Question

a. Find a 98% confidence interval for the true mean of a population if a sample of
52 results in a mean of 100. Assume the population standard deviation is 12.
b. Assume now that the same results occurred, the population was normal, and the
sample size was reduced to 10.
c. Repeat problem 2b assuming that the population standard deviation was unknown, and
“s” was 12.

Answer 1

a)

98% Confidence Interval :-
X̅ ± Z( α /2) σ / √ ( n )
Z(α/2) = Z (0.02 /2) = 2.326
100 ± Z (0.02/2 ) * 12/√(52)
Lower Limit = 100 - Z(0.02/2) 12/√(52)
Lower Limit = 96.13
Upper Limit = 100 + Z(0.02/2) 12/√(52)
Upper Limit = 103.87
98% Confidence interval is ( 96.13 , 103.87 )

b)

98 Confidence Interval :-
X̅ ± Z( α /2) σ / √ ( n )
Z(α/2) = Z (0.02 /2) = 2.326
100 ± Z (0.02/2 ) * 12/√(10)
Lower Limit = 100 - Z(0.02/2) 12/√(10)
Lower Limit = 91.17
Upper Limit = 100 + Z(0.02/2) 12/√(10)
Upper Limit = 108.83
98% Confidence interval is ( 91.17 , 108.83 )

c)

Confidence Interval
X̅ ± t(α/2, n-1) S/√(n)
t(α/2, n-1) = t(0.02 /2, 10- 1 ) = 2.821
100 ± t(0.02/2, 10 -1) * 12/√(10)
Lower Limit = 100 - t(0.02/2, 10 -1) 12/√(10)
Lower Limit = 89.30
Upper Limit = 100 + t(0.02/2, 10 -1) 12/√(10)
Upper Limit = 110.70
98% Confidence interval is ( 89.30 , 110.70 )