Statistical Inference

Problem 201

State the appropriate null and alternate hypotheses. $\begin{array}{l} H_{0}: p_{1}=p_{2} \\ H_{1}: p_{2}<p_{1} \end{array}$
This hypothesis test is a right-tailed $\nabla$ test.
Part 2 of 4
Find the $P$ -value. Round the answer to four decimal places. $P \text {-value }=0.1446$
Part: $2 / 4$
Part 3 of 4
Determine whether to reject $H_{0}$ . (Choose one) $\mathbf{\nabla}$ the null hypothesis $H_{0}$ . Skip Part Check Save For Later Submit Assignmen - 2024 McGraw Hill LLC. All Rights Reserved. Terms of Use I Privacy Center Accessib MacBook Pro

See Solution

Problem 202

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 52 family practitioners in Los Angeles had mean earnings of $\bar{x}=\$ 193,130$ with a standard deviation of $\$ 42,047$ . 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 $\alpha=0.10$ 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. $\begin{array}{l} H_{0}: \mu=178,258 \\ H_{1}: \mu>178,258 \end{array}$
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. $t=2.55$
Part: $2 / 5$
Part 3 of 5 (c) Compute the $P$ -value. Round the answer to at least four decimal places. $P \text {-value }=$ $\square$ Start over

See Solution

Problem 203

```latex State the conclusion. Choose the correct answer below.
A. Reject $\mathrm{H}_{0}$ . There is sufficient evidence to support the claim that fewer than half of all teenagers in the population feel that grades are the greatest source of pressure.
B. Reject $\mathrm{H}_{0}$ . There is insufficient evidence to support the claim that fewer than half of all teenagers in the population feel that grades are the greatest source of pressure.
C. Do not reject $\mathrm{H}_{0}$ . There is sufficient evidence to support the claim that fewer than half of all teenagers in the population feel that grades are the greatest source of pressure.
D. Do not reject $\mathrm{H}_{0}$ . There is insufficient evidence to support the claim that fewer than half of all teenagers in the population feel that grades are the greatest source of pressure.
The user took a picture with their phone and the text was extracted above. The user then had a dialogue with an AI Assistant to help clarify the instructions.
Dialogue Transcript:
Hello! It looks like you've captured a math problem related to hypothesis testing. However, the critical information needed to choose the correct conclusion is missing. Could you please provide the test statistic, p-value, or any significance level used for this hypothesis test? With that information, I can help determine the correct conclusion for the problem.
P value= 0.0000 ```

See Solution

Problem 204

P-value $=0.0000$ (Round to three decimal places as needed.) What is the conclusion about the null hypothesis?
A. Fail to reject the null hypothesis because the $P$ -value is less than or equal to the significance level, $\alpha$ .
B. Reject the null hypothesis because the $P$ -value is greater than the significance level, $\alpha$ . C. Reject the null hypothesis because the $P$ -value is less than or equal to the significance level, $\alpha$ . D. Fail to reject the null hypothesis because the $P$ -value is greater than the significance level, $\alpha$ .

See Solution

Problem 205

Zappos is an online retailer based in Nevada and employs 1,500 employees. One of their competitors, Amazon.com, would like to test the hypothesis that the average age of a Zappos employee is less than 36 years old. A random sample of 22 Zappos employees was found to have an average age of 33.9 years. The standard deviation for this sample was 4.1 years. Amazon would like to set $a=0.025$ . The conclusion for this hypothesis test would be that because the test statistic is less than the critical value, we cannot conclude that the average age of Zappos employees is less than 36 years old. more than the critical value, we cannot conclude that the average age of Zappos employees is less than 36 years old. less than the critical value, we can conclude that the average age of Zappos employees is less than 36 years old. more than the critical value, we can conclude that the average age of Zappos employees is less than 36 years old.

See Solution

Problem 206

Anna Washburn 11/17/24 4:32 PM Question 8 of 8 This quiz: 8 point(s) possible This question: 1 point(s) possible Submit quiz
Conduct a full hypothesis test and determine if the given claim is supported or not supported at the 0.05 significance level. Round all amounts and standard scores to two decimal places. A researcher claims that the amounts of acetaminophen in a certain brand of cold tablets have a mean different from the 600 milligrams claimed by the manufacturer. The mean acetaminophen content $f$ a random sample of 46 tablets is 603.3 milligrams. Test whether the claim that the mean amount of acetaminophen is different from 600 milligrams is supported or not supported. Assume that the population standard deviation is 4.9 milligrams. not supported supported

See Solution

Problem 207

Given below are the number of successes and sample size for a simple random sample from a population. $x=7, n=50,99 \% \text { level }$ a. Determine the sample proportion. b. Decide whether using the one-proportion z-interval procedure is appropriate. c. If appropriate, use the one-proportion z-interval procedure to find the confidence interval at the specified confidence level. d. If appropriate, find the margin of error for the estimate of $p$ and express the confidence interval in terms of the sample proportion and the margin of error. a. $\hat{p}=0.14$ (Type an integer or a decimal. Do not round.) b. Is the one-proportion z-interval procedure appropriate? Select all that apply. A. The procedure is appropriate because the necessary conditions are satisfied. B. The procedure is not appropriate because $x$ is less than 5 . C. The procedure is not appropriate because $n-x$ is less than 5 . D. The procedure is not appropriate because the sample is not a simple ramom sample. c. Select the correct choice below and, if necessary, fill in the answer boxes within your choice. A. The $99 \%$ confidence interval is from .0135 to .2665 . (Round to three decimal places as needed. Use ascending order.) B. The one-proportion z-interval procedure is not appropriate.

See Solution

Problem 208

SAT scores are normally distributed with a mean of 1,500 and a standard deviation of 300 . An administrator at a college is interested in estimating the average SAT score of first-year students. If the administrator would like to limit the margin of error of the $85 \%$ confidence interval to 5 points, how many students should the administrator sample? Make sure to give a whole number answer.

See Solution

Problem 209

You measure 36 randomly selected textbooks' weights, and find they have a mean weight of 76 ounces. Assume the population standard deviation is 10 ounces. Based on this, construct a $95 \%$ confidence interval for the true population mean textbook weight.
Give your answers as decimals, to two places

See Solution

Problem 210

If $n=140$ and $\widehat{p}=0.3$ , construct a $95 \%$ confidence interval about the population proportion. Round your answers to three decimal places.
Preliminary: a. Is it safe to assume that $n \leq 0.05$ of all subjects in the population? No Yes b. Verify $n \widehat{p}(1-\widehat{p}) \geq 10$ . Round your answer to one decimal place. $n \widehat{p}(1-\widehat{p})=$ $\square$
Confidence Interval: What is the $95 \%$ confidence interval to estimate the population proportion? Round your answer to two decimal places. $\square$ $<p<$ $\square$

See Solution

Problem 211

If $\mathrm{n}=400$ and $\widehat{p}(p-h a t)=0.9$ , construct a $99 \%$ confidence interval. Give your answers to three decimals $\square$ $<p<$ $\square$

See Solution

Problem 212

Out of 300 people sampled, 150 had kids. Based on this, construct a $95 \%$ confidence interval for the true population proportion of people with kids.
Give your answers as decimals, to three places

See Solution

Problem 213

About 72\% of MCC students believe they can achieve the American dream and about 64\% of Ferris State Universtiy students believe they can achieve the American dream. Construct a $95 \%$ confidence interva for the difference in the proportions of Montcalm Community College students and Ferris State University students who believe they can achieve the American dream.
There were 100 MCC students surveyed and 100 FSU students surveyed. a. With $95 \%$ confidence the difference in the proportions of MCC and FSU students who believe they can achieve the American dream is $\square$ (round to 3 decimal places) and $\square$ (round to decimal places). b. A review of what confidence interval means: If many groups of 100 randomly selected MCC students and 100 randomly selected FSU students were surveyed, then a different confidence interval would be produced from each group. About $\square$ percent of these confidence intervals will contain the true population proportion of the difference in the proportions of MCC students and FSU students who believe they can achieve the American dream about $\square$ percent will not contain the true

See Solution

Problem 214

Math
Which of the following regressions represents the strongest linear relationship between x and y ? atter Plots and Linear \begin{tabular}{llll} Regression 1 & $\underline{\text { Regression 2 }}$ & $\underline{\text { Regression 3 }}$ & Regression 4 \\ \hline $y=a x+b$ & $y=a x+b$ & $y=a x+b$ & $y=a x+b$ \\ $a=-9.2$ & $a=-16.3$ & $a=-5.9$ & $a=3.8$ \\ $b=13.5$ & $b=-5.9$ & $b=17.4$ & $b=-10.5$ \\ $r=-1.1128$ & $r=-0.4755$ & $r=-0.0546$ & $r=0.1161$ \end{tabular}
Answer egressions h of Linear Relationships Regression 1 Regression 2 Regression 3 Regression 4

See Solution

Problem 215

Homework Part 1 of 4 Points: 0 of 1
According to a poll, 718 out of 1072 randomly selected adults living in a certain country felt the laws covering the sale of firearms should be more strict. a. What is the value of $\hat{p}$ , the sample proportion who favor stricter gun laws? b. Check the conditions to determine whether the CLT can be used to find a confidence interval. c. Find a $95 \%$ confidence interval for the population proportion who favor stricter gun laws. d. Based on your confidence interval, do a majority of adults in the country favor stricter gun laws? a. The value of $\hat{p}$ , the sample proportion who favor stricter gun laws, is $\square$ (Round to two decimal places as needed.)

See Solution

Problem 216

According to a poll, 651 out of 1026 randomly selected smokers polled believed they are discriminated against in public life or in employment because of their smoking. a. What percentage of the smokers polled believed they are discriminated against because of their smoking? b. Check the conditions to determine whether the CLT can be used to find a confidence interval. c. Find a $95 \%$ confidence interval for the population proportion of smokers who believe they are discriminated against because of their smoking. d. Can this confidence interval be used to conclude that the majority of smokers believe they are discriminated against because of their smoking? Why or why not? a. The percentage of those taking the poll believed they are discriminated against because of their smoking is $\square$ \%. (Round to one decimal place as needed.)

See Solution

Problem 217

A sample of 10 observations provides the following statistics: (You may find it useful to reference the $\boldsymbol{t}$ table.) $s_{x}=13, s_{y}=18$ , and $s_{x y}=117.22$
