Question

A simple random sample of size n is drawn. The sample mean, x, is found to be 35.1, and the sample standard deviation, s, is found to be 8.7

• a) Construct a 90% confidence interval for μ if the sample size, n, is 100.

• b) Construct a 90% confidence interval for μ if the sample size, n, is 40. How does decreasing the sample size affect the margin of error, E?

• c) Construct a 96% confidence interval for μ if the sample size, n, is 40. How does increasing the level of confidence affect the margin of error, E?

• d) If the sample size is n = 25, what conditions must be satisfied to compute the confidence interval?

Answer 1

Part a)

Confidence Interval
X̅ ± t(α/2, n-1) S/√(n)
t(α/2, n-1) = t(0.1 /2, 100- 1 ) = 1.66
35.1 ± t(0.1/2, 100 -1) * 8.7/√(100)
Lower Limit = 35.1 - t(0.1/2, 100 -1) 8.7/√(100)
Lower Limit = 33.6558
Upper Limit = 35.1 + t(0.1/2, 100 -1) 8.7/√(100)
Upper Limit = 36.5442
90% Confidence interval is ( 33.6558 , 36.5442 )
Margin of Error = t(α/2, n-1) S/√(n) = 1.4442

Part b)

Confidence Interval
X̅ ± t(α/2, n-1) S/√(n)
t(α/2, n-1) = t(0.1 /2, 40- 1 ) = 1.685
35.1 ± t(0.1/2, 40 -1) * 8.7/√(40)
Lower Limit = 35.1 - t(0.1/2, 40 -1) 8.7/√(40)
Lower Limit = 32.7821
Upper Limit = 35.1 + t(0.1/2, 40 -1) 8.7/√(40)
Upper Limit = 37.4179
90% Confidence interval is ( 32.7821 , 37.4179 )

Margin of Error = t(α/2, n-1) S/√(n) = 2.3179

As sample size decreases, margin of error increases.

part c)

Confidence Interval
X̅ ± t(α/2, n-1) S/√(n)
t(α/2, n-1) = t(0.04 /2, 40- 1 ) = 2.125
35.1 ± t(0.04/2, 40 -1) * 8.7/√(40)
Lower Limit = 35.1 - t(0.04/2, 40 -1) 8.7/√(40)
Lower Limit = 32.1769
Upper Limit = 35.1 + t(0.04/2, 40 -1) 8.7/√(40)
Upper Limit = 38.0231
96% Confidence interval is ( 32.1769 , 38.0231 )

Margin of Error = t(α/2, n-1) S/√(n) = 2.9231

As level of confidence level decrease, margin of error increases.

Part d)

For confidence interval, one should consider large sample size, but if population is normally distributed then smaller sample size is also consider.