Question

Farmer's Corporation made a report that the mean annual household income of its readers is $119,155 (Thebill, March 2017). Assume this estimate of the mean annual household income is based on a sample of 80 households, and, based on past studies the population standard deviation is known to be σ = $30,000.00.

1. Developed a 90% confidence interval estimate of the population mean

2. Developed a 95% confidence interval estimate of the population mean.

3. Developed a 99% confidence interval estimate of the population mean

4. Discuss what happens to the width of the confidence interval as the confidence level is increased. Does this result seem reasonable? Explain.

Answer 1

a)

90% Confidence Interval :-
X̅ ± Z( α /2) σ / √ ( n )
Z(α/2) = Z (0.1 /2) = 1.645
119155 ± Z (0.1/2 ) * 30000/√(80)
Lower Limit = 119155 - Z(0.1/2) 30000/√(80)
Lower Limit = 113638
Upper Limit = 119155 + Z(0.1/2) 30000/√(80)
Upper Limit = 124672
90% Confidence interval is ( 113638 , 124672 )

b)

95% Confidence Interval :-
X̅ ± Z( α /2) σ / √ ( n )
Z(α/2) = Z (0.05 /2) = 1.96
119155 ± Z (0.05/2 ) * 30000/√(80)
Lower Limit = 119155 - Z(0.05/2) 30000/√(80)
Lower Limit = 112581
Upper Limit = 119155 + Z(0.05/2) 30000/√(80)
Upper Limit = 125729
95% Confidence interval is ( 112581 , 125729 )

c)

99% Confidence Interval :-
X̅ ± Z( α /2) σ / √ ( n )
Z(α/2) = Z (0.01 /2) = 2.576
119155 ± Z (0.01/2 ) * 30000/√(80)
Lower Limit = 119155 - Z(0.01/2) 30000/√(80)
Lower Limit = 110515
Upper Limit = 119155 + Z(0.01/2) 30000/√(80)
Upper Limit = 127795
99% Confidence interval is ( 110515 , 127795)

d)

As confidence level increased, the width of confidence interval increases.

The result seems reasonable, because as confidence level increases, the accuracy of population

parameter falls in the interval also increased so width increases.