2. Determine the 99.5% CI for the population mean and explain why the Z or T sta

Question

2.   Determine the 99.5% CI for the population mean and explain why the Z or T statistic applies. If the question can’t be answered with the given information, explain why that is so.

2.1   Random sample size 25, mean 50, and standard deviation 16; unknown population distribution

2.2   Same as 2.1 but with known normal population and known population standard deviation

2.3   Same as 2.1 but sample size 250 with unknown standard deviation and population believed to be non-symmetric with a considerable skew, extreme outliers, and possible multiple modes.

Explanation / Answer

2.1)here as we do not know population distribution; t stat should be applied,

std error of mean =std deviation/(n)1/2 =3.2

for 99.5% and (n-1=24) degree of freedom; t=3.0905

threfore 99.5% confidence interval =sample mean -/+ t*std error =40.1104 ; 59.8896

2.2) for we know population std deviation and distribution is normal; z stat should be applied,

for 99.5% CI, z=2.8070

threfore 99.5% confidence interval =sample mean -/+ z*std error =41.0175 ; 58.9825

2.3) here as shape is non normal and outliers; we should apply t distribution

std error of mean =std deviation/(n)1/2 =1.0119

for 99.5% and (n-1=249) degree of freedom; t=2.8323

threfore 99.5% confidence interval =sample mean -/+ t*std error =47.1339 ; 52.8661

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