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

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{((n-1) s^2)/(\chi^2_(\alpha/2, n-1))}< \sigma<\sqrt{((n-1) s^2)/(\chi^2_(1-\alpha/2, n-1))}\\\\\\\sqrt{(26(21.4)^2)/(41.92317010)}<\sigma<\sqrt{(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`