confidence intrerval

Question

In a random sample of 400

 students at a university, 328

 stated that they were nonsmokers. Based on this sample, compute a 99%

 confidence interval for the proportion of all students at the university who are nonsmokers. Then find the lower limit and upper limit of the 99%

 confidence interval.

Carry your intermediate computations to at least three decimal places. Round your answers to two decimal places. (If necessary, consult a list of formulas.)

Lower limit:
Upper limit:

Answer 1

Hi Aubree,

We need to check and make sure we can use the Central Limit Theorem to assume normality. This requires np and n(1-p) to exceed 5 where:

n=sample size

p=probability we are given

p=328/400=0.82

np=0.82*400=328

1-p=0.18

n(1-p)=72

We're good to go. The formula for a confidence interval for a proportion is:

CI=p-hat +/- sqrt(p-hat(1-p-hat)/n)

Let's break this down:

p-hat=probability estimate

1-p-hat=1-probability estimate

n=sample size

Now, let's identify our values:

p-hat=328/400

p-hat=0.82

1-p-hat=0.18

n=400

CI=0.82 +/-sqrt[(0.82*0.18)/(400)]=

CI=0.82 +/- 0.019

To two decimals, this works out to:

CI= 0.82 +/- 0.02=

(0.80,0.84)

Lower Bound=0.80

Upper Bound=0.84

I hope this helps.