Questions: Over the years, a sushi restaurant has had a mean customer satisfaction rating of 71.2 with a variance of 25.4. The owner wants to see if using a new menu will have any effect on the variance, σ^2. The owner surveys a random sample of 14 customers who ordered from the new menu. For the customers surveyed, the variance of ratings is 46.2. Assume the ratings for customers who order from the new menu follow a normal distribution. Is there enough evidence to conclude that the population variance, σ^2, differs from 25.4? To answer, complete the parts below to perform a hypothesis test. Use the 0.10 level of significance. (a) State the null hypothesis H0 and the alternative hypothesis H1 that you would use for the test. H0: H1: σ s < ≤ > ≥ = ≠ × 5 (b) Perform a chi-square test. Here is some information to help you with your test. - χ0.95^2 is the value that cuts off an area of 0.05 in the left tail of the distribution, and χ0.05^2 is the value that cuts off an area of 0.05 in the right tail of the distribution. - The value of the test statistic is given by χ^2 = (n-1) s^2 / σ^2.

Over the years, a sushi restaurant has had a mean customer satisfaction rating of 71.2 with a variance of 25.4. The owner wants to see if using a new menu will have any effect on the variance, σ^2. The owner surveys a random sample of 14 customers who ordered from the new menu. For the customers surveyed, the variance of ratings is 46.2. Assume the ratings for customers who order from the new menu follow a normal distribution.

Is there enough evidence to conclude that the population variance, σ^2, differs from 25.4? To answer, complete the parts below to perform a hypothesis test. Use the 0.10 level of significance.
(a) State the null hypothesis H0 and the alternative hypothesis H1 that you would use for the test.

H0:
H1: 

σ s  
<  ≤  > 
≥  =  ≠  

× 5


(b) Perform a chi-square test. Here is some information to help you with your test.
- χ0.95^2 is the value that cuts off an area of 0.05 in the left tail of the distribution, and χ0.05^2 is the value that cuts off an area of 0.05 in the right tail of the distribution.
- The value of the test statistic is given by χ^2 = (n-1) s^2 / σ^2.
Transcript text: Over the years, a sushi restaurant has had a mean customer satisfaction rating of 71.2 with a variance of 25.4 . The owner wants to see if using a new menu will have any effect on the variance, $\sigma^{2}$. The owner surveys a random sample of 14 customers who ordered from the new menu. For the customers surveyed, the variance of ratings is 46.2 . Assume the ratings for customers who order from the new menu follow a normal distribution. Is there enough evidence to conclude that the population variance, $\sigma^{2}$, differs from 25.4 ? To answer, complete the parts below to perform a hypothesis test. Use the 0.10 level of significance. (a) State the null hypothesis $H_{0}$ and the alternative hypothesis $H_{1}$ that you would use for the test. \[ \begin{array}{l} H_{0}: \\ H_{1}: \square \end{array} \] \begin{tabular}{|c|c|c|} \hline$\sigma$ & $s$ & $\square^{\square}$ \\ $\square<\square$ & $\square \leq \square$ & $\square>\square$ \\ $\square \geq \square$ & $\square=\square$ & $\square \neq \square$ \\ \hline$\times$ & 5 \\ \hline \end{tabular} (b) Perform a chi-square test. Here is some information to help you with your test. - $\chi_{0.95}^{2}$ is the value that cuts off an area of 0.05 in the left tail of the distribution, and $\chi_{0.05}^{2}$ is the value that cuts off an area of 0.05 in the right tail of the distribution. - The value of the test statistic is given by $\chi^{2}=\frac{(n-1) s^{2}}{\sigma^{2}}$. Bookmarks Chi-square Distribution
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Solution

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Solution Steps

Step 1: State the Hypotheses

We are testing the following hypotheses:

  • Null Hypothesis \( H_0: \sigma^2 = 25.4 \)
  • Alternative Hypothesis \( H_1: \sigma^2 \neq 25.4 \)
Step 2: Calculate the Test Statistic

The test statistic for the chi-square test is calculated using the formula:

\[ \chi^2 = \frac{(n-1) s^2}{\sigma_0^2} \]

Substituting the values:

\[ \chi^2 = \frac{(14 - 1) \cdot 46.2}{25.4} = \frac{13 \cdot 46.2}{25.4} = 23.6457 \]

Step 3: Determine the P-value

To find the P-value, we calculate the cumulative distribution function (CDF) of the chi-square distribution for the test statistic with \( n-1 = 13 \) degrees of freedom:

\[ P = P(\chi^2(13) > 23.6457) = 0.0691 \]

Step 4: Determine the Critical Values

For a two-tailed test at the significance level \( \alpha = 0.10 \), the critical values are:

\[ \text{Critical Values} = [5.8919, 22.362] \]

Step 5: Make a Decision

Since the test statistic \( \chi^2 = 23.6457 \) is greater than the upper critical value \( 22.362 \) and the P-value \( 0.0691 \) is less than \( \alpha = 0.10 \), we reject the null hypothesis.

Final Answer

There is enough evidence to conclude that the population variance differs from \( 25.4 \).

\(\boxed{\text{Reject } H_0}\)

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