proportion will fall between the lower bound and the upper bound. C. $99 \%$ of sample proportions will fall between the lower bound and the upper bound. D. One has $99 \%$ confidence that the interval from the lower bound to the upper bound actually does contain the true value of the population proportion. Clear all Final chect

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Problem 183

22. An instructor has given a short quiz consisting of two parts. For a randomly selected student, let $X=$ the number of points eamed on the first part and $Y=$ the number of points eamed on the second part. Suppose that the joint pmf of $X$ and $Y$ is given in the accompanying table. \begin{tabular}{|c|c|c|c|c|c|} \hline \multicolumn{2}{|l|}{\multirow[b]{2}{*}{ $p(x, y)$ }} & \multicolumn{4}{|c|}{ $y$ } \\ \hline & & 0 & 5 & 10 & 15 \\ \hline & 0 & . 02 & . 06 & . 02 & . 10 \\ \hline $x$ & 5 & . 04 & . 15 & . 20 & . 10 \\ \hline & 10 & .01 & .15 & . 14 & . 01 \\ \hline \end{tabular} a. If the score recorded in the grade book is the total number of points earned on the two parts, what is the expected recorded score $E(X+Y)$ ? b. If the maximum of the two scores is recorded, what is the expected recorded score?

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Problem 184

Weight loss: In a study to determine whether counseling could help people lose weight, a sample of people experienced a group-based behavioral intervention, which involved weekly meetings with a trained interventionist for a period of six months. The following data are the numbers of pounds lost for 14 people. Assume the population is approximately normal. Perform a hypothesis test to determine whether the mean weight loss is greater than 14 pounds. Use the $\alpha=0.10$ level of significance and the $P$ -value method with the TI-84 Plus calculator. \begin{tabular}{ccccccc} \hline 19.6 & 25.8 & 4.9 & 21.0 & 18 & 9.6 & 14.4 \\ 18.0 & 34.7 & 30.6 & 9.4 & 32.5 & 20.8 & 16.4 \\ \hline \end{tabular} Send data to Excel
Part: $0 / 5$
Part 1 of 5 (a) State the appropriate null and alternate hypotheses. $\begin{array}{l} H_{0}: \square \\ H_{1}: \square \end{array}$
This hypothesis test is a (Choose one) $\boldsymbol{\nabla}$ test. $\square$

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Problem 185

Previously, 5\% of mothers smoked more than 21 cigarettes during their pregnancy. An obstetrician believes that the percentage of mothers who smoke 21 cigarettes or more is less than $5 \%$ today. She randomly selects 115 pregnant mothers and finds that 4 of them smoked 21 or more cigarettes during pregnancy. Test the researcher's statement at the $\alpha=0.1$ level of significance.
What are the null and alternative hypotheses? $H_{0}: p=0.05$ versus $H_{1}: p<0.05$ (Type integers or decimals. Do not round.) Because $n p_{0}\left(1-p_{0}\right)=5.5<10$ , the normal model may not be used to approximate the $P$ -value. (Round to one decimal place as needed.) Find the P -value. P -value $=$ $\square$ (Round to three decimal places as needed.)

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Problem 186

Keep cool: Following are prices, in dollars, of a random sample of ten 7.5-cubic-foot refrigerators. A consumer organization reports that the mean price of 7.5 -cubic-foot refrigerators is less than $\$ 370$ . Do the data provide convincing evidence of this claim? Use the $\alpha=0.01$ level of significance and the $P$ -method with the
Critical Values for the Student's t Distribution Table. \begin{tabular}{lllll} \hline 350 & 414 & 360 & 313 & 353 \\ 318 & 369 & 383 & 329 & 339 \\ \hline \end{tabular} Send data to Excel
Part 1 of 6
Following is a dotplot for these data. Is it reasonable to assume that the conditions for performing a hypothesis test are satisfied? Explain.
The dotplot shows that there are no $\quad$ outliers. The dotplot shows that there is no evidence of strong skewness.
We $\square$ can assume that the population is approximately normal.
It $\square$ is reasonable to assume that the conditions are satisfied.
Part: $1 / 6$
Part 2 of 6
State the appropriate null and alternate hypotheses. $H_{0}$ : $\square$ $\square$ < $\square$ $\square$ $\square$ $\square$ $\square$ $H_{1}$ : $\square$ 묵 $\square$ $\mu$

