Question

In a survey of 1010 ?adults, 808 disapprove of the job the legislature is doing.

Construct? 90%, 95%, and? 99% confidence intervals for the population proportion. Interpret the results and compare the widths of the confidence intervals.

Answer 1

CI for	90%
n	1010
p	0.80
z-value of 90% CI	1.6449
SE = sqrt(p*(1-p)/n)	0.01259
ME = z*SE	0.02070
Lower Limit = p - ME	0.77930
Upper Limit = p + ME	0.82070
90% CI	(0.7793 , 0.8207 )
CI width	0.0414

CI for	95%
n	1010
p	0.80
z-value of 95% CI	1.9600
SE = sqrt(p*(1-p)/n)	0.01259
ME = z*SE	0.02467
Lower Limit = p - ME	0.77533
Upper Limit = p + ME	0.82467
95% CI	(0.7753 , 0.8247 )
CI width	0.0493

CI for	99%
n	1010
p	0.80
z-value of 99% CI	2.5758
SE = sqrt(p*(1-p)/n)	0.01259
ME = z*SE	0.03242
Lower Limit = p - ME	0.76758
Upper Limit = p + ME	0.83242
99% CI	(0.7676 , 0.8324 )
CI width	0.0648

As CI level increases width of confidence interval increases