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)

Not a function, you would get a parabola open to the right. y = √x and y = -√x

Step-by-step explanation: