A simple random sample with n=56 provided a sample mean of 23.5 and a sample standard...

A simple random sample with n=56 provided a sample mean of 23.5 and a sample standard deviation of 4.2

Develop a 90 % confidence interval for the population mean (to 1 decimal).

Develop a 95 % confidence interval for the population mean (to 1 decimal)

Develop a 99 % confidence interval for the population mean (to 1 decimal).

Expert Solution

Part a)

Confidence Interval
X̅ ± t(α/2, n-1) S/√(n)
t(α/2, n-1) = t(0.1 /2, 56- 1 ) = 1.673
23.5 ± t(0.1/2, 56 -1) * 4.2/√(56)
Lower Limit = 23.5 - t(0.1/2, 56 -1) 4.2/√(56)
Lower Limit = 22.6
Upper Limit = 23.5 + t(0.1/2, 56 -1) 4.2/√(56)
Upper Limit = 24.4
90% Confidence interval is ( 22.6 , 24.4)

Part b)

Confidence Interval
X̅ ± t(α/2, n-1) S/√(n)
t(α/2, n-1) = t(0.05 /2, 56- 1 ) = 2.004
23.5 ± t(0.05/2, 56 -1) * 4.2/√(56)
Lower Limit = 23.5 - t(0.05/2, 56 -1) 4.2/√(56)
Lower Limit = 22.4
Upper Limit = 23.5 + t(0.05/2, 56 -1) 4.2/√(56)
Upper Limit = 24.6
95% Confidence interval is ( 22.4 , 24.6 )

Part c)

Confidence Interval
X̅ ± t(α/2, n-1) S/√(n)
t(α/2, n-1) = t(0.01 /2, 56- 1 ) = 2.668
23.5 ± t(0.01/2, 56 -1) * 4.2/√(56)
Lower Limit = 23.5 - t(0.01/2, 56 -1) 4.2/√(56)
Lower Limit = 22.0
Upper Limit = 23.5 + t(0.01/2, 56 -1) 4.2/√(56)
Upper Limit = 25.0
99% Confidence interval is ( 22.0 , 25.0 )

orchestra answered 1 year ago

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