Math / Data & Statistics

QuestionFree dessert: 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. Can you conclude that the mean number of diners decreased while the free dessert offer was in effect? Use the $\alpha=0.01$ level of significance and the $P$ -value method with the
Critical Values for the Student's $t$ Distribution Table. \begin{tabular}{llllll} \hline 170 & 133 & 150 & 111 & 171 & 103 \\ 101 & 110 & 133 & 179 & 151 & 112 \\ \hline \end{tabular} Send data to Excel
Part 1 of 6
Following is a boxplot for these data. Is it reasonable to assume that the conditions for performing a hypothesis test are satisfied? Explain.
The boxplot shows that there are outliers. The boxplot shows that there is no evidence of strong skewness.
We can $\square$ assume that the population is approximately normal.
It $\square$ reasonable to assume that the conditions are satisfied.
Part 2 of 6
State the appropriate null and alternate hypotheses. $\begin{array}{l} H_{0}=\mu=150 \\ H_{1}: \mu<150 \end{array}$
This hypothesis test is a left-tailed $\square$ test. 5

Studdy Solution

STEP 1

1. We have a random sample of 12 days with the number of diners recorded.
2. The significance level is $\alpha = 0.01$ .
3. We will use the Student's $t$ distribution for hypothesis testing.
4. The null hypothesis is that the mean number of diners is 150.
5. The alternative hypothesis is that the mean number of diners is less than 150.

STEP 2

1. Evaluate the assumptions for the hypothesis test.
2. State the null and alternative hypotheses.
3. Calculate the sample mean and standard deviation.
4. Perform the $t$ -test.
5. Determine the $P$ -value.
6. Make a conclusion based on the $P$ -value and significance level.

STEP 3

Evaluate the assumptions for the hypothesis test:
- The boxplot shows outliers, but no evidence of strong skewness. - We can assume the population is approximately normal due to the Central Limit Theorem, given the sample size is reasonably large (n=12).
It is reasonable to assume that the conditions for performing a hypothesis test are satisfied.

STEP 4

State the null and alternative hypotheses:
\[ H_0: \mu = 150 $ \[ H_1: \mu < 150 $
This hypothesis test is a left-tailed test.

STEP 5

Calculate the sample mean $\bar{x}$ and standard deviation $s$ .
Sample data: $170, 133, 150, 111, 171, 103, 101, 110, 133, 179, 151, 112$ .
Calculate $\bar{x}$ and $s$ .

STEP 6

Calculate the sample mean $\bar{x}$ :
$\bar{x} = \frac{170 + 133 + 150 + 111 + 171 + 103 + 101 + 110 + 133 + 179 + 151 + 112}{12}$
$\bar{x} = \frac{1624}{12} = 135.33$
Calculate the sample standard deviation $s$ :
$s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}}$

STEP 7

Calculate the sample standard deviation $s$ :
$s = \sqrt{\frac{(170-135.33)^2 + (133-135.33)^2 + \ldots + (112-135.33)^2}{11}}$
Calculate the value of $s$ .

STEP 8

Perform the $t$ -test:
Calculate the test statistic $t$ :
$t = \frac{\bar{x} - \mu}{s/\sqrt{n}}$
Substitute $\bar{x} = 135.33$ , $\mu = 150$ , $s$ , and $n = 12$ .

STEP 9

Calculate the test statistic $t$ :
$t = \frac{135.33 - 150}{s/\sqrt{12}}$
Calculate the value of $t$ .

STEP 10

Determine the $P$ -value using the $t$ -distribution table with $n-1 = 11$ degrees of freedom.
Compare the calculated $t$ value to the critical value at $\alpha = 0.01$ .

STEP 11

Find the $P$ -value from the $t$ -distribution table.

STEP 12

Make a conclusion based on the $P$ -value and significance level $\alpha = 0.01$ .
If $P$ -value < $\alpha$ , reject $H_0$ .

STEP 13

Conclude whether the mean number of diners decreased.
The conclusion will depend on the calculated $P$ -value and comparison to $\alpha = 0.01$ .

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Studdy Solution

STEP 1

STEP 2

1. Evaluate the assumptions for the hypothesis test.
2. State the null and alternative hypotheses.
3. Calculate the sample mean and standard deviation.
4. Perform the $t$ -test.
5. Determine the $P$ -value.
6. Make a conclusion based on the $P$ -value and significance level.

STEP 3

STEP 4

State the null and alternative hypotheses:
\[ H_0: \mu = 150 $ \[ H_1: \mu < 150 $
This hypothesis test is a left-tailed test.

STEP 5

Calculate the sample mean $\bar{x}$ and standard deviation $s$ .
Sample data: $170, 133, 150, 111, 171, 103, 101, 110, 133, 179, 151, 112$ .
Calculate $\bar{x}$ and $s$ .

STEP 6

Calculate the sample mean $\bar{x}$ :
$\bar{x} = \frac{170 + 133 + 150 + 111 + 171 + 103 + 101 + 110 + 133 + 179 + 151 + 112}{12}$
$\bar{x} = \frac{1624}{12} = 135.33$
Calculate the sample standard deviation $s$ :
$s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}}$

STEP 7

Calculate the sample standard deviation $s$ :
$s = \sqrt{\frac{(170-135.33)^2 + (133-135.33)^2 + \ldots + (112-135.33)^2}{11}}$
Calculate the value of $s$ .

