Use the traditional method to test the given hypothesis. Assume that the population is normally distributed and that the sample has been randomly selected. Select the appropriate response.

When 12 bolts are tested for hardness, their indexes have a standard deviation of 41.7. Test the claim that the standard deviation of the hardness indexes for all such bolts is greater than 30.0. Use a 0.025 level of significance.
Select the correct test statistic and conclusion based on the null hypothesis.

A) Test statistic: χ^2 = 20.158. Reject the null hypothesis.
B) Test statistic: χ^2 = 21.253. Fail to reject the null hypothesis.
C) Test statistic: χ^2 = 12.532. Fail to reject the null hypothesis.
D) Test statistic: χ^2 = 8.982. Reject the null hypothesis.

1 answer

To test the claim that the standard deviation of the hardness indexes for all bolts is greater than 30.0, we will be using the Chi-Square test for the variance.

Step 1: Set up the Hypotheses

  • Null Hypothesis (\(H_0\)): \(\sigma \leq 30\) (the standard deviation is less than or equal to 30)
  • Alternative Hypothesis (\(H_a\)): \(\sigma > 30\) (the standard deviation is greater than 30)

Step 2: Determine Sample Statistics

  • Sample size (\(n\)) = 12
  • Sample standard deviation (\(s\)) = 41.7
  • Hypothesized standard deviation (\(\sigma_0\)) = 30

Step 3: Calculate Sample Variance

\[ s^2 = (41.7)^2 = 1736.89 \]

Step 4: Compute the Test Statistic

The test statistic for a Chi-Square test for variance is calculated using the formula:

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

Plugging in the values:

\[ \chi^2 = \frac{(12-1) \cdot 1736.89}{30^2} \]

Calculating this step by step:

\[ \chi^2 = \frac{11 \cdot 1736.89}{900} = \frac{19106.79}{900} \approx 21.2297 \]

Step 5: Determine the Critical Value and Conclusion

Since this is a one-tailed test (because we are testing whether the standard deviation is greater), we will find the critical value from the Chi-Square distribution table with \(n - 1 = 11\) degrees of freedom at a significance level of \(\alpha = 0.025\).

The critical value for \(\chi^2_{0.025}(11)\) is approximately 21.920 (this value can vary slightly depending on the Chi-Square distribution table used).

Now we compare our test statistic to the critical value:

  • If \(\chi^2\) > critical value, we reject the null hypothesis.
  • If \(\chi^2\) ≤ critical value, we fail to reject the null hypothesis.

Since our calculated \(\chi^2 \approx 21.2297\) is less than the critical value of 21.920, we fail to reject the null hypothesis.

Final Result

The correct answer, based on our analysis, is:

B) Test statistic: χ² = 21.253. Fail to reject the null hypothesis.

(Note: We rounded our calculated value slightly differently than in option B, but if we assume typographical differences or degrees of approximation in options, then B is the closest representation of our conclusion about failing to reject the null hypothesis.)