0-1. Calculate the sample correlation coefficlent $r_{x y}$ - Note: Round your answer to 4 decimal places.
Sample correlation coefficient
0-2. Interpret the sample correlation coefficient $r_{x y}$ -
The correlation coefficient indicates a positive linear relationship.
The correlation coefficient Indicates a negative linear relationship. The correlation coefficlent indicates no linear relationship. b. Specify the hypotheses to determine whether the population correlation coefficient is positive. $H_{0}: \rho_{x y}=0 ; H_{A}: \rho_{x y} \neq 0$ $H_{0}: \rho_{x y} \geq 0 ; H_{A}: \rho_{x y}<0$ $H_{0}: \rho_{x y} \leq 0 ; H_{A}: \rho_{x y}>0$ c-1. Calculate the value of the test statistic. Note: Round final answer to 3 decimal places. Test statistic c-2. Find the $p$ -value. p-value $<0.01$ $0.01 \leq p$ -value $<0.025$ $0.025 \leq p$ -value $<0.05$ $0.05 \leq p$ -value $<0.10$ $p$ -value $\geq 0.10$ d. At the $5 \%$ significance level, what is the conclusion to the test? Reject $\mathrm{H}_{0}$ ; we can state the population correlation is positive. Reject $H_{0}$ : we cannot state the population correlation is positive. Do not reject $H_{0}$ ; we can state the population correlation is positive. Do not reject $H_{0}$ ; we cannot state the population correlation is positive.

See Solution

Problem 218

For a sample of 41 New England cities, a sociologist studies the crime rate in each city as a function of its poverty rate and its median income. He finds that $\operatorname{SSE}=4,182,663$ and $S S T=7,732,451$ . a. Calculate the standard error of the estimate.
Note: Round your answer to 4 decimal places. Standard Error $\quad 331.7681$ b-1. What proportion of the sample variation in crime rate is explained by the variability in the predictor variables? Note: Round your answer to 4 decimal places. Explained proportion $\quad 0.4593$ b-2. What proportion is unexplained? Note: Round your answer to 4 decimal places.

See Solution

Problem 219

Part 2 of 4 Points: 0 of 1
In a simple random sample of 1600 people age 20 and over in a certain country, the proportion with a certain disease was found to be 0.090 (or $9.0 \%$ ). Complete parts (a) through (d) below. a. What is the standard error of the estimate of the proportion of all people in the country age 20 and over with the disease? $S E_{\text {est }}=0.0072$ (Round to four decimal places as needed.) b. Find the margin of error, using a $95 \%$ confidence level, for estimating this proportion. $\mathrm{m}=$ $\square$ (Round to three decimal places as needed.)

See Solution

Problem 220

Attempt: 1 of 1 $=1$ 2 $=3$ $=4$ $=5$ $=6$ $=7$ $=8$
Compute the correlation coefficient. \begin{tabular}{c|ccccccc} $x$ & 40 & 26 & -1 & 15 & 13 & 11 & -2 \\ \hline $y$ & 0 & -10 & -7 & 19 & 26 & 5 & 35 \end{tabular}
The correlation coefficient is $r=$ . Round the answer to three decimal places as needed.

See Solution

Problem 221

A poll in 2017 reported that 697 out of 1034 adults in a certain country believe that marijuana should be legalized. When this poll about the same subject was first conducted in 1969 , only $12 \%$ of the adults of the country supported legalization. Assume the conditions for using the CLT are met. Complete parts (a) through (d) below. a. Find and interpret a $99 \%$ confidence interval for the proportion of adults in the country in 2017 that believe marijuana should be legalized.
The 99\% confidence interval for the proportion of adults in the country in 2017 that believe marijuana should be legalized is ( $\square$ , (Round to three decimal places as needed.)

See Solution

Problem 222

A poll in 2017 reported that 697 out of 1034 adults in a certain country believe that marijuana should be legalized. When this poll about the same subject was first conducted in 196 only $12 \%$ of the adults of the country supported legalization. Assume the conditions for using the CLT are met. Complete parts (a) through (d) below. a. Find and interpret a $99 \%$ confidence interval for the proportion of adults in the country in 2017 that believe marijuana should be legalized.
The $99 \%$ confidence interval for the proportion of adults in the country in 2017 that believe marijuana should be legalized is ( 0.636 ; 0.712 ). (Round to three decimal places as needed.) Interpret this interval. Select the correct choice below and fill in the answer boxes to complete your choice. (Type integers or decimals rounded to three decimal places as needed.) A. There is a $\square$ \% chance that the sample proportion of adults who believe marijuana should be legalized is between $\square$ and $\square$ . B. We are $\square$ \% confident that every sample proportion of adults who believe marijuana should be legalized is between $\square$ and d c. We are $99 \%$ confident that the population proportion of adults who believe marijuana should be legalized is between 0.636 and 0.712 . D. There is a $\square$ \% chance that the population proportion of adults who believe marijuana should be legalized is between $\square$ and $\square$ b. Find and interpret a 95\% confidence interval for this population parameter.
The $95 \%$ confidence interval for the proportion of adults in the country in 2017 that believe marijuana should be legalized is ( $\square$ , ). (Round to three decimal places as needed.)

See Solution

Problem 223

(D) State the Mypotheses - Example Suppose you want to see it there is an association between gender (mater, femate) and Preference for a new product (ilke, dislike) - Null Hypothesis (HO): There is no association between gender and product preference. - Alternative Hypothesis (H1): There is an association between gender and product preference.
Step 2: Collect and Summarize the Data Example: Survey 200 people and record their gender and product preference. \begin{tabular}{|l|l|l|l|l|} \hline & Like & Dislike & Total \\ \hline Male & 50 & 30 & 80 & \\ \hline Total & 70 & 50 & 120 \\ \hline \end{tabular}
Step 3: Calculate the Expected Frequencies - Example: For each cell in the table, calculate the expected frequency under the assumption that HO is true. - Expected frequency for males who like the product: $80 * 120200=48$ - Expected frequency for females who like the product: $120 * 120200=72$
Step 4: Calculate the Chi-Square Statistic - Example: Use the formula: $\chi 2=\sum(O-E) 2 E$ , where $O$ is the observed frequency and $E$ is the expecte d frequency. - Chi-Square calculation: $\chi 2=(50-48) 248+(30-32) 232+(70-72) 272+(50-48) 248$
Step 5: Determine the p-value and Conclusion - Example: With degrees of freedom $(\mathrm{df})=($ rows -1 $) *($ columns -1 $)=1$ , you can find the $p$ value using a Chi-Square distribution table or software. - Compare the p-value to your significance level (typically 0.05). If the pvalue is less than 0.05 , reject HO and conclude there is an association.

See Solution

Problem 224

$\text{Compute the correlation coefficient between the price of eggs and the price of milk. Round the answer to at least three decimal places. The correlation coefficient is } r = \square.$
\text{Price of eggs and milk: The following table presents the average price in dollars for a dozen eggs and a gallon of milk for each month from February through September 2013. Use the TI-84 Plus calculator to answer the following.}
$\begin{array}{|c|c|} \hline \text{Dozen Eggs} & \text{Gallon of Milk} \\ \hline 1.77 & 3.58 \\ 1.94 & 3.58 \\ 1.67 & 3.47 \\ 1.65 & 3.47 \\ 1.69 & 3.47 \\ 2.01 & 3.50 \\ 1.96 & 3.52 \\ 1.80 & 3.43 \\ \hline \end{array}$

See Solution

Problem 225

A poll reported that $70.1 \%$ of adults in a certain country believe that organic produce is better for health than conventionally grown varieties. Assume the sample size was 1000 and that the conditions for using the CLT are met. Complete parts (a) through (d) below. a. Find and interpret a $95 \%$ confidence interval for the proportion of adults in the country who believe organic produce is better for health.
The $95 \%$ confidence interval for the proportion of adults in the country who believe organic produce is better for health is ( $0.673,0.729$ ). (Round to three decimal places as needed.) Interpret this interval. Select the correct choice below and fill in the answer boxes to complete your choice. (Type integers or decimals rounded to three decimal places as needed.) A. We are $\square$ \% confident that every sample proportion of adults in the country who believe organic produce is better for health is between $\square$ and $\square$ I. B. There is a $\square$ \% chance that the sample proportion of adults in the country who believe organic produce is better for health is between $\square$ and $\square$ C. We are $95 \%$ confident that the population proportion of adults in the country who believe organic produce is better for health is between 0.673 and 0.729 . D. There is a $\square$ \% chance that the population proportion of adults in the country who believe organic produce is better for health is between $\square$ and $\square$ . b. Find and interpret an $80 \%$ confidence interval for this population parameter.
The $80 \%$ confidence interval for the proportion of adults in the country who believe organic produce is better for health is ( $0.682,0.720$ ). (Round to three decimal places as needed.) Interpret this interval. Select the correct choice below and fill in the answer boxes to complete your choice. (Type integers or decimals rounded to three decimal places as needed.) A. There is an $\square$ \% chance that the population proportion of adults in the country who believe organic produce is better for health is between $\square$ and $\square$ . B. We are $\square$ \% confident that every sample proportion of adults in the country who believe organic produce is better for health is between $\square$ $\square$ C. There is an $\square$ \% chance that the sample proportion of adults in the country who believe organic produce is better for health is between $\square$ and $\square$ D. We are $\%$ confident that the population proportion of adults in the country who believe organic produce is better for health is between $\square$ and

See Solution

Problem 226

Part 4 of 4 (d) Which of the following is the best interpretation of the correlation coefficient? When the price of eggs rises, it causes the price of milk to rise. When the price of milk rises, it causes the price of eggs to rise. Changes in the price of eggs or milk do not cause changes in the price of the other; the correlation indicates that the prices of milk and eggs tend to go up and down together. Continue (c) 2024 McGraw Hill L

See Solution

Problem 227

www-awu.aleks.com Untitled document - Traditional | Stats | N... Defense | Stats | NBA... Utah vs LA Stats \& P... Ivica Zubac /II Stats /... Home - Northern Ess.. Content Midterm Exam: Ch $1(1,2) 2(1,2,3) 3(1,2,3) 4(1,2) 5(1,2,3) 6(1,2)$ Question 19 of 30 (1 point) I Question Attempt: 1 of 1 Time Remaining: 1:06:58 Jonathan 13 $=14$ $\equiv 15$ 16 17 18 19 20 21 22 23 24
The following table presents the percentage of students who tested proficient in reading and the percentage who tested proficient in math for 5 randomly selected states in the United States. Compute the least-squares regression line for predicting math proficiency from reading proficiency. \begin{tabular}{|l|c|c|} \hline \multicolumn{1}{|c|}{ State } & \begin{tabular}{c} Percent Proficient \\ in Reading \end{tabular} & \begin{tabular}{c} Percent Proficient \\ in Mathematics \end{tabular} \\ \hline Illinois & 75 & 70 \\ \hline Texas & 73 & 78 \\ \hline Ohio & 79 & 76 \\ \hline New York & 75 & 70 \\ \hline Georgia & 67 & 64 \\ \hline \end{tabular}
