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charle [14.2K]
3 years ago
6

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.
Mathematics
1 answer:
OleMash [197]3 years ago
8 0

Answer:

The 95% confidence interval of the true mean.

(29.4261 ,36.9739)

Step-by-step explanation:

<u>Step :- (i)</u>

Given sample size 'n' =15

sample of the mean x⁻ = 33.2

The standard deviation of the sample 'S' = 8.3

<u>95% of confidence intervals</u>

<u></u>(x^{-} - t_{\alpha } \frac{S}{\sqrt{n} } ,x^{-} + t_{\alpha }\frac{S}{\sqrt{n} } )<u></u>

<u>Step:-(ii)</u>

<u>The degrees of freedom γ=n-1 = 15-1=14</u>

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)

<u>Conclusion</u>:-

the 95% confidence interval of the true mean.

(29.4261 ,36.9739)

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