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
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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
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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]
<span>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.</span>
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