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.

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

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

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)

i think that you should ask google for the answer to that problem because if we don't know it google does

-362

step-by-step explanation:

(12)(4)+2(12)-34(12)+2(12)-3

  \/         \/         \/         \/

  48 +   24 -   408 + 24 - 3

      \   /               \   /

          72     -     432 - 3 = -363