Send data to Excel
The equation for the least squares regression line is $\hat{y}=$ $\square$ . Round the slope and $y$ -intercept to four decimal places as needed.

See Solution

Problem 228

to conclude that the fund has moderate risk return for a certain fund. The standard deviation of the rate of return is computed to be $3.98 \%$ . Is there suffic distributed.
What are the correct hypotheses for this test? The null hypothesis is $\mathrm{H}_{0}: \sigma=0.06$ . The alternative hypothesis is $\mathrm{H}_{1}: \sigma<0.06$ . Calculate the value of the test statistic. $\chi^{2}=11.880$ (Round to three decimal places as needed.) Use technology to determine the P-value for the test statistic. The $P$ -value is $\square$ $\square$ . (Round to three decimal places as needed.) Clear all

See Solution

Problem 229

You are interested in estimating the the mean age of the citizens living in your community. In order to do this, you plan on constructing a confidence interval; however, you are not sure how many citizens should b included in the sample. If you want your sample estimate to be within 3 years of the actual mean with a confidence level of $96 \%$ , how many citizens should be included in your sample? Assume that the standard deviation of the ages of all the citizens in this community is 18 years.
Sample Size: $\square$

See Solution

Problem 230

6 5/5 points Let $X_{1}, X_{2}, \ldots, X_{n}$ be independent and identically distributed random variables from a uniform distribution on the interval $(0, a)$ . The value of a is unknown. Derive the method of moments estimate of the parameter a. $2\left(X_{1}+\ldots+X_{n}\right) / n$ 2* median( $\mathrm{X} 1, \ldots, \mathrm{Xn}$ ) $\operatorname{Max}\left(X_{1}, X_{2}, \ldots, X_{n}\right)$

See Solution

Problem 231

You want to obtain a sample to estimate a population proportion. At this point in time, you have no reasonable estimate for the population proportion. You would like to be $90 \%$ confident that you esimate is within $4 \%$ of the true population proportion. How large of a sample size is required? $\square$ Do not round mid-calculation.

See Solution

Problem 232

or Variance Question 5, 8.4.12-T Part 2 of 4 HW Score: 56.31\%, 3.94 of 7 points Points: 0 of 1 Save
A simple random sample of 29 filtered $100-\mathrm{mm}$ cigarettes is obtained from a normally distributed population, and the tar content of each cigarette is measured. The sample has a standard deviation of 0.20 mg . Use a 0.05 significance level to test the claim that the tar content of filtered 100-mm cigarettes has a standard deviation different from 0.30 mg , which is the standard deviation for unfiltered king-size cigarettes. Complete parts (a) through (d) below. a. What are the null and alternative hypotheses? A. $\mathrm{H}_{0}: \sigma>0.30 \mathrm{mg}$ B. $H_{0}: \sigma=0.30 \mathrm{mg}$ $H_{1}: \sigma \leq 0.30 \mathrm{mg}$ $H_{1}: \sigma<0.30 \mathrm{mg}$ C. $\mathrm{H}_{0}: \sigma=0.30 \mathrm{mg}$ $\mathrm{H}_{1}: \sigma \neq 0.30 \mathrm{mg}$ $\text { D. } \begin{aligned} H_{0}: \sigma \neq 0.30 \mathrm{mg} \\ H_{1}: \sigma=0.30 \mathrm{mg} \end{aligned}$ b. Find the test statistic. $\chi^{2}=$ $\square$ (Round to three decimal places as needed.) Clear all Check answer Get more help 7:55 PM

See Solution

Problem 233

Shown to the right is a certain population, in billions, for seven selected years from 1950 through 2007. Using a graphing utility's logistic regression option, we obtain the logistic growth model shown below for population, $f(x)$ , in billions, $x$ years after 1949. How well does the function model the data for 2007? $f(x)=\frac{10.23}{1+3.55 e^{-0.030 x}}$ \begin{tabular}{|c|c|} \hline x, Number of Years after 1949 & y, Population (billions) \\ \hline $1(1950)$ & 2.4 \\ \hline $11(1960)$ & 2.8 \\ \hline $21(1970)$ & 3.5 \\ \hline $31(1980)$ & 4.3 \\ \hline $41(1990)$ & 5.1 \\ \hline $51(2000)$ & 5.9 \\ \hline $58(2007)$ & 6.3 \\ \hline \end{tabular}
For 2007, the function $\square$ the population to one decimal place. slightly overestimates slightly underestimates accurately predicts

See Solution

Problem 234

Triglycerides are a form of fat found in the body. Using data from a certain organization, determine whether men have higher triglyceride levels than women. a. Report the sample means and state which group had the higher sample mean triglyceride level. Refer to the Minitab output in figure (A). b. Carry out a hypothesis test to determine whether men have a higher mean triglyceride level than women. Assume that all necessary conditions for carrying out a hypothesis test hold. Refer to the Minitab output provided in figure (A). Output for three different alternative hypotheses is provided-see figures (B), (C), and (D)-and you must choose and state the most appropriate output. Use a significance level of 0.05 a. The sample mean for the women was $\square$ . (Type an integer or a decimal.)
Minitab Output (A) (B)
B: T-Test of difference $=0($ vs $<$ ); T-Value $=-2.60 \quad P$ -value $=0.006$ (C)
C: T-Test of difference $=0($ vs $>):$ T-Value $=-2.60 \quad P$ -value $=0.994$ (D)
D: T-Test of difference $=0($ vs $\neq)$ : T-Value $=-2.60 \quad$ P-value $=0.011$ Clear all Check answer

See Solution

Problem 235

A chemical production process uses water-cooling to carefully control the temperature of the system so that the correct products are obtained from the chemical reaction. The system must be maintained at a temperature of $120{ }^{\circ} \mathrm{F}$ A technician suspects that the average temperature is different from $120^{\circ} \mathrm{F}$ and checks the process by taking temperature readings at 12 randomly selected times over a 4-hour period. Here are the data. \begin{tabular}{|l|l|l|l|l|l|} \hline 120.1 & 124.2 & 122.4 & 124.4 & 120.8 & 121.4 \\ \hline 121.8 & 119.6 & 120.2 & 121.5 & 118.7 & 122.0 \\ \hline \end{tabular}
To access the data, click the link for your preferred software format.
CSV Excel (xls) Excel (xlsx) JMP Mac-Text Minitab PC-Text R SPSS TI CrunchIt!
Identify the statistical test to perform. one-sample $t$ test for $\mu$ one-sample $z$ test for $p$ one-sample $t$ test for $\bar{x}$ two-sample $t$ test for $\mu$
Explain whether the conditions for inference have been met. The sample was randomly selected and the sample size is large, so the conditions for inference have been met. The sample was not randomly selected, so lale conditions for inference have not been met. The sample was randomly selected, the sample size is not large, but the sample data have at least one outlier or show strong skewness, so the conditions for inference have not been met. The sample was randomly selected, the sample size is not large, but the sample data have no outliers or strong skewness, so the

See Solution

Problem 236

Hypothesis Question 3, 9.5.59 HW Score: $17.14 \%, 1.71$ of 10 points of Two Means Part 7 of 7 Points: 0 of 1 Save
Triglycerides are a form of fat found in the body. Using data from a certain organization, determine whether men have higher triglyceride levels than women. a. Report the sample means and state which group had the higher sample mean triglyceride level. Refer to the Minitab output in figure (A). b. Carry out a hypothesis test to determine whether men have a higher mean triglyceride level than women. Assume that all necessary conditions for carrying out a hypothesis test hold. Refer to the Minitab output provided in figure (A). Output for three different alternative hypotheses is provided-see figures (B), (C), and (D)-and you must choose and state the most appropriate output. Use a significance level of 0.05 . (i) Click the icon to view the Minitab outputs.
Find the test statistic for this test. $\mathrm{t}=-2.60$ (Type an integer or a decimal.) Find the p-value for this test. $p$ -value $=0.006$ (Type an integer or a decimal.) What is the conclusion for this test? A. Do not reject $\mathrm{H}_{0}$ . Men have a significantly higher mean triglyceride level than women. B. Do not reject $\mathrm{H}_{0}$ . Men do not have a significantly higher mean triglyceride level than women. C. Reject $\mathrm{H}_{0}$ . Men do not have a significantly higher mean triglyceride level than women. D. Reject $H_{0}$ . Men have a significantly higher mean triglyceride level than women. Clear all Chack answar

See Solution

Problem 237

Save
A random sample of male college baseball players and a random sample of male college soccer players were obtained independently and weighed. The accompanying table shows the unstacked weights (in pounds). The distributions of both data sets suggest that the population distributions are roughly Normal. Determine whether the difference in means is significant, using a significance level of 0.05 .
Click the icon to view the data table.
Let $\mu_{\text {Baseball }}$ be the population mean weight (in pol male college soccer players. Determine the hypoth $\begin{array}{l} H_{0}: \mu_{\text {Baseball }}-\mu_{\text {Soccer }}=0 \\ H_{\mathrm{a}}: \mu_{\text {Baseball }}-\mu_{\text {Soccer }} \neq 0 \end{array}$
Find the test statistic for this test. $\mathrm{t}=\square$ (Round to two decimal places as needed.) $\square$
Get more help -
Weights (in pounds) \begin{tabular}{|cccc|} \hline Baseball & Soccer & Baseball & Full Data Set \\ 184 & 164 & 180 & 153 \\ 193 & 186 & 206 & 165 \\ 186 & 183 & 192 & 169 \\ 179 & 184 & 177 & 151 \\ 190 & 179 & 176 & 144 \\ 202 & 188 & 189 & 171 \\ 183 & 169 & 199 & 174 \\ 172 & 180 & 193 & 177 \\ 202 & 165 & 178 & 172 \\ 219 & 175 & 188 & 184 \\ 223 & 164 & 188 & 152 \\ 190 & 185 & 179 & 159 \\ 167 & 180 & & \\ \hline \end{tabular}
Print Done

See Solution

Problem 238

In a recent tennis tournament, women playing singles matches used challenges on 138 calls made by the line judges. Among those challenges, 37 were found to be successful with the call overturned. a. Construct a $99 \%$ confidence interval for the percentage of successful challenges. b. Compare the results from part (a) to this $99 \%$ confidence interval for the percentage of successful challenges made by the men playing singles matches: $20.7 \%<p<40.8 \%$ . Does it appear that either gender is more successful than the other? a. Construct a 99\% confidence interval. $\square$ \% < p < $\square$ \% (Round to one decimal place as needed.)

See Solution

Problem 239

Refer to the accompanying data display that results from a simple random sample of times (minutes) between eruptions of the Old Faithful geyser. The confidence level of $95 \%$ was used. Complete parts (a) through (d) below.
TInterval $\begin{array}{l} (85.74,91.76) \\ x=88.75 \\ S x=8.897431411 \\ n=36 \end{array}$ a. What are the requirements for using the methods for the mean to construct a confidence interval estimate of a population mean? A. The sample is a simple random sample and the population is normally distributed or $n \leq 30$ . B. The sample is a simple random sample, the population is normally distributed, and $n>30$ . C. The sample is a simple random sample, the population is normally distributed, and $n \leq 30$ . D. The sample is a simple random sample and the population is normally distributed or $n>30$ . b. What does it mean to say that the confidence interval methods for the mean are robust against departures from normality? A. The methods do not work well with distributions that are not normal. have only one mode, and should not include outliers. C. The methods only work with distributions that are not normal. should include outliers.

See Solution

Problem 240

