Question

(6 pts) Suppose you have selected a random sample of n 7 measurements from a normal distribution. Compare the standard normal z values with the corresponding t values if you were forming the following confidence intervals. (a) 80% confidence interval (b) 90% confidence interval (c) 95% confidence interval
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Answer #1

Solution:-

a)

For 80% confidence interval

z = + 1.282

D.F = n - 1

D.F = 6

t = + 1.44

b)

For 90% confidence interval

z = 1.645

D.F = n - 1

D.F = 6

t = + 1.943

c)

For 95% confidence interval

z = 1.96

D.F = n - 1

D.F = 6

t = + 2.447

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