To test the claim that the last class has less variation than previous classes, we can use an F-test for variances.
Given:
- Standard deviation of previous classes (\( s_1 \)) = 14.1
- Variance of previous classes (\( \sigma_1^2 \)) = \( 14.1^2 = 198.81 \)
- Standard deviation of the last class (\( s_2 \)) = 9.3
- Variance of the last class (\( \sigma_2^2 \)) = \( 9.3^2 = 86.49 \)
- Sample size of previous classes (\( n_1 \)) is unknown but since it's not specified, we will assume it is significantly large (approaching infinity for critical value determination), thus we generally ignore it in critical value lookup.
- Sample size of the last class (\( n_2 \)) = 27
We will conduct the F-test using the following formula: \[ F = \frac{s_1^2}{s_2^2} = \frac{198.81}{86.49} \]
Now, let's calculate that: \[ F = \frac{198.81}{86.49} \approx 2.300 \]
Our null hypothesis \( H_0 \) is that the variance of the last class is equal to the variance of previous classes (i.e., \( \sigma_2^2 = \sigma_1^2 \)), while the alternative hypothesis \( H_a \) is that the variance of the last class is less (i.e., \( \sigma_2^2 < \sigma_1^2 \)).
Finding the Critical Value:
For the F-test:
- We look for \( F( \alpha, df_1, df_2) \).
- Here \( df_1 \) (degrees of freedom for the numerator) is \( n_1 - 1 \) and \( df_2 \) (degrees of freedom for the denominator) is \( n_2 - 1 = 27 - 1 = 26 \).
- At a significance level \( \alpha = 0.01 \) for a one-tailed test, we can look for the critical value \( F_{0.01}(df_1, 26) \).
Using an F-distribution table or calculator:
- The critical value of \( F(0.01, \infty, 26) \) has no set degree of freedom for \( df_1 \).
- More reasonably for this practical scenario, we can check it with \( df_1 = 26 \) for convenience, the value might come closer.
With degrees of freedom (using an F-table or calculator):
- The critical value corresponding to \( df_1 = \infty \) (the numerator) and \( df_2 = 26 \) (denominator) should be looked up.
Conclusion:
From the options given, we see critical values:
- A) 12.198, 45.642
- B) 11.160
- C) 12.198
- D) 11.160, 48.290
Given the theoretical understanding, generally, in normal tests for \(\alpha = 0.01\), the critical value we are likely to pick closer might be option C: \( 12.198 \).
Thus, the best critical value needed to test the claim is C) 12.198.