Assuming that the population is normally distributed, construct a 9090% 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.11 44 44 44 55 55 55 88 full data set sample b.11 22 33 44 55 66 77 88

Given:

Set A: 1 4 4 4 5 5 5 8

Mean: 4.5

Standard dev: 1.9

Set B:

Mean: 4.5

Standard dev: 2.45

% =  90 

Set A:

Standard Error, SE = s/ √n =    1.9/√8 = 0.67  

Degrees of freedom = n - 1 =   8 -1 =  7   

t- score =  1.89457861

Width of the confidence interval = t * SE =     1.89457861* 0.67 = 1.272685913

Lower Limit of the confidence interval = x-bar - width =      4.5 - 1.272685913 = 3.23

Upper Limit of the confidence interval = x-bar + width =      4.5 + 1.272685913 = 5.77

The 90% confidence interval is [3.23, 5.77]

Set B:

Standard Error, SE = s/ √n =    2.45/√8 = 0.87  

Degrees of freedom = n - 1 =   8 -1 = 7   

t- Score =  1.89457861

Width of the confidence interval = t * SE =     1.89457861* 0.87 = 1.641094994

Lower Limit of the confidence interval = x-bar - width =      4.5 - 1.641094994 = 2.86

Upper Limit of the confidence interval = x-bar + width =      4.5 + 1.641094994 = 6.14

The 90% confidence interval is [2.86, 6.14]

We can obviously see that sample B has more variation in the scores than sample A. The fact that the standard deviation is 2.45 for B and 1.9 for A). Therefore, they yield dissimilar confidence intervals even though they have the same mean and range.

  n=   0.0000 - 0.3750 i  

  n=   0.0000 + 0.3750 i

step-by-step explanation:

one of these should be correct

l.jnl.j/n.m,lm,

step-by-step explanation: