Question

(1 point) A random sample of 100 observations produced a mean of 2 = 32 from a population with a normal distribution and a standard deviation 4.1. (a) Find a 95% confidence interval for a 25.33 SAS 26.47 (b) Find a 90% confidence interval for 25.48 SHS 26.37 (c) Find a 99% confidence interval for 25.15 SHS 26.65

Answer 1

Part a)
Confidence Interval :-
X̅ ± Z( α /2) σ / √ ( n )
Z(α/2) = Z (0.05 /2) = 1.96
32 ± Z (0.05/2 ) * 4.1/√(100)
Lower Limit = 32 - Z(0.05/2) 4.1/√(100)
Lower Limit = 31.20
Upper Limit = 32 + Z(0.05/2) 4.1/√(100)
Upper Limit = 32.80
95% Confidence interval is ( 31.20 , 32.80 )

Part b)
Confidence Interval :-
X̅ ± Z( α /2) σ / √ ( n )
Z(α/2) = Z (0.1 /2) = 1.645
32 ± Z (0.1/2 ) * 4.1/√(100)
Lower Limit = 32 - Z(0.1/2) 4.1/√(100)
Lower Limit = 31.33
Upper Limit = 32 + Z(0.1/2) 4.1/√(100)
Upper Limit = 32.67
90% Confidence interval is ( 31.33 , 32.67 )

Part c)
Confidence Interval :-
X̅ ± Z( α /2) σ / √ ( n )
Z(α/2) = Z (0.01 /2) = 2.576
32 ± Z (0.01/2 ) * 4.1/√(100)
Lower Limit = 32 - Z(0.01/2) 4.1/√(100)
Lower Limit = 30.94
Upper Limit = 32 + Z(0.01/2) 4.1/√(100)
Upper Limit = 33.06
99% Confidence interval is ( 30.94 , 33.06 )