In a random sample of students at a university, stated that they were nonsmokers. Based on this sample, compute a confidence interval for the proportion of all students at the university who are nonsmokers. Then find the lower limit and upper limit of the confidence interval.

(0.8165 ; 0.8819)

Lower boundary = 0.8165

Upper boundary = 0.8819

Step-by-step explanation:

Given :

Sample proportion. Phat = x/ n = 276/ 325 = 0.8492

Confidence interval :

Phat ± margin of error

Margin of Error = Zα/2* [√Phat(1 - Phat) / n]

Phat ± Zα/2* [√Phat(1 - Phat) / n]

The 90% Z critical value is = 1.645

0.8492 ± 1.645*[√0.8492(1 - 0.8492) / 325)

0.8492 ± 1.645*[√0.8492(0.1508) / 325]

0.8492 ± 1.645*√0.0003940288

0.8492 ± 0.0326535

Lower boundary = 0.8492 - 0.0326535 = 0.8165

Upper boundary = 0.8492 + 0.0326535 = 0.8819

Confidence interval = (0.8165 ; 0.8819)

the answer is:

the answer is:

to solve the problem, we need to remember the quotient of power property, it's defined by the following relation:

if we have a quotienf of powers that have the same base, we need to keep the same base and subtract the exponent of the denominator power to the exponent of the numerator power.

so, we are given the expression:

then, calculating we have:

hence, the answer is:

have a nice day!