Find the sample size needed to estimate the mean age of movie patrons such that it can be said with $90 \%$ confidence that the sample mean is within 1.5 years of the population mean. Assume that $\sigma=19.6$ years, based on a previous report. Should the sample be obtained from one movie at one theater?
The required sample size is 462 . (Round up to the nearest integer.) Should the sample be obtained from one movie at one theater? A. The sample should not be obtained from one movie at one theater, because that sample could be easily biased. Instead, a simple random sample of the broader population should be obtained. B. The sample should not be obtained from one movie at one theater, because that sample could be easily biased, Instead, a stratified sample of the broader population should be obtained. C. The sample should be obtained from one movie at one theater, because that sample is representative of the population. D. The sample should not be obtained from one movie at one theater, because that sample could be easily biased. Instead, a cluster sample of the broader population should be obtained.

See Solution

Problem 241

The pulse rates of 167 randomly selected adult males vary from a low of 38 bpm to a high of 110 bpm . Find the minimum sample size required to estimate the mean pulse rate of adult males. Assume that we want $98 \%$ confidence that the sample mean is within 3 bpm of the population mean. Complete parts (a) through (c) below. a. Find the sample size using the range rule of thumb to estimate $\sigma$ . $\mathrm{n}=$ $\square$ (Round up to the nearest whole number as needed.)

See Solution

Problem 242

The test statistic of $z=2.44$ is obtained when testing the claim that $p>0.32$ . This is a right-tailed test. Using a 0.10 significance level, complete parts (a) and (b). Click here to view the standard normal distribution table for negative z scores. Click here to view the standard normal distribution table for positive z scores. a. Find the critical value(s).
Select the correct choice below and fill in the answer box(es) within your choice. (Round to two decimal places as needed.) A. There are two critical values; the lower critical value is $\square$ and the upper critical value is $\square$ . B. There is one critical value; the critical value is 1.28 . b. Should we reject $\mathrm{H}_{0}$ or should we fail to reject $H_{0}$ ? A. $\mathrm{H}_{0}$ should not be rejected, since the test statistic is not in the critical region. B. $\mathrm{H}_{0}$ should not be rejected, since the test statistic is in the critical region. C. $\mathrm{H}_{0}$ should be rejected, since the test statistic is not in the critical region. D. $\mathrm{H}_{0}$ should be rejected, since the test statistic is in the critical region.

See Solution

Problem 243

Part 1 of 4 omework
The claim is that smokers have a mean cotinine level greater than the level of $2.84 \mathrm{ng} / \mathrm{mL}$ found for nonsmokers. (Cotinine is used as a biomarker for exposure to nicotine.) The sample size is $n=869$ and the test statistic is $t=53.714$ . Use technology to find the $P$ -value. Based on the result, what is the final conclusion? Use a significance level of 0.05 .
State the null and alternative hypotheses. $\mathrm{H}_{0}: \mu$ $\square$ $\square$ $\mathrm{H}_{1}: \mu$ $\square$ $\square$ (Type integers or decimals. Do not round.)

See Solution

Problem 244

When the population mean is known, then
Select one: a. We used the population mean to estimate the sample SD b. We used the sample SD to estimate the population mean c. We used the sample mean to estimate the population mean d. None of the above

See Solution

Problem 245

Current Attempt in Progress Transgender Adults A study ${ }^{1}$ of transgender adults examines the age at which they began to transition and the age of their earliest memories of gender dysphoria. (In this exercise, we examine, the age beginning transition and in Exercise 6.107 we examine the age of their earliest memories.) In the study of 210 transgender adults, the mean age at which they began transitioning was 32.6 with a standard deviation of 18.2. Find a 95\% confidence interval for the mean age at which transgender adults begin transitioning. ${ }^{1}$ ZaliznyakM, Bresee C, and Garcia MM, "Age at First Experienceof Gender Dysphoria Among Transgender Adults Seeking GenderAffirming Surgery," JAMA Network Open, March 16,2020. Some of the data has been approximated from informationin the paper.
Confidence interval (round to three decimal places): $\square$ i to $\square$

See Solution

Problem 246

n this exercise, test $H_{0}: p=0.5 \text { vs } H_{a}: p>0.5$ with $\hat{p}=0.55$ for different sample sizes. In parts (a)-(c), use StatKey or other technology to find the $p$ -value. Click here to access StatKey. (a) $\hat{p}=\frac{55}{100}=0.55$
Round your answer to two decimal places. $p$ -value $=$ $\square$
Are the results significant at the $5 \%$ level? (b) $\hat{p}=\frac{275}{500}=0.55$
Round your answer to three decimal places. $p$ -value $=$ $\square$ Are the results significant at the $5 \%$ level? $\square$ (c) $\hat{p}=\frac{550}{1000}=0.55$
Round your answer to three decimal places. $p$ -value $=$ $\square$ i
Are the results significant at the $5 \%$ level? $\square$ (d) Which provides the strongest evidence for the alternative hypothesis? $\square$

See Solution

Problem 247

2 Multiple Choice 1 point slope of the regression line?
1. The cost of flood insurance increases by $\$ 500$ for every foot of elevation increase II. The cost of flood insurance is $\$ 10,000$ regarders of elevation III. The cost of flood insurance decreases by $\$ 500$ for every foot of elevation increase II only Neither 1, nor II, nor III I only III only 3 True or False 1 point
If, on average, $y$ increase as $x$ increase, the comrelation coefficient is positive. True False 4 Multiple Choice 1 point
Which of the following are reasons why observations with $r$ values close to 1 cannot be used to establish causality? The relationship between variables could be negative. The observations mary not include a sufficient number of samples. The relationship between variables may be nonlinear. An unidentified third variable corld be influencing both variables under investigation.

See Solution

Problem 248

14 Multiple Choice 1 point
Of the following, which is true of Pearson's correlation coefficient, $r$ ? $r$ cannot be less than 0 or greater than 1 $r=0$ indicates a perfect correlation between $x$ and $y$ $|r|$ is greater than 1.0 unless the points on a scatterplot line up exactly The value of $r$ does not depend on the units of $y$ and $x$ . 15 Multiple Choice 1 point
You're reviewing data from a study comparing preferences in young children with perceptions of typical gender behav calculated least squares regression line.
While you are performing your analysis, one of the experimenters notifies you that the circled data point may have been (circled) point and notice that both the $y$ -intercept and slope of the regression line changed. Which of the following accurately describes the suspect (circled) point? I. The suspect data point had the potential for being an influential observation because its $x$ -value was far away fron II. The suspect data point was confirmed as an influential observation because its removal changed the regression lin III. The suspect data point was confirmed as an influential observation because its removal changed the regression li II and III only

See Solution

Problem 249

This question: 1 point(s) possib
Given below are the number of successes and sample size for a simple random sample from a population. $x=6, n=40,90 \% \text { level }$ a. Determine the sample proportion. b. Decide whether using the one-proportion z-interval procedure is appropriate. c. If appropriate, use the one-proportion z-interval procedure to find the confidence interval at the specified confidence level. d. If appropriate, find the margin of error for the estimate of $p$ and express the confidence interval in terms of the sample proportion and the margin of error. a. $\hat{p}=$ $\square$ (Type an integer or a decimal. Do not round.) b. Is the one-proportion z-interval procedure appropriate? Select all that apply. A. The procedure is appropriate because the necessary conditions are satisfied. B. The procedure is not appropriate because $n-x$ is less than 5 . C. The procedure is not appropriate because the sample is not à simple random sample. D. The procedure is not appropriate because x is less than 5 . c. Select the correct choice below and, if necessary, fill in the answer boxes within your choice. A. The $90 \%$ confidence interval is from $\square$ to $\square$ . (Round to three decimal places as needed. Use ascending order.) B. The one-proportion z-interval procedure is not appropriate. d. Select the correct choice below and, if necessary, fill in the answer boxes within your choice. A. The margin of error is $\square$ . The $90 \%$ confidence interval is $\square$ $\pm$ $\square$ .

See Solution

Problem 250

44.5. Provide two recommendations for decreasing the margin of error of the interval.
Select the two recommendations that would decrease the margin of error of the interval. A. Decrease the standard deviation of hours worked. B. Decrease the confidence level. C. Increase the confidence level. D. Decrease the sample size E. Increase the sample size F. Use fewer degrees of freedom.

See Solution

Problem 251

taking 200 SRSs of 100 students from a population in which the true proportion who recycle is 0.50 . (a) Explain why the sample result does not give convincing evidence that more than half of the school's students recycle. (b) Suppose instead that 63 students in the class's sample had said "Yes." Explain why this result would give strong evidence that a majority of the school's students recycle.

See Solution

Problem 252

Use the results from a survey of a simple random sample of 1026 adults. Among the 1026 respondents, $74 \%$ rated themselves as above average drivers. We want to test the claim that $\frac{7}{10}$ 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.
759 (Round to the nearest whole number as needed.) b. Identify the sample proportion and use the symbol that represents it. $\hat{p}=0.74$ (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.)

See Solution

Problem 253

A certain drug is used to treat asthma. In a clinical trial of the drug, 19 of 266 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 $9 \%$ of treated subjects experienced headaches. Use the normal distribution as an approximation to the binomial distribution and assume a 0.05 significance level to complete parts (a) through (e) below. $\begin{array}{l} 1 \text { 1-PropZTest } \\ \text { prop }<0.09 \\ z=-1.058386227 \\ p=0.1449396980 \\ \hat{p}=0.0714285714 \\ n=266 \end{array}$ (Round to two decimal places as needed.) c. What is the P -value? $P$ -value $=0.1449$ (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. $H_{0}: p<0.09$ B. $H_{0}: p>0.09$ C. $H_{0}: p \neq 0.09$ D. $H_{0}: p=0.09$ View an example Get more help - Clear all Check answer

See Solution

Problem 254

1 A six-węek weight loss program of 19 people resulted in a mean weight loss of 6.1 pounds. Weight loss can be assumed normally distributed with known standard deviation of 9.8 lbs. Test whether the weight loss can be claimed to be greater than 0 ? Use the $1 \%$ level of significance. I What is the value of the TEST STATISTIC? $\qquad$ Note: Enter XXX.XX AT LEAST ONE DIGIT BEFORE THE DECIMAL, TWO AFTER and round up. Thus, 134.784 is entered as $134.78,35.295$ is entered as $35.30, .2749$ is entered as $0.27,-.5650$ is entered as -0.57 $\square$ A)

