Given a random sample of size 16 drawn from a normal distribution with unknown m

Question

Given a random sample of size 16 drawn from a normal distribution with unknown mean and variance of 25. Suppose the sample mean is 8. a) Construct the 95% CI for the population mean. b) Based on the CI in a), does the claim that the population mean is 6 seem to be reasonable? Explain. c) Suppose another random sample of size 36 drawn from the same normal distribution and the sample mean is 8. Construct a 95% CI for the population mean. d) Compare CI in a) and c), and comment. e) What is another way to change the width of the confidence interval? How could you decrease the width of the confidence interval while keeping the sample size constant at 16?

Explanation / Answer

Since population variance is given, we should be using the z-score formula for sample mean.

For 95% Confidence Interval, Z = 1.96

a)

Sample size (n) = 16

Population Variance = 25

Population Standard Deviation () = 5

Sample mean (X) = 8

Lower limit = X- (Z * /sqrt(n)) = 8 - (1.96 * 5/SQRT(16))                 = 5.55

Upper limit = X + (Z * /sqrt(n)) = =8 + (1.96 * 5/SQRT(16))             = 10.45

b)

Yes, the claim that the population mean is 6 seems to be reasonable as 6 lies between 5.55 and 10.45 i.e. lie in the confidence interval.

c)

Sample size = 36

Sample mean = 8

Lower limit = X- (Z * /sqrt(n)) = 8 - (1.96 * 5/SQRT(36))                 = 6.37

Upper limit = X+ (Z * /sqrt(n))= =8 + (1.96 * 5/SQRT(36))             = 9.63

d)

Increasing the sample size decreases the confidence interval width because it decreases the standard error, which is getting proved by comparing (a) and (c). Also, the suggested population mean of 6 in step (b) will no longer be reasonable for case (c)

e)

By changing the confidence level, we can change the width of confidence interval. Lower the confidence level, lower the confidence interval width.

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