Questions: For a standardized exam at your school, the mean score is 106.0 with a variance of 404.9. You have heard that student athletes are held to strict academic standards. Because of that, you want to know if the variance in test scores among student athletes, σ^2, is lower. To find out, you survey a random sample of 10 student athletes. You find the sample variance is 175.6. Assume test scores for student athletes follow a normal distribution. Is there sufficient evidence to conclude that the population variance, σ^2, is less than 404.9? To answer, complete the parts below to perform a hypothesis test. Use the 0.05 level of significance. (a) State the null hypothesis H0 and the alternative hypothesis H1 that you would use for the test. H0: H1: (b) Perform a chi-square test and find the p-value. Here is some information to help you with your chi-square test. - The value of the test statistic is given by χ^2=(n-1) s^2/?.

For a standardized exam at your school, the mean score is 106.0 with a variance of 404.9. You have heard that student athletes are held to strict academic standards. Because of that, you want to know if the variance in test scores among student athletes, σ^2, is lower. To find out, you survey a random sample of 10 student athletes. You find the sample variance is 175.6. Assume test scores for student athletes follow a normal distribution.

Is there sufficient evidence to conclude that the population variance, σ^2, is less than 404.9? To answer, complete the parts below to perform a hypothesis test. Use the 0.05 level of significance.
(a) State the null hypothesis H0 and the alternative hypothesis H1 that you would use for the test.

H0: 
H1: 

(b) Perform a chi-square test and find the p-value. Here is some information to help you with your chi-square test.
- The value of the test statistic is given by χ^2=(n-1) s^2/?.
Transcript text: For a standardized exam at your school, the mean score is 106.0 with a variance of 404.9. You have heard that student athletes are held to strict academic standards. Because of that, you want to know if the variance in test scores among student athletes, $\sigma^{2}$, is lower. To find out, you survey a random sample of 10 student athletes. You find the sample variance is 175.6. Assume test scores for student athletes follow a normal distribution. Is there sufficient evidence to conclude that the population variance, $\sigma^{2}$, is less than 404.9? To answer, complete the parts below to perform a hypothesis test. Use the 0.05 level of significance. (a) State the null hypothesis $H_{0}$ and the alternative hypothesis $H_{1}$ that you would use for the test. \[ \begin{array}{l} H_{0}: \square \\ H_{1}: \square \end{array} \] (b) Perform a chi-square test and find the $\rho$-value. Here is some information to help you with your chi-square test. - The value of the test statistic is given by $\chi^{2}=\frac{(n-1) s^{2}}{?}$.
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Solution

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Solution Steps

Step 1: State the Hypotheses

We are conducting a hypothesis test to determine if the population variance of test scores among student athletes is less than the known population variance of 404.9.

  • Null Hypothesis (\(H_0\)): \[ \sigma^2 = 404.9 \]
  • Alternative Hypothesis (\(H_1\)): \[ \sigma^2 < 404.9 \]
Step 2: Calculate the Test Statistic

The test statistic for the chi-square test is calculated using the formula: \[ \chi^2 = \frac{(n-1) s^2}{\sigma_0^2} \] where:

  • \(n = 10\) (sample size)
  • \(s^2 = 175.6\) (sample variance)
  • \(\sigma_0^2 = 404.9\) (hypothesized population variance)

Substituting the values: \[ \chi^2 = \frac{(10 - 1) \cdot 175.6}{404.9} = \frac{9 \cdot 175.6}{404.9} = 3.9032 \]

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 \(9\) degrees of freedom: \[ P = P(\chi^2(9) < 3.9032) = 0.0823 \]

Step 4: Compare P-value with Significance Level

The significance level (\(\alpha\)) is set at \(0.05\). We compare the P-value with \(\alpha\):

  • P-value: \(0.0823\)
  • \(\alpha = 0.05\)

Since \(0.0823 > 0.05\), we fail to reject the null hypothesis.

Step 5: Conclusion

Based on the results of the hypothesis test, we conclude that there is not sufficient evidence to support the claim that the population variance of test scores among student athletes is less than \(404.9\).

Final Answer

\[ \boxed{\text{Fail to reject } H_0} \]

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