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Problem 187

Keep cool: Following are prices, in dollars, of a random sample of ten 7.5 -cubic-foot refrigerators. A consumer organization reports that the mean price of 7.5 -cubicfoot refrigerators is less than $\$ 370.00$ the data provide convincing evidence of this claim? Use the $a=0.01$ level of significance and the $P$ -method with the - Critical Values for the Student's $t$ Distribution Table. \begin{tabular}{lllll} \hline 350 & 414 & 360 & 313 & 353 \\ 318 & 369 & 383 & 329 & 339 \\ \hline \end{tabular} Send data to Excel
Part 1 of 6
Following is a dotplot for these data. Is it reasonable to assume that the conditions for performing a hypothesis test are satisfied? Explain.
The dotplot shows that there are no outliers. The dotplot shows that there is no evidence of strong skewness. We $\square$ can $\square$ It is assume that the population is approximately normal, reasonable to assume that the conditions are satisfied.
Part: $1 / 6$
Part 2 of 6
State the appropriate null and alternate hypotheses. $\begin{array}{l} H_{0}: \mu=370 \\ H_{1}=\mu<370 \end{array}$ $\square$
Part: $2 / 6$
Part 3 of 6
Compute the value of the test statistic. Round the answer to three decimal places. $t=-2.449$ $\square$
Part: $3 / 6$
Part 4 of 6
Select the correct interval for the $P$ -value. $P$ -value $>0.10$ $0.025<P$ -value $\leq 0.05$ $0.05<P$ -value $\leq 0.10$ $P$ -value $\leq 0.025$
Part: $4 / 6$
Part 5 of 6
Determine whether to reject $H_{0}$ . $\qquad$ the null hypothesis $H_{0}$ .

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Problem 188

(Expafica)
Keep cool: Following are prices, in doilars, of a random sample of ten 7.5 -cubic-foot refrigerators. A consumer organization reports that the mean price of 7.5 -cubic-foot refrigerators is less than $\$ 370$ . Do the data provide convincing evidence of this claim? Use the $\alpha=0.01$ level of significance and the $P$ -method with the - Critical Values for the Student's t Distribution Table. \begin{tabular}{lllll} 350 & 414 & 360 & 313 & 353 \\ 318 & 369 & 383 & 329 & 339 \\ \hline \end{tabular} abo （ Send data to Excel
Part 1 of 6
Following is a dotplot for these data. Is it reasonable to assume that the conditions for performing a hypothesis test are satisfied? Explain.
The dotpiot shows that there are no outllers. The dotplot shows that there is no evidence of strong skewness. We $\square$ assume that the population is approximately normal,
It 1 $\square$ reasonable to assume that the conditions are satisfled
Part: $1 / 6$
Part 2 of 6
State the appropriate null and alternate hypotheses. $\begin{array}{l} H_{0}: \mu=370 \\ H_{1}: \mu<370 \end{array}$ \begin{tabular}{|c|c|c|} \hline $\square<\square$ & $\square>\square$ & $\square=\square$ \\ $\square \neq \square$ & $\mu$ & \\ \hline $\times$ & 0 \\ \hline \end{tabular}
Part: $2 / 6$
Part 3 of 6
Compute the value of the test statistic. Round the answer to three decimal places. $t=-2.449$ $\square$ $\times$ 6 $\times$ 5 $\qquad$

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Problem 189

7 a) Explain the concepts of regression and correlation analyses. [2 marks] b) The following data refer to the ages and prices of various bus models on the market: \begin{tabular}{|l|l|l|l|l|l|l|l|l|} \hline Age (In Years) & 5 & 6 & 3 & 2 & 4 & 7 & 8 & 9 \\ \hline Bus Price (\000) & 16 & 14 & 22 & 25 & 18 & 12 & 10 & 8 \\ \hline \end{tabular} b.1) Determine the independent and dependent variables of the data. [2 marks] b.2) Show the data on a scatter plot and comment on the distribution. [3 marks] b.3) Estimate the Ordinary Least Squares (OLS) model connecting the two variables. Interpret your estimated linear regression coefficients. [5 marks] b.4) Estimate the price of a bus that is 12 years old. [2 marks] b.5) Calculate Pearson's product moment correlation coefficient and the coefficient of determination. Appraise your findings. [6 marks] b.6) Test the claim that the age has no effect on the price of a bus at 5 \%$ level. [ 5 marks]

