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

anan

8080​%

confidence interval estimate of the population standard deviation.

6363

6464

6464

5555

6464

5252

6161

6060

6161

6868

6060

6666

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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} ]

We are given

Confidence level = 80%

Sample size = n = 12

Degrees of freedom = n – 1 = 11

Sample standard deviation = S = 4.4823

χ²_{α/2, n – 1} = 95.4762

χ²_{1 -α/2, n– 1} = 63.3799

(By using chi square table)

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

Sqrt[(12 – 1)* 4.4823^2 / 95.4762] < σ < sqrt[(12 – 1)* 4.4823^2/ 63.3799]

Sqrt(16.6239) < σ < sqrt(25.0425)

4.0772 < σ <   5.0042

Lower limit = 4.0772

Upper limit = 5.0042