Question

A sample of 15 commuters in Chicago showed the average of the commuting times was 33.2 minutes. If s = 8.3 minutes, find the 95% confi dence interval of the true mean.

Answer 1

The 95% confidence interval of the true mean.

(29.4261 ,36.9739)

Explanation:

Step :- (i)

Given sample size 'n' =15

sample of the mean x⁻ = 33.2

The standard deviation of the sample 'S' = 8.3

95% of confidence intervals

`(x^(-) - t_(\alpha ) (S)/(√(n) ) ,x^(-) + t_(\alpha )(S)/(√(n) ) )`

Step:-(ii)

The degrees of freedom γ=n-1 = 15-1=14

The tabulated value t = 1.761 at 0.05 level of significance.

now substitute all possible values, we get

`(33.2 - 1.761(8.3)/(√(15) ) ,33.2+ 1.761(8)/(√(15) ) )`

After calculation , we get

(33.2-3.7739 , 33.2+3.7739

(29.4261 ,36.9739)

Conclusion:-

the 95% confidence interval of the true mean.

(29.4261 ,36.9739)