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Problem 190

7 a) Explain the concepts of regression and correlation analyses. [2 marks] b) The following data refer to the ages and prices of various bus models on the market: \begin{tabular}{|l|l|l|l|l|l|l|l|l|} \hline Age (In Years) & 5 & 6 & 3 & 2 & 4 & 7 & 8 & 9 \\ \hline Bus Price (\000) & 16 & 14 & 22 & 25 & 18 & 12 & 10 & 8 \\ \hline \end{tabular} b.1) Determine the independent and dependent variables of the data. [2 marks] b.2) Show the data on a scatter plot and comment on the distribution. [3 marks] b.3) Estimate the Ordinary Least Squares (OLS) model connecting the two variables. Interpret your estimated linear regression coefficients. [5 marks] b.4) Estimate the price of a bus that is 12 years old. [2 marks] b.5) Calculate Pearson's product moment correlation coefficient and the coefficient of determination. Appraise your findings. [6 marks] b.6) Test the claim that the age has no effect on the price of a bus at 5 \%$ level. [ 5 marks]

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Problem 191

According to the February 2008 Federal Trade Commission report on consumer fraud and identity theft, $23 \%$ of all complaints in 2007 were for identity theft. In that year, Alaska had 321 complaints of identity theft out of 1,432 consumer complaints. Does this data provide enough evidence to show that Alaska had a lower proportion of identity theft than $23 \%$ ?
The hypotheses are: $\begin{array}{l} \mathrm{H}_{0}: p=23 \% \\ \mathrm{H}_{1}: p<23 \% \end{array}$
What is a type I error in the context of this problem?

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Problem 192

In a random sample of 29 residents living in major cities on the West Coast (Group 1) and 29 reaults can be seen in the table below. The population standard deviation of the age in Weat Coast cities is known to be 10.95 years and in East Coast cities is known to be 9.67 years. Assume the populations are normally distributed. Run a test at a 0.05 level of aignificence to test if west coast cities are, on average, older. \begin{tabular}{|c|c|} \hline \begin{tabular}{l} West Const \\ (Group 1) \end{tabular} & \begin{tabular}{l} East Coast \\ (Group 2) \end{tabular} \\ \hline 25 & 35 \\ \hline 47 & 45 \\ \hline 18 & 37 \\ \hline 38 & 20 \\ \hline 30 & 19 \\ \hline 52 & 26 \\ \hline 52 & 79 \\ \hline 61 & 46 \\ \hline 43 & 29 \\ \hline 22 & 55 \\ \hline 34 & 25 \\ \hline 35 & 35 \\ \hline 55 & 36 \\ \hline 60 & 53 \\ \hline 68 & 41 \\ \hline 20 & 50 \\ \hline 34 & 32 \\ \hline 36 & 38 \\ \hline 37 & 26 \\ \hline 42 & 44 \\ \hline 60 & 19 \\ \hline \end{tabular} \begin{tabular}{|l|l|} \hline 71 & 28 \\ \hline 54 & 27 \\ \hline 20 & 18 \\ \hline 45 & 30 \\ \hline 52 & 21 \\ \hline 34 & 22 \\ \hline 58 & 43 \\ \hline 64 & 61 \\ \hline \end{tabular}
Enter the test atatistic - round to ecimal places.
Test Statistic $z=$

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Problem 193

Problem 1: During an investigation into the association between smoking and lung cancer, two populations of adults aged $30-60$ years were collected. The group was tracked over a 10-year period and divided into "smokers" and "non-smokers." Out the 1000 individual smokers, 150 developed lung cancer. Of the 1000 individual non-smokers, 30 developed lung cancer. (a) Create a risk data table and (b) calculate relative risk, (c) attributable risk, and (d) odds ratio.

