Question

A sociologist develops a test to measure attitudes about public transportation, and 27 randomly selected subjects are given the test. Their mean score is 76.2 and their standard deviation is 21.4. Construct the 95% confidence interval for the standard deviation, σ, of the scores of all subjects.a. 17.5 < σ < 27.8b. 17.2 < σ < 27.2c. 285.61 < σ <858.49d. 16.6 < σ < 28.6e. 16.9 < σ < 29.3

Answer 1

Answer: e. 16.9<\sigma<29.3

Step-by-step explanation:

As per given , we have

n= 27 , $\overline{x}=76.2$  , s=21.4

Critical value using Chi-square table:-

$\chi^2_{\alpha/2, n-1}=\chi^2_{0.025, 26}=41.92317010\\\\ \chi^2_{1-\alpha/2, n-1}=\chi^2_{0.975, 26}=13.84390498$

Confidence interval for standard deviation:

$\sqrt{\dfrac{(n-1) s^2}{\chi^2_{\alpha/2, n-1}}}< \sigma<\sqrt{\dfrac{(n-1) s^2}{\chi^2_{1-\alpha/2, n-1}}}\\\\\\\sqrt{\dfrac{26(21.4)^2}{41.92317010}}<\sigma<\sqrt{\dfrac{26(21.4)^2}{13.84390498}}\\\\\\ 16.8528513346<\sigma<29.3272365692\\\\\approx 16.9<\sigma<29.3$

Hence, the 95% confidence interval for the standard deviation, σ, of the scores of all subjects : $16.9<\sigma<29.3$