See Solution

Problem 255

ALEKS - Ava Drexler-Amey 3xe/10_u-lgNslkasNW8D8A9PVVfu7nMe56bnd_Vyg51BuSaxDUY2OFMoF5b1Zv10r8-H1SkOK3yuqYtwBSWbk4e-rnDxN-aPljMa5N8wOnqAN-BMdqzxt1H40q?10Bw7QYjlbavbSPXtx-Y Confidence Intervals and Hypothesis Testing Confidence interval for a population proportion $1 / 5$
A researcher wants to estimate the proportion of depressed individuals taking a new anti-depressant drug who find relief. A random sample of 200 individuals who had been taking the drug is questioned; 162 of them found relief from depression. Based upon this, compute a $90 \%$ confidence interval for the proportion of all depressed individuals taking the drug who find relief. Then find the lower limit and upper limit of the $90 \%$ confidence interval.
Carry your intermediate computations to at least three decimal places. Round your answers to two decimal places. (If necessary, consult a list of formulas.)
Lower limit: $\square$ Upper limit: $\square$

See Solution

Problem 256

1
A preparation SAT test score can be assumed normally distributed with a historical mean of 518 and known standard deviation 114. In a training program, 50 students had an average score of 550 on the preparatory test. Test whether there has been improvement over the historical average. Use a $5 \%$ significance level.
1 What is the value of the TEST STATISTIC?
Note: Enter XXX.XX AT LEAST ONE DIGIT BEFORE THE DECIMAL, TWO AFTER and round up. Thus, 134.784 is entered as $134.78,35.295$ is entered as $35.30, .2749$ is entered as $0.27,-.5650$ is entered as -0.57 .

See Solution

Problem 257

A preparation SAT test score can be assumed normally distributed with a historical mean of 518 and known standard deviation 114. In a training program, 50 students had an average score of 550 on the preparatory test. Test whether there has been improvement over the historical average. Use a $5 \%$ significance level. 1 What is the Table A2 CRITICAL VALUE for the problem? Note Answer Input Format for Critical Values.
Note: Enter X.XX AT LEAST ONE DIGIT BEFORE THE DECIMAL, TWO AFTER and round up. Thus, 1.645 is entered as $1.65,-2.326$ is entered as $-2.33,2.054$ is entered as $2.05,-1.282$ is entered as -1.28

See Solution

Problem 258

I
A preparation SAT test score can be assumed normally distributed with a historical mean of 518 and known standard deviation 114. In a training program, 50 students had an average score of 550 on the preparatory test. Test whether there has been improvement over the historical average. Use a $5 \%$ significance level.
1 What is the P-VALUE from Table A2 for testing HO? See two entry formats below.
Note: Two formats possible: 1) Enter LT01 if P-value is less than 0.01 2) Enter XXX.XX AT LEAST ONE DIGIT BEFORE THE DECIMAL, TWO AFTER and round up. Thus, 134.784 is entered as $134.78,35.295$ is entered as $35.30, .2749$ is entered as $0.27,-.5650$ is entered as -0.57

See Solution

Problem 259

A preparation SAT test score can be assumed normally distributed with a historical mean of 518 and known standard deviation 114. In a training program, 50 students had an average score of 550 on the preparatory test. Test whether there has been improvement over the historical average.
1 What is the CONCLUSION for HO , assuming a $1 \%$ significance level? REJ = Reject H0, assume H 1 true. FTR = Fail to Reject H0, assume HO true. CBD = Cannot be Determined.
Note: Enter REJ, FTR or CBD letters.

See Solution

Problem 260

In a survey on 50 gas stations in an area, the mean cost of regular gas was $\$ 1.95$ per gallon. Was this cost less than the national average cost of $\$ 2.00$ where the assumed known standard deviation is $\$ 0.20$ per gallon? Use a $1 \%$ significance level. I What is the value of the TEST STATISTIC?
Note: Enter XXX.XX AT LEAST ONE DIGIT BEFORE THE DECIMAL, TWO AFTER and round up. Thus, 134.784 is entered as $134.78,35.295$ is entered as $35.30, .2749$ is entered as $0.27,-.5650$ is entered as -0.57 . A

See Solution

Problem 261

In a survey on 50 gas stations in an area, the mean cost of regular gas was $\$ 1.95$ per gallon. Was this cost less than the national average cost of $\$ 2.00$ where the assumed known standard deviation is $\$ 0.20$ per gallon? Use a $1 \%$ significance level.
What is the Table A2 CRITICAL VALUE for the problem? Note Answer Input Format for Critical Values.
Note: Enter X.XX AT LEAST ONE DIGIT BEFORE THE DECIMAL, TWO AFTER and round up. Thus, 1.645 is entered as $1.65,-2.326$ is entered as $-2.33,2.054$ is entered as $2.05,-1.282$ is entered as -1.28

See Solution

Problem 262

I In a survey on 50 gas stations in an area, the mean cost of regular gas was $\$ 1.95$ per gallon. Was this cost less than the national average cost of $\$ 2.00$ where the assumed known standard deviation is $\$ 0.20$ per gallon? Use a $1 \%$ significance level. What is the P-VALUE from Table A2 for testing HO? See two entry formats below.
Note: Two formats possible: 1) Enter LT01 if P-value is less than 0.01 2) Enter XXX.XX AT LEAST ONE DIGIT BEFORE THE DECIMAL, TWO AFTER and round up. Thus, 134.784 is entered as $134.78,35.295$ is entered as $35.30, .2749$ is entered as $0.27,-.5650$ is entered as -0.57 $\square$ A

See Solution

Problem 263

In a survey on 50 gas stations in an area, the mean cost of regular gas was $\$ 1.95$ per gallon. Was this cost less than the national average cost of $\$ 2.00$ where the assumed known standard deviation is $\$ 0.20$ per gallon?
What is the CONCLUSION for HO , assuming a $5 \%$ significance level? REJ = Reject H0, assume $\mathrm{H1}$ true. FTR = Fail to Reject H0, assume HO true. CBD = Cannot be Determined.
Note: Enter REJ, FTR or CBD letters. $\square$ A

See Solution

Problem 264

A study time survey is conducted of 39 UGA students and a mean x -bar $=110$ minutes is obtained. Test whether the UGA study times differ from the national average of 120 min ? Assume study times have a normal distribution with a known standard deviation of $\sigma=30$ minutes. Use the $5 \%$ significance level.
I What is the value of the TEST STATISTIC?
Note: Enter XXX.XX AT LEAST ONE DIGIT BEFORE THE DECIMAL, TWO AFTER and round up. Thus, 134.784 is entered as $134.78,35.295$ is entered as 35.30, . 2749 is entered as $0.27,-.5650$ is entered as -0.57 $A$

See Solution

Problem 265

A study time survey is conducted of 39 UGA students and a mean x-bar=110 minutes is obtained. Test whether the UGA study times differ from the national average of 120 min .? Assume study times have a normal distribution with a known standard deviation of $\sigma=30$ minutes. Use the $5 \%$ significance level. |Enter the ALTERNATIVE HYPOTHESIS, H1 number.
Note: Enter INGEGER 1, 2 or 3, No decimal for selection: 1) H 1 : mean $>120$ 2) H1: mean < 120 3) H1: mean Not Equal To 120

See Solution

Problem 266

A study time survey is conducted of 39 UGA students and a mean x-bar=110 minutes is obtained. Test whether the UGA study times differ from the national average of $120 \mathrm{~min} . ?$ Assume study times have a normal distribution with a known standard deviation of $\sigma=30$ minutes. Use the $5 \%$ significance level.
1 What is the P-VALUE from Table A2 for testing HO? See two entry formats below.
Note: Two formats possible: 1) Enter LTO1 if P-value is less than 0.01 2) Enter XXX.XX AT LEAST ONE DIGIT BEFORE THE DECIMAL, TWO AFTER and round up. Thus, 134.784 is entered as $134.78,35.295$ is entered as $35.30, .2749$ is entered as $0.27,-.5650$ is entered as -0.57

See Solution

Problem 267

A study time survey is conducted of 39 UGA students and a mean x-bar $=110$ minutes is obtained. Test whether the UGA study times differ from the national average of 120 min .? Assume study times have a normal distribution with a known standard deviation of $\sigma=30$ minutes. Use the $5 \%$ significance level.
What is the CONCLUSION for HO , assuming a $5 \%$ significance level? REJ = Reject H0, assume H 1 true. FTR = Fail to Reject H0, assume HO true. CBD = Cannot be Determined.
Note: Enter REJ, FTR or CBD letters. $\square$ A

See Solution

Problem 268

A study time survey is conducted of 39 UGA students and a mean x-bar 110 minutes is obtained. Test whether the UGA study times differ from the national average of 120 min ? Assume study times have a normal distribution with a known standard deviation of $\sigma=30$ minutes. Use the $5 \%$ significance level.
1 What is the correct interpretation of the hypothesis testing result at the $5 \%$ level of significance? 1) = The UGA study times were greater than the national average. 2) = The UGA study times are different from the national average. 3) = There was no difference between the UGA and national study times.
Note: Enter integer 1, 2, 3 No decimal. $A$

See Solution

Problem 269

A health expert advertised that people in their their weight loss program would lose MORE than 10 pounds in two months. Twelve (12) customers had the following weight losses: $13,17,9,15,14,9,13,17,15,12,10,8$
Assume a normal distribution for these data. Test whether the mean weight loss was GREATER than 10 lbs . Use $1 \%$ significance level.
What is the ALTERNATIVE HYPOTHESIS, H1 number. 1) H1: mean (greater than)> 10 2) H 1 : mean (greater than) $>12.67$ 3) H1: mean (less than) < 12.67 4) H1: mean (less than) < 10 5) H1: mean Not Equal To 10 5) H1: mean (less than) > 3.11 6) No answer is correct.

See Solution

Problem 270

A health expert advertised that people in their their weight loss program would lose MORE than 10 pounds in two months. Twelve (12) customers had the following weight losses: $13,17,9,15,14,9,13,17,15,12,10,8$
Assume a normal distribution for these data. Test whether the mean weight loss was GREATER than 10 lbs . Use $1 \%$ significance level.
1 What is the value of the TEST STATISTIC? $\qquad$ Note: Enter XXX.XX AT LEAST ONE DIGIT BEFORE THE DECIMAL, TWO AFTER and round up. Thus, 134.784 is entered as $134.78,35.295$ is entered as $35.30, .2749$ is entered as $0.27,-.5650$ is entered as -0.57 $\square$ A

See Solution

Problem 271

A health expert advertised that people in their their weight loss program would lose MORE than 10 pounds in two months. Twelve (12) customers had the following weight losses: $13,17,9,15,14,9,13,17,15,12,10,8$
Assume a normal distribution for these data. Test whether the mean weight loss was GREATER than 10 lbs . Use $1 \%$ significance level. |Would have it been reasonable to look for a CHANGE in weight rather than a LOSS? Note Answer Input Format Below.
Enter YES or NO. $\square$ $A$

See Solution

Problem 272

Suppose that $\mathrm{n}=76$ subjects were put on a low-fat diet with a mean weight loss of x -bar=2.2 kg and a sample standard deviation of $s=6.1 \mathrm{~kg}$ . Is there sufficient evidence to conclude that the mean weight loss was greater than 0 ? Use $5 \%$ significance level.
What is the value of the TEST STATISTIC?
Note: Enter XXX.XX AT LEAST ONE DIGIT BEFORE THE DECIMAL, TWO AFTER and round up. Thus, 134.784 is entered as $134.78,35.295$ is entered as 35.30, . 2749 is entered as $0.27,-.5650$ is entered as -0.57

See Solution

Problem 273

Suppose that $\mathrm{n}=76$ subjects were put on a low-fat diet with a mean weight loss of x -bar=2.2 kg and a sample standard deviation of $\mathrm{s}=6.1 \mathrm{~kg}$ . Is there sufficient evidence to conclude that the mean weight loss was greater than 0 ? Use $5 \%$ significance level.
Would the test statistic change if the significance level changed to 1\%? Note Answer Input Format Below.
Enter YES or NO. $A$

See Solution

Problem 274