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Problem 194

Baby weights: Following are weights in pounds for random samples of 20 newborn baby boys and baby girls born in Denver in 2011. Box plots indicate that the samples come from populations that are approximately normal. Let $\mu_{1}$ denote the mean weight of boys. Can you conclude that the mean weights differ between boys and girls? Use the $\alpha=0.05$ level and the $P$ -value method with the TI-84 Plus calculator. \begin{tabular}{lllllllll} \hline \multicolumn{1}{c}{ Boys } \\ \hline 8.1 & 7.9 & 8.3 & 7.3 & 6.4 & 8.4 & 8.5 & 6.9 & 6.3 \\ 7.4 & 7.8 & 7.5 & 6.9 & 7.8 & 8.6 & 7.7 & 7.4 & 7.7 \\ 8.1 & 6.4 & & & & & & & \\ \hline \end{tabular} \begin{tabular}{lllllllll} \hline \multicolumn{1}{c}{ Girls } \\ \hline 7.4 & 6.0 & 6.7 & 8.2 & 7.5 & 5.7 & 6.6 & 6.4 & 8.5 \\ 7.2 & 6.9 & 8.2 & 6.5 & 6.7 & 7.2 & 6.3 & 5.9 & 8.1 \\ 8.2 & 6.7 & & & & & & & \\ \hline \end{tabular} Send data to Excel
Part: $0 / 4$
Part 1 of 4 State the null and alternate hypotheses. $\mathrm{H}_{0}$ : $\square$ $\square$ $=\square$ $\square=$ 1

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Problem 195

Save
A clinical trial was conducted to test the effectiveness of a drug for treating insomnia in older subjects. Before treatment, 25 subjects had a mean wake time of 105.0 min . After treatment, the 25 subjects had a mean wake time of 100.7 min and a standard deviation of 20.8 min . Assume that the 25 sample values appear to be from a normally distributed population and construct a $90 \%$ confidence interval estimate of the mean wake time for a population with drug treatments. What does the result suggest about the mean wake time of 105.0 min before the treatment? Does the drug appear to be effective?
Construct the 90\% confidence interval estimate of the mean wake time for a population with the treatment. $\square$ $\square$ $\min <\mu<$ min (Round to one decimal place as needed.)

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Problem 196

Here are summary statistics for randomly selected weights of newborn girls: $n=36, \bar{x}=3216.7 \mathrm{~g}, \mathrm{~s}=688.5 \mathrm{~g}$ . Use a confidence level of $95 \%$ to complete parts (a) through (d) below. a. Identify the critical value $t_{\alpha / 2}$ used for finding the margin of error. $t_{\alpha / 2}=2.03$ (Round to two decimal places as needed.) b. Find the margin of error. $E=$ $\square$ g (Round to one decimal place as needed.)

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Problem 197

Calcium is essential to tree growth. In 1990 , the concentration of calcium in precipitation in Chautauqua, New York, was 0.11 milligram per liter $\left(\frac{\mathrm{mg}}{\mathrm{L}}\right)$ . A random sample of 8 precipitation dates in 2018 results in the following data: $\begin{array}{llllllll} 0.234 & 0.313 & 0.108 & 0.065 & 0.087 & 0.070 & 0.262 & 0.126 \end{array}$
A normal probability plot suggests the data could come from a population that is normally distributed. A boxplot does not show any outliers. Does the sample evidence suggest that calcium concentrations have changed since 1990? Use the $\mathrm{a}=0.1$ level of significance.
What are the null and alternative hypotheses? $\begin{array}{l} H_{0}: \mu=0.11 \\ H_{1}: \mu=0.11 \end{array}$ (Type integers or decimals. Do not round.) Find the test statistic. $t_{0}=1.40^{\circ}\left(R_{0} \quad \text { imal places as needed. }\right)$
Find the P -valt P -value $=$ $\square$ Is there suffici $\square$ three decimal places as needed.)
Since P -value $\square$ $\alpha$ , the null hypothesis and conclude that there $\square$ sufficient evidence that the calcium level in rainwater has changed. Clear all Check answer

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Problem 198

A math teacher claims that she has developed a review course that increases the scores of students on the math portion of a college entrance exam. Based on data from the College Board, scores are normally distributed with $\mu=519$ . The teacher obtains a random sample of 1800 students, puts them through the review class, and finds that the mean math' score of the 1800 students is 525 with a standard deviation of 112. Complete parts (a) through (d) below.
Click the icon to view the t-Distribution Area in the Right Tail. (b) Test the hypothesis at the $\alpha=0.10$ level of significance. Is a mean math score of 525 statistically significantly higher than 519 ? Conduct a hypothesis test using the P -value approach.
Find the test statistic. $t_{0}=2.27$ (Round to two decimal places as needed.) Approximate the P -value corresponding to the critical value of t . The P -value is between 0.02 and 0.01 . Is the sample mean statistically significantly higher? A. Yes, because the $P$ -value range is below $\alpha=0.10$ B. No, because the P -value range is below $\alpha=0.10$ . C. Yes, because the P -value range is above $\alpha=0.10$ . D. No, because the P -value range is above $\alpha=0.10$ .