STEP 8

Perform the $t$ -test:
Calculate the test statistic $t$ :
$t = \frac{\bar{x} - \mu}{s/\sqrt{n}}$
Substitute $\bar{x} = 135.33$ , $\mu = 150$ , $s$ , and $n = 12$ .

STEP 9

Calculate the test statistic $t$ :
$t = \frac{135.33 - 150}{s/\sqrt{12}}$
Calculate the value of $t$ .

STEP 10

Determine the $P$ -value using the $t$ -distribution table with $n-1 = 11$ degrees of freedom.
Compare the calculated $t$ value to the critical value at $\alpha = 0.01$ .

STEP 11

Find the $P$ -value from the $t$ -distribution table.

STEP 12

Make a conclusion based on the $P$ -value and significance level $\alpha = 0.01$ .
If $P$ -value < $\alpha$ , reject $H_0$ .

STEP 13

Conclude whether the mean number of diners decreased.
The conclusion will depend on the calculated $P$ -value and comparison to $\alpha = 0.01$ .

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Studdy Solution

STEP 1

STEP 2

1. Evaluate the assumptions for the hypothesis test.2. State the null and alternative hypotheses.3. Calculate the sample mean and standard deviation.4. Perform the ttt-test.5. Determine the PPP-value.6. Make a conclusion based on the PPP-value and significance level.

STEP 3

STEP 4

State the null and alternative hypotheses:\[ H_0: \mu = 150 $ \[ H_1: \mu < 150 $This hypothesis test is a left-tailed test.

STEP 5

Calculate the sample mean xˉ\bar{x}xˉ and standard deviation sss.Sample data: 170,133,150,111,171,103,101,110,133,179,151,112170, 133, 150, 111, 171, 103, 101, 110, 133, 179, 151, 112170,133,150,111,171,103,101,110,133,179,151,112.Calculate xˉ\bar{x}xˉ and sss.

STEP 6

STEP 7

Calculate the sample standard deviation sss:s=(170−135.33)2+(133−135.33)2+…+(112−135.33)211s = \sqrt{\frac{(170-135.33)^2 + (133-135.33)^2 + \ldots + (112-135.33)^2}{11}}s=11(170−135.33)2+(133−135.33)2+…+(112−135.33)2​​Calculate the value of sss.

STEP 8

Perform the ttt-test:Calculate the test statistic ttt:t=xˉ−μs/nt = \frac{\bar{x} - \mu}{s/\sqrt{n}}t=s/n​xˉ−μ​Substitute xˉ=135.33\bar{x} = 135.33xˉ=135.33, μ=150\mu = 150μ=150, sss, and n=12n = 12n=12.

STEP 9

Calculate the test statistic ttt:t=135.33−150s/12t = \frac{135.33 - 150}{s/\sqrt{12}}t=s/12​135.33−150​Calculate the value of ttt.

STEP 10

Determine the PPP-value using the ttt-distribution table with n−1=11n-1 = 11n−1=11 degrees of freedom.Compare the calculated ttt value to the critical value at α=0.01\alpha = 0.01α=0.01.

STEP 11

Find the PPP-value from the ttt-distribution table.

STEP 12

Make a conclusion based on the PPP-value and significance level α=0.01\alpha = 0.01α=0.01.If PPP-value < α\alphaα, reject H0H_0H0​.

STEP 13

Conclude whether the mean number of diners decreased. The conclusion will depend on the calculated PPP-value and comparison to α=0.01\alpha = 0.01α=0.01.

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1. Evaluate the assumptions for the hypothesis test.
2. State the null and alternative hypotheses.
3. Calculate the sample mean and standard deviation.
4. Perform the $t$ -test.
5. Determine the $P$ -value.
6. Make a conclusion based on the $P$ -value and significance level.

State the null and alternative hypotheses:
\[ H_0: \mu = 150 $ \[ H_1: \mu < 150 $
This hypothesis test is a left-tailed test.

Calculate the sample mean $\bar{x}$ and standard deviation $s$ .
Sample data: $170, 133, 150, 111, 171, 103, 101, 110, 133, 179, 151, 112$ .
Calculate $\bar{x}$ and $s$ .

Calculate the sample standard deviation $s$ :
$s = \sqrt{\frac{(170-135.33)^2 + (133-135.33)^2 + \ldots + (112-135.33)^2}{11}}$
Calculate the value of $s$ .

Perform the $t$ -test:
Calculate the test statistic $t$ :
$t = \frac{\bar{x} - \mu}{s/\sqrt{n}}$
Substitute $\bar{x} = 135.33$ , $\mu = 150$ , $s$ , and $n = 12$ .

Calculate the test statistic $t$ :
$t = \frac{135.33 - 150}{s/\sqrt{12}}$
Calculate the value of $t$ .

Determine the $P$ -value using the $t$ -distribution table with $n-1 = 11$ degrees of freedom.
Compare the calculated $t$ value to the critical value at $\alpha = 0.01$ .

Find the $P$ -value from the $t$ -distribution table.

Make a conclusion based on the $P$ -value and significance level $\alpha = 0.01$ .
If $P$ -value < $\alpha$ , reject $H_0$ .

Conclude whether the mean number of diners decreased.
The conclusion will depend on the calculated $P$ -value and comparison to $\alpha = 0.01$ .