Consider a 95 confidence interval with a standard deviation

Consider a 95% confidence interval with a standard deviation of 7 and a sample size of 12. If the standard deviation was increased to 9 what would happen to the width of the interval? If the sample size was changed to 6 what would happen to the width of the interval? If the confidence decreased to 90% what would happen to the width of the interval? Why is a prediction limit wider than a confidence limit given the same data and same alpha?

Solution

Z= 1.96

sigma = 7

n = 12

+/- 1.96 * 7 / srqt (12)
+/- 3.96

a)

+/- 1.96 * 9 / srqt (12)

+/-5.09

increase

b)

+/- 1.96 * 9 / srqt (6)

+/-7.20

increase

c)

1-0.90 = 0.10/2 = 0.05

Z=1.64

+/- 1.64 * 9 / srqt (12)

+/- 4.26

increase

d)

I can gladly help you but you should post it in a new question