Question

The number of successes and the sample size are given for a simple random sample from a population. Use the one-proportion z-interval procedure to find the required confidence interval.

n = 76, x = 31; 98% level

		0.298 to 0.518
		0.276 to 0.540
		0.297 to 0.519
		0.277 to 0.539

Use the one-proportion z-interval procedure to find the required confidence interval.

A researcher wishes to estimate the proportion of adults in the city of Darby who are vegetarian. In a random sample of 1821 adults from this city, the proportion that are vegetarian is 0.044. Find a 99% confidence interval for the proportion of all adults in the city of Darby that are vegetarians.

		0.0316 to 0.0564
		0.0392 to 0.0488
		0.0138 to 0.0742
		0.0363 to 0.0517

Use the one-proportion z-interval procedure to find the required confidence interval.

Of 146 patients selected at random from a clinic, 35 were found to have high blood pressure. Construct a 95% confidence interval for the  percentage of all patients at this clinic that have high blood pressure.

		14.9% to 33.1%
		15.7% to 32.2%
		18.1% to 29.8%
		17.0% to 30.9%

Answer 1

Solution-1:

in ti83calculator go to

STATS>1-propzint

98% confidence interval for p is from 0.27675 to 0.53904

0.277 to 0.539

Solution-2:

p^=sample proportion=0.044

n=1821

z crit for 99%=2.576

99% confidence interval for p is

p^-z*sqrt(p^*(1-p^)/n),p^+z*sqrt(p^*(1-p^)/n)

0.044-2.576*sqrt(0.044*(1-0.044)/1821),0.044+2.576*sqrt(0.044*(1-0.044)/1821)

0.03161927,0.05638073

0.0316 ,0.0564

0.0316 to 0.0564

Solution-3:

p^=x/n=35/146=0.239726

z crit for 95%=1.96

95% confidence interval for p is

p^-z*sqrt(p^*(1-p^)/n),p^+z*sqrt(p^*(1-p^)/n)

0.239726-1.96*sqrt(0.239726*(1-0.239726)/146),0.239726+1.96*sqrt(0.239726*(1-0.239726)/146)

0.1704756, 0.3089764

0.170,0.309

0.170*100,0.309*100

=17.0% to 30.9%

ANSWER"

17.0% to 30.9%