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

Answer:

The 95% confidence interval of the true mean.

(29.4261 ,36.9739)

Step-by-step 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

$(x^{-} - t_{\alpha } \frac{S}{\sqrt{n} } ,x^{-} + t_{\alpha }\frac{S}{\sqrt{n} } )$

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

$(33.2 - 1.761\frac{8.3}{\sqrt{15} } ,33.2+ 1.761\frac{8}{\sqrt{15} } )$

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)