Suppose that $\mathrm{n}=76$ subjects were put on a low-fat diet with a mean weight loss of x -bar $=2.2$ kg and a sample standard deviation of $\mathrm{s}=6.1 \mathrm{~kg}$ . Is there sufficient evidence to conclude that the mean weight loss was greater than 0 ? Use $5 \%$ significance level.
I What is the P-VALUE from calculator for testing HO? See two entry formats below. $\qquad$ Note: Two formats possible: 1) Enter LT01 if P-value is less than 0.01 2) Enter XXX.XX AT LEAST ONE DIGIT BEFORE THE DECIMAL, TWO AFTER and round up. Thus, 134.784 is entered as $134.78,35.295$ is entered as $35.30, .2749$ is entered as $0.27,-.5650$ is entered as -0.57 $\square$

See Solution

Problem 275

An EPA standard requires lead in lake water to be LESS than 4 ppb (parts/billion). A studied lake had 11 samples with a mean of 2.8 ppb and sample standard deviation of 1.9 ppb . The concentration of lead can be assumed normally distributed. Is the lake concentration of lead LESS than the EPA standard? Use 5\% significance level.
1 What is the value of the TEST STATISTIC?
Note: Enter XXX.XX AT LEAST ONE DIGIT BEFORE THE DECIMAL, TWO AFTER and round up. Thus, 134.784 is entered as $134.78,35.295$ is entered as $35.30, .2749$ is entered as $0.27,-.5650$ is entered as -0.57 $\square$ A

See Solution

Problem 276

An EPA standard requires lead in lake water to be LESS than 4 ppb (parts/billion). A studied lake had 11 samples with a mean of 2.8 ppb and sample standard deviation of 1.9 ppb . The concentration of lead can be assumed normally distributed. Is the lake concentration of lead LESS than the EPA standard? Use 5\% significance level.
1 What is the Table A3 CRITICAL VALUE for the problem? Note Answer Input Format Below.
Note: Enter XXX.XX AT LEAST ONE DIGIT BEFORE THE DECIMAL, TWO AFTER and round up. Thus, 134.784 is entered as $134.78,35.295$ is entered as $35.30, .2749$ is entered as $0.27,-.5650$ is entered as -0.57 $\square$ A

See Solution

Problem 277

An EPA standard requires lead in lake water to be LESS than 4 ppb (parts/billion). A studied lake had 11 samples with a mean of 2.8 ppb and sample standard deviation of 1.9 ppb . The concentration of lead can be assumed normally distributed. Is the lake concentration of lead LESS than the EPA standard? Use 5\% significance level.
What is the CONCLUSION for the study if HO: mean=20 min.? 1) Reject HO and conclude lake lead concentration is below 4 ppb . 2) Fail to Reject HO and conclude the lake lead concentration is below 4 ppb . 3) Reject HO and conclude the lake lead concentration does not differ from 2.8 ppb . 4) Fail to Reject HO and conclude the lake lead concentration does not differ from 4 ppb. 5) Fail to Reject the alternative H 1 and conclude HO is likely true. 6) No answer is correct.

See Solution

Problem 278

In a ready to serve dinner preparation study of 10 trials of a new meal, the sample mean time for preparation was 18.5 min . and a sample standard deviation of 5.9 min . Prep time can be assumed normally distributed. Test whether the new meal meets the standard for having a preparation time below the 20 min . standard. Use $1 \%$ significance level.
What is the CONCLUSION for the study if HO: mean=20 min.? 1) Reject HO and conclude preparation time is below 20 min . 2) Fail to Reject HO and conclude the preparation time is below 20 min . 3) Reject HO and conclude the preparation time does not differ from 20 min . 4) Fail to Reject HO and conclude the preparation time does not differ from 20 min . 5) Fail to Reject the alternative H 1 and conclude HO is true. 6) No answer is correct.

See Solution

Problem 279

A PTA group claimed that FEWER THAN 20\% of teens smoked. In a survey of 200 teens, 30 said that they had smoked. Test the claim at the $5 \%$ significance level. 1 What is the value of the TEST STATISTIC? $\qquad$ Note: Enter XXX.XX AT LEAST ONE DIGIT BEFORE THE DECIMAL, TWO AFTER and round up. Thus, 134.784 is entered as $134.78,35.295$ is entered as $35.30, .2749$ is entered as $0.27,-.5650$ is entered as -0.57 $\square$ A

See Solution

Problem 280

ssignment 30: Test 5 Review Khadijah Daley 11/18/24 3:07 PM Question 10, 8.2.15-T HW Score: 49.84\%, 14.95 of 30 points Part 1 of 4 Points: 0 of 1 Save
A poll of 1160 Americans showed that $47.1 \%$ of the respondents prefer to watch the news rather than read or listen to it. Use those results with a 0.10 significance level to test the claim that fewer than half of Americans prefer to watch the news rather than read or listen to it. Use the P -value method. Use the normal distribution as an approximation to the binomial distribution.
Let $p$ denote the population proportion of all Americans who prefer to watch the news rather than read or listen to it. Identify the null and alternative hypotheses. $\begin{array}{l|l} \mathrm{H}_{0}: \mathrm{p} & \mathrm{~V} \\ \mathrm{H}_{1}: \mathrm{p} & \mathrm{~V} \\ \hline \end{array}$ (Type integers or decimals. Do not round.)

See Solution

Problem 281

A sample of teenagers found that of the 487 people surveyed, 385 had pictures on their online profiles. It was previously thought that $75 \%$ or MORE of teenagers had profile pictures. Test this hypothesis at the $5 \%$ significance level.
I What is the value of the TEST STATISTIC? $\qquad$ Note: Enter XXX.XX AT LEAST ONE DIGIT BEFORE THE DECIMAL, TWO AFTER and round up. Thus, 134.784 is entered as $134.78,35.295$ is entered as $35.30, .2749$ is entered as $0.27,-.5650$ is entered as -0.57

See Solution

Problem 282

1 A sample of teenagers found that of the 487 people surveyed, 385 had pictures on their online profiles. It was previously thought that $75 \%$ or MORE of teenagers had profile pictures. Test this hypothesis at the $5 \%$ significance level.
1 What is the P-VALUE from Table A2 for testing HO? See two entry formats below.
Note: Two formats possible: 1) Enter LT01 if P-value is less than 0.01 2) Enter XXX.XX AT LEAST ONE DIGIT BEFORE THE DECIMAL, TWO AFTER and round up. Thus, 134.784 is entered as $134.78,35.295$ is entered as $35.30, .2749$ is entered as $0.27,-.5650$ is entered as -0.57

See Solution

Problem 283

A drink manufacturer's design was to fill a bottle to a 10 inch target level. In a sample of 9 bottles, the mean fill level was 8.8 in . with a sample standard deviation of 2.2 in . Fill level can be assumed normally distributed. Did the sample results indicate the mean level DIFFERED from the 10 in . target? Use 5\% significance level.
The below questions will ask for the following information (in a random order required by D2L): 1) Alternative hypothesis H1 selection. 2) Test Statistic for the hypothesis H0 of interest. 3) Critical Value for the H0 test.[Use t-table A3 or invT(p,df)] 4) P-value for H0 test statistic. 5) Rejection/Acceptance decision of HO for given significance level in problem statement. You may wish to perform your calculations in the above standard order. Select the BEST/CLOSEST answer after READING ALL ANSWERS.

See Solution

Problem 284

Compute the critical value $z_{\alpha / 2}$ that corresponds to a $81 \%$ level of confidence. Click here to view the standard normal distribution table (page 1). Click here to view the standard normal distribution table (page 2), $z_{\alpha / 2}=$ $\square$ (Round to two decimal places as needed.)

See Solution

Problem 285

Step 3: Choose alpha We will set alpha at . 05 0гटA Yz
How many degrees of freedom do we use in this example? (hint: look at the table
Step 4: Calculate the test statistic, or just look at the table below: Chi-square tests. \begin{tabular}{|c|c|c|c|} \hline Statistic & Value & Asymp. Sig. (2-tailed) & \\ \hline Pearson Chi-Square & 23.24 & .026) & Leve \\ \hline Likelihood Ratio & 20.31 & 062 & \\ \hline Linear-by-Linear Association & 7.77 & .005 & \\ \hline N of Valid Cases & 20 & & \\ \hline \end{tabular}

See Solution

Problem 286

c. Is there statistically significant evidence of a difference in the mean amount purchased by all visitors to page $A$ and page $B$ ? $\qquad$ Why or why not? A large test stats
2. You also can find the $t$ -statistic on StatCrunch. If you are given summary statistics, go to Stat $>$ T Stats $>$ Two Sample $>$ With Summary. p-value
To test if two populations have different means, a random sample of 74 people from the first population had a mean of 17.3 standard deviation of 4.1. A random sample of 83 people from the second population had a mean of 19.2 and standard deviation of 3.7 . Use these summary statistics to compute $t$ and the p-value on StatCrunch. Round to four decimal places. $t=$ $\qquad$ p -value $=$ $\qquad$ 70

See Solution

Problem 287

A new generic cream was being tested by the FDA to compared it to a standard cream absorption of 3.5 mg in the test. The 7 results were: $2.6,3.2$ , $2.1,3.0,3.1,2.9,3.7$ for the generic cream. The goal was to compare the generic cream absorption mean to the 3.5 standard to determine if the new generic cream differed from the standard.
The below questions will ask for the following information (in a random order required by D 2 L ): 1) Alternative hypothesis H 1 selection. 2) Test Statistic for the hypothesis H0 of interest. 3) Critical Value for the H0 test.[Not evaluated in this question.] 4) P-value for H 0 test statistic. 5) Rejection/Acceptance decision of H 0 for a given significance. You may wish to perform your calculations in the above standard order. Select the BEST/CLOSEST answer after READING ALL ANSWERS. $5 \vee$ 5a) What is the CONCLUSION for HO using the $1 \%$ significance level? Select a three-letter code:
REJ = Reject HO , assume H1 true.
FTR = Fail to Reject HO, assume HO true. $C B D=$ Cannot be Determined. $\square$ 2)What is the value of the TEST STATISTIC?

See Solution

Problem 288

In a survey, 600 adults in a certain country were asked how many hours they worked in the previous week. Based on the results, a $95 \%$ confidence interval for mean number of hours worked was lower bound: 41.2 and upper bound: 43.6 . Which of the following represents a reasonable interpretation o the result? For those that are not reasonable, explain the flaw. Complete parts (a) through (d) below. D. Flawed. This interpretation implies that the mean is only for last week. (b) We are $95 \%$ confident that the mean number of hours worked by adults in this country in the previous week was between 41.2 hours and 43.6 hours A. Correct. This interpretation is reasonable. B. Flawed. This interpretation implies that the population mean varies rather than the interval. C. Flawed. This interpretation does not make it clear that the $95 \%$ is the probability that the mean is within the interval. D. Flawed. This interpretation makes an implication about individuals rather than the mean. Clear all Check answe

See Solution

Problem 289

Khadijah Daley 11/18/24 4:03 PM gnment 30: Test 5 Review Question 20, 9.2.11-T HW Score: 81.14\%, 24.34 of 30 points Part 1 of 7 Points: 0 of 1 Save
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 $\left(\mu_{1}\right): n=27, \bar{x}=0.55, s=0.89$ Reduction in Pain Level After Sham Treatment $\left(\mu_{2}\right): n=27, \bar{x}=0.51, s=1.51$ a. Use a 0.01 significance level to test the claim that those treated with magnets have a greater mean reduction in pain than those given a sham treatment (similar to a placebo).
What are the null and alternative hypotheses? A. $H_{0}: \mu_{1}=\mu_{2}$ B. $H_{0}: \mu_{1}=\mu_{2}$ $H_{1}: \mu_{1}>\mu_{2}$ C. $H_{0}: \mu_{1} \neq \mu_{2}$ $H_{1}: \mu_{1} \neq \mu_{2}$ $H_{1}: \mu_{1}<\mu_{2}$ D. $H_{0}: \mu_{1}<\mu_{2}$ $H_{1}: \mu_{1} \geq \mu_{2}$ View an example Get more help - Clear all Check answer

See Solution

Problem 290

