To determine whether the sample data provide sufficient evidence to support the claim that the population standard deviation is greater than 17 at a 0.10 level of significance, we need to perform a hypothesis test for a single standard deviation.
The appropriate null and alternative hypotheses can be stated as follows:
- Null hypothesis (\(H_0\)): \(\sigma \leq 17\) (The population standard deviation is less than or equal to 17)
- Alternative hypothesis (\(H_1\)): \(\sigma > 17\) (The population standard deviation is greater than 17)
Next, we perform the hypothesis test using the Chi-Square (\(\chi^2\)) distribution.
Given:
- Sample size (\(n\)) = 25
- Sample standard deviation (\(s\)) = 8
- Population standard deviation claim (\(\sigma_0\)) = 17
- Level of significance (\(\alpha\)) = 0.10
The chi-square test statistic for variance can be calculated using the formula:
\[
\chi^2 = \frac{(n-1)s^2}{\sigma_0^2}
\]
Plugging in the values:
\[
\chi^2 = \frac{(25-1) \cdot 8^2}{17^2}
\]
\[
\chi^2 = \frac{24 \cdot 64}{289}
\]
\[
\chi^2 = \frac{1536}{289}
\]
\[
\chi^2 \approx 5.31
\]
Now, we need to find the critical value from the chi-square distribution table for \(n - 1 = 24\) degrees of freedom at the 0.10 level of significance (upper tail critical value for a one-tailed test).
Using a chi-square distribution table or a calculator, we find that the critical value (\(\chi^2_{0.10, 24}\)) is approximately 34.382.
Finally, we compare the test statistic to the critical value:
- If \(\chi^2 \geq 34.382\), we reject the null hypothesis.
- If \(\chi^2 < 34.382\), we fail to reject the null hypothesis.
Since \(\chi^2 \approx 5.31\) is much less than 34.382, we fail to reject the null hypothesis.
Therefore, based on the sample data and at the 0.10 level of significance, there is insufficient evidence to support the claim that the population standard deviation is greater than 17.