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Problem 199

\begin{problem} A math teacher claims that she has developed a review course that increases the scores of students on the math portion of a college entrance exam. Based on data from the College Board, scores are normally distributed with $\mu=519$ . The teacher obtains a random sample of 1800 students, puts them through the review class, and finds that the mean math score of the 1800 students is 525 with a standard deviation of 112. Complete parts (a) through (d) below.
(c) Do you think that a mean math score of 525 versus 519 will affect the decision of a school admissions administrator? In other words, does the increase in the score have any practical significance? \begin{enumerate} \item[A.] No, because the score became only $1.16\%$ greater. \item[B.] No, because every increase in score is practically significant. \item[C.] Yes, because the score became more than $116\%$ greater. \item[D.] Yes, because it is statistically significant. \end{enumerate}
(d) Test the hypothesis with $n=375$ students. Assume that the sample mean is still 525 and the sample standard deviation is still 112. Is a sample mean of 525 significantly more than 519 at the $\alpha=0.15$ significance level?
Find the test statistic $t_0 = 1.04$ (Round to two decimal places).
Approximate the P-value using the P-value approach.
The P-value is between: \begin{enumerate} \item[1.] 0.15 and 0.10 \item[2.] 0.005 and 0.0025 \item[3.] 0.001 and 0.0005 \item[4.] 0.20 and 0.15 \end{enumerate} \end{problem}

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Problem 200

IQ scores: Scores on an IQ test are normally distributed. A sample of 25 IQ scores had standard deviation $s=8$ . The developer of the test claims that the population standard deviation is greater than $\sigma=17$ . Do these data provide sufficient evidence to support this claim? Use the $\alpha=0.10$ level of significance.
Part: $0 / 5$ $\square$
Part 1 of 5
State the appropriate null and alternate hypotheses. $\begin{array}{l} H_{0}: \sigma(\text { Choose one) } \nabla \\ H_{1}: \sigma(\text { Choose one }) \nabla \end{array}$ Start over
This hypothesis test is a $\square$ (Choose one) test.

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Statistical Inference

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Problem 128

If n=140n=140n=140 and pundefined=0.55\widehat{p}=0.55p​=0.55, construct a 99%99 \%99% confidence interval. Give your answers to three decimals □\square□ <p<<p<<p< □\square□

Problem 129

If n=520n=520n=520 and p^=0.62\hat{p}=0.62p^​=0.62, construct a 99%99 \%99% confidence interval.Give your answers to three decimals. □\square□ <p<<p<<p< □\square□

Problem 130

What is P-value? Give a formal definition.Problem 7) State the Central Limit Theorem.

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2. Find the required sample size for 95%95 \%95% confidence level with a margin of error of 4%4 \%4% and standard deviation =0.35=0.35=0.35.

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Problem 169

Destiny performs a hypothesis test on a population with μ=115\mu=115μ=115 and σ=30\sigma=30σ=30. Her sample size is 25 with a mean of 120.4. Which of the following correctly depicts the zzz-statistic for this data? 0.04 0.90 3.53 3.93

Problem 170

Problem 171

College student spends studying each week. They take a simple random sample of 91 students and compute a sample mean of 5.4 hours per week with a standard deviation of 1.2 hours. Find the 95%95 \%95% confidence interval for the population mean. Follow the PANIC acronym and answer each part.

Problem 172

Problem 173

If $n=140$ and $\widehat{p}=0.55$ , construct a $99 \%$ confidence interval. Give your answers to three decimals $\square$ $<p<$ $\square$

If $n=520$ and $\hat{p}=0.62$ , construct a $99 \%$ confidence interval.
Give your answers to three decimals. $\square$ $<p<$ $\square$

What is P-value? Give a formal definition.
Problem 7) State the Central Limit Theorem.

2. Find the required sample size for $95 \%$ confidence level with a margin of error of $4 \%$ and standard deviation $=0.35$ .

Destiny performs a hypothesis test on a population with $\mu=115$ and $\sigma=30$ . Her sample size is 25 with a mean of 120.4. Which of the following correctly depicts the $z$ -statistic for this data? 0.04 0.90 3.53 3.93

College student spends studying each week. They take a simple random sample of 91 students and compute a sample mean of 5.4 hours per week with a standard deviation of 1.2 hours. Find the $95 \%$ confidence interval for the population mean. Follow the PANIC acronym and answer each part.