Question

Assuming that the population is normally​ distributed, construct a 90​% confidence interval for the population mean for each of the samples below. Explain why these two samples produce different confidence intervals even though they have the same mean and range. Sample​ A: 1   4   4   4   5   5   5   8 Full data set Sample​ B: 1   2   3   4   5   6   7   8 Construct a 90​% confidence interval for the population mean for sample A. ​(Type integers or decimals rounded to two decimal places as​ needed.)

Answer 1

Solution:

Sample A mean = (1+4+4+4+5+5+5+8)/8 = 36/8=4.5

Sample B mean =(1+2+3+4+5+6+7+8)/8 = 36/8 = 4.5

Sample A standard deviation = sqrt(summation(Xi-mean)^2 /(n-1)) = sqrt(((1-4.5)^2 +(4-4.5)^2+(4-4.5)^2+(4-4.5)^2+(5-4.5)^2+(5-4.5)^2+(5-4.5)^2+(8-4.5)^2)/(8-1)) = 1.9272

Sample B standard deviation = sqrt(((1-4.5)^2+(2-4.5)^2+(3-4.5)^2+(4-4.5)^2+(5-4.5)^2+(6-4.5)^2+(7-4.5)^2+(8-4.5)^2)/(8-1))) = 2.4495

90% confidence interval can be calculated as

For sample A

Mean +/- talpha/2*SD/sqrt(n)

4.5+/- 1.8945*1.9272/sqrt(8) = 4.5+/- 1.8945*1.9272/2.835

= 4.5+/- 1.2878 = 3.21 to 5.79

For sample B

4.5 +/- 1.8945*2.4495/2.835

4.5+/- 1.6369

So 90% confidence interval for Sample B

2.86 To 6.14

As we can see that both samples have same mean and range but both sample has different sample standard deviation.

So 90% confidence interval is different as sample standard deviation is different.