1) Null = 70 and the alternate is != 70.

At .05 level of significance we cannot reject the null since T -1.8  does not lie in the region z > 19.6 or z < -1.96

P value is .0719 which means there is a 7.19% chance that the value is true.

If the standard deviation was specified at 1.75 we would reject the null.

If the sample mean is 69 lbs and the standard deviation is 3.5 we would reject the null as t -2 lies in the rejection region.

 

 

2) If x̅ = 85, σ = standard deviation = 8, and n=64, set up 95% confidence interval estimate of the population mean μ.

a 95% confidence interval the population mean is 83.04 < u < 86.96

 





I think I did this right. without seeing another version of the program I don't know if I got the right output. 

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