Question

Listed below are speeds​ (mi/h) measured from traffic on a busy highway. This simple random sample was obtained at​ 3:30 P.M. on a weekday. Use the sample data to construct a 90​% confidence interval estimate of the population standard deviation. 62 61 61 55 61 53 60 59 60 70 58 68 LOADING... Click the icon to view the table of​ Chi-Square critical values.

Answer 1

Solution:

Confidence interval for population standard deviation is given as below:

Sqrt[(n – 1)*S² / χ²_{α/2, n– 1} ] < σ < sqrt[(n – 1)*S² / χ²_{1 -α/2, n– 1} ]

From given data, we have

Confidence level = 90%

Sample size = n = 12

Degrees of freedom = n – 1 = 11

Sample standard deviation = S = 4.716187

χ²_{α/2, n – 1} = 19.6751

χ²_{1 -α/2, n– 1} = 4.5748

(By using chi square table)

Sqrt[(12 – 1)* 4.716187^2 / 19.6751] < σ < sqrt[(12 – 1)* 4.716187^2 / 4.5748]

Sqrt{12.4353 ] < σ < sqrt[53.4812]

3.5264 < σ < 7.3131

Lower limit = 3.5264

Upper limit = 7.3131