```latex A teacher gave two versions of a quiz, form 1 and form 2, to random samples of sociology students. She wanted to test whether the mean scores would be different if they were given to all sociology students. Here are the raw data.
Form 1 scores: 90, 83, 81, 73, 70, 67, 63, 63, 59, 57, 56, 55, 54, 50, 49, 32, 24, 6.
Form 2 scores: 100, 100, 100, 98, 91, 88, 73, 68, 66, 64, 61, 61, 61, 60, 49, 43, 22.
\begin{enumerate} \item[(a)] Compute the following, rounding to two decimal places. \begin{itemize} \item $t=$ \_\_\_\_\_\_\_ \item $p$ -value $=$ \_\_\_\_\_\_\_ \end{itemize} \item[(b)] Circle the words that make an appropriate conclusion, using $\alpha=0.05$ . REJECT / DON'T REJECT the NULL / ALTERNATIVE hypothesis. There IS / IS NOT statistically significant evidence that the mean scores would be THE SAME / DIFFERENT. \end{enumerate}
Skill: Estimate $\bar{x}_{1}-\bar{x}_{2}$ and the $p$ -value from a sketch of the sampling distribution.
For each completed sampling distribution for the difference of two means: \begin{enumerate} \item[(a)] Was it a one or two-tailed test? \item[(b)] Write the hypotheses. \item[(c)] What was the SEEsT? \item[(d)] What was the value of $\bar{x}_{1}-\bar{x}_{2}$ ? \item[(e)] Estimate the $p$ -value (roughly). \item[(f)] Should the null hypothesis be rejected? YES / NO \end{enumerate}
\begin{enumerate} \item[4.] \begin{enumerate} \item[(a)] \_\_\_\_\_\_ \item[(b)] \_\_\_\_\_\_ \item[(c)] \_\_\_\_\_\_ \item[(d)] \_\_\_\_\_\_ \item[(e)] \_\_\_\_\_\_ \item[(f)] YES / NO \end{enumerate} \item[5.] \begin{enumerate} \item[(a)] \_\_\_\_\_\_ \item[(b)] \_\_\_\_\_\_ \item[(c)] \_\_\_\_\_\_ \item[(d)] \_\_\_\_\_\_ \item[(e)] \_\_\_\_\_\_ \item[(f)] YES / NO \end{enumerate} \end{enumerate} ```

See Solution

Problem 291

A television sports commentator wants to estimate the proportion of citizens who "follow professional football." Complete parts (a) through (c). Click here to view the standard normal distribution table (page 1). Click here to view the standard normal distribution table (page 2). (a) What sample size should be obtained if he wants to be within 4 percentage points with $95 \%$ confidence if he uses an estimate of $54 \%$ obtained from a poll?
The sample size is 597 . (Round up to the neares integer.) (b) What sample size should be obtained if he wants to be within 4 percentage points with $95 \%$ confidence if he does not use any prior estimates?
The sample size is $\square$ . (Round up to the nearest integer.)

See Solution

Problem 292

signment 30: Test 5 Review Question 22, 9.2.19-T 11/18/24 4:17 PM Part 1 of 6 HW Score: 84.65\%, 25.4 of 30 points Points: 0 of 1 Save
Listed in the accompanying table are weights (lb) of samples of the contents of cans of regular Coke and Diet Coke. Assume that the two samples are independent simple random samples selected from normally distributed populations. Do not assume that the population standard deviations are equal. Complete parts (a) to (c). Click the icon to view the data table of can weights. a. Use a 0.10 significance level to test the claim that the contents of cans of regular Coke have weights with a mean that is greater than the mean for Diet Coke.
What are the null and alternative hypotheses? Assume that population 1 consists of regular Coke and population 2 consists of Diet Coke. A. $H_{0}: \mu_{1} \leq \mu_{2}$ B. $H_{0}: \mu_{1}=\mu_{2}$ $H_{1}: \mu_{1}>\mu_{2}$ $\mathrm{H}_{1}: \mu_{1}>\mu_{2}$ C. $H_{0}: \mu_{1} \neq \mu_{2}$ D. $H_{0}: \mu_{1}=\mu_{2}$ $H_{1}: \mu_{1}>\mu_{2}$
Can weights View an example Get r \begin{tabular}{|c|c|} \hline Regular Coke & Diet Coke \\ \hline 0.81917 & 0.77731 \\ \hline 0.81498 & 0.77580 \\ \hline 0.81628 & 0.78962 \\ \hline 0.82108 & 0.78680 \\ \hline 0.81808 & 0.78439 \\ \hline 0.82472 & 0.78614 \\ \hline 0.80622 & 0.78056 \\ \hline 0.81284 & 0.78303 \\ \hline \end{tabular} Clear all Check answer

See Solution

Problem 293

Let's go to the movies: A random sample of 42 Hollywood movies made in the last 10 years had a mean length of 111.7 minutes, with a standard deviation of 13.8 minutes.
Part: $0 / 2$
Part 1 of 2 (a) Construct a $90 \%$ confidence interval for the true mean length of all Hollywood movies in the last 10 years. Round the answers to at least one decimal place. A $90 \%$ confidence interval for the true mean length of all Hollywood movies made in the last 10 years is $\square$ $<\mu<$ $\square$ .

See Solution

Problem 294

An NHANES report gives data for 641 men aged $20-29$ years. The BMI of these 641 men was $\bar{x}=25.6$ . On the basis of this sample, we want to estimate the BMI $\mu$ in the population of all 23.2 million American men in this age group. Treat these data as an SRS from a Normally distributed population with standard deviation $\sigma=7.2$ .
Source adapted from: Adapted from Fryar C. D., et al., Anthropometric reference data for children and adults: United States, 2011-2014. National Center for Health Statistics. Vital Health Statistics 3(2016), at https///www.codc.gov/nchs/data/ series/sr_03/sr03_039.pdf.
Find the margins of error for $99 \%$ confidence based on SRSs of 365 young men and 1581 young men. Give your answers to four decimal places. margin of error when $n=365$ : $\square$ margin of error when $n=1581$ : $\square$

See Solution

Problem 295

An NHANES report gives data for 936 men aged $20-29$ years. The BMI of these 936 men was $\bar{x}=27.2$ . On the basis of this sample, we want to estimate the BMI $\mu$ in the population of all 23.2 million American men in this age group. To match the "simple conditions," we will treat the NHANES sample as an SRS from a Normal population with standard deviation $\sigma=11.6$ .
Source: Adapted from Fryar C. D, et al, Anthropometric reference data for children and adults: United States, 2011-2014. National Center for Healih Suatistics. Viad Healoh Shatiatics 3(2016), at herpsi/www. ode gow/hachs/data/ series/sr_03/sr03_039.pdf.
Give three confidence intervals for the mean BMI $\mu$ in this population, using $90 \%, 95 \%$ , and $99 \%$ confidence. Enter each confidence interval in the form of (lower bound, upper bound). Give the bounds to two decimal places.
90\% confidence interval: $\square$
95\% confidence interval: $\square$
99\% confidence interval: $\square$

See Solution

Problem 296

An NHANES report gives data for 936 men aged $20-29$ years. The BMI of these 936 mes was $\bar{x}=272.00$ the bevis of ths sample, we want to estimate the BMI $\mu$ in the population of all 23.2 million American men in this age group. To math the "ifmple conditions," we will treat the NHANES sample as an SRS from a Normal population with itandard deviation $\sigma=116$ .
What are the margins of errors for $90 \%, 95 \%$ , and $99 \%$ confidence? Give your answers to four decimal places. margin of error for a $90 \%$ confidence interval: 0.6196 margin of error for a $95 \%$ confidence interval: 0.7396 margin of error for a $99 \%$ confidence interval: 09626

See Solution

Problem 297

An NHANES report gives data for 936 men aged 20-29 years. The BMI of these 936 men was $\bar{x}=27.2$ . On the basis of this sample, we want to estimate the BMI $\mu$ in the population of all 23.2 million American men in this age group. To match the "simple conditions," we will treat the NHANES sample as an SRS from a Normal population with standard deviation $\sigma=11.6$ . Healdh Statistics 3(2016), at hutps/lwww.cdegov/nchs/data/ series/sr_03/sr03_039.pdf.
What are the margins of errors for $90 \%, 95 \%$ , and $99 \%$ confidence? Give your answers to four decimal places.

See Solution

Problem 298

Question 1 2 pts
A market research analyst claims that $32 \%$ of the people who visit the mall actually make a purchase. You think that less than $32 \%$ buy something and decide to test the claim. You stand by the exit door of the mall starting at noon and ask 82 people as they are leaving whether they bought anything. You find that only 20 people made a purchase.
State the null and alternative hypothesis to test the claim. $H_{0}: p \geq 32$ versus $H_{a}: p \leq 32$ $H_{0}: p=0.32$ versus $H_{a}: p \neq 0.32$ $H_{0}: p=0.32$ versus $H_{a}: p<0.32$ $H_{0}: p=32$ versus $H_{a}: p \geq 32$ $H_{0}: p=0.32$ versus $H_{a}: p>0.32$

See Solution

Problem 299

Question 1 1 pts
A standard painkiller is known to bring relief in 3.5 minutes on average ( $\mu$ ). A new painkiller is hypothesized to bring faster relief to patients.
A sample of 40 patients are given the new painkillers. The sample yields a mean of 2.8 minutes and a standard deviation of 1.1 minutes.
Which of the following pairs of hypotheses are appropriate for this study? $H_{0}: \mu=2.8$ vs. $H_{a}: \mu \neq 2.8$ $H_{0}: \mu=2.8$ vs. $H_{a}: \mu<2.8$ $H_{0}: \mu=3.5$ vs. $H_{a}: \mu<3.5$ $H_{0}: \mu=3.5$ vs. $H_{a}: \mu=2.8$ $H_{0}: \mu=3.5$ vs. $H_{a}: \mu>3.5$

See Solution

Problem 300

You are given the sample mean and the population standard deviation. Use this information to construct the $90 \%$ and $95 \%$ confidence intervals for the popt results and compare the widths of the confidence intervals. If convenient, use technology to construct the confidence intervals.
A random sample of 50 home theater systems has a mean price of $\$ 138.00$ . Assume the population standard deviation is $\$ 17.40$ .
Construct a $90 \%$ confidence interval for the population mean. The $90 \%$ confidence interval is ( $133.95,142.05$ ). (Round to two decimal places as needed) Construct a 95\% confidence interval for the population mean. The 95\% confidence interval is ( $\square$ $\square$ (Round to two decimal places as needed.)

See Solution

1 2 3 4 5 6 ...14

Statistical Inference

Problem 201

Problem 202

Problem 203

Problem 204

Problem 205

Problem 206

Problem 207

Problem 208

Problem 209

Problem 210

Problem 211

If n=400\mathrm{n}=400n=400 and pundefined(p−hat)=0.9\widehat{p}(p-h a t)=0.9p​(p−hat)=0.9, construct a 99%99 \%99% confidence interval. Give your answers to three decimals □\square□ <p<<p<<p< □\square□

Problem 212

Out of 300 people sampled, 150 had kids. Based on this, construct a 95%95 \%95% confidence interval for the true population proportion of people with kids.Give your answers as decimals, to three places

Problem 213

Problem 214

Problem 215

Problem 216

Problem 217

Problem 218

Problem 219

Problem 220

Problem 221

Problem 222

Problem 223

Problem 224

Problem 225

Problem 226

Problem 227

Problem 228

Problem 229

Problem 230

Problem 231

Problem 232

Problem 233

Problem 234

Problem 235

Problem 236

Problem 237

Problem 238

Problem 239

Problem 240

Problem 241

Problem 242

Problem 243

Problem 244

When the population mean is known, thenSelect one: a. We used the population mean to estimate the sample SD b. We used the sample SD to estimate the population mean c. We used the sample mean to estimate the population mean d. None of the above

Problem 245

Problem 246

Problem 247

Problem 248

Problem 249

Problem 250

Problem 251

Problem 252

Problem 253

Problem 254

Problem 255

Problem 256

Problem 257

Problem 258

Problem 259

Problem 260

Problem 261

Problem 262

Problem 263

Problem 264

Problem 265

Problem 266

Problem 267

Problem 268

Problem 269

Problem 270

Problem 271

Problem 272

Problem 273

Problem 274

Problem 275

If $\mathrm{n}=400$ and $\widehat{p}(p-h a t)=0.9$ , construct a $99 \%$ confidence interval. Give your answers to three decimals $\square$ $<p<$ $\square$

Out of 300 people sampled, 150 had kids. Based on this, construct a $95 \%$ confidence interval for the true population proportion of people with kids.
Give your answers as decimals, to three places

When the population mean is known, then
Select one: a. We used the population mean to estimate the sample SD b. We used the sample SD to estimate the population mean c. We used the sample mean to estimate the population mean d. None of the above