The association for Dressings and Sauce wanted to find out t

The association for Dressings and Sauce wanted to find out the percentage of American adults that eat salad regularly. A random sample of 1800 adult Americans was surveyed and 1498 of them said they eat salad at least once a week. Find a 95% confidince interval for the population proportion of American adults who eat salad at least once a week.

A(.79535, . 86909)

B(.80668, . 85776)

C(.81496, . 84948)

D(.82259, .84185)

Solution

CI = p ± Z a/2 Sqrt(p*(1-p)/n)))
x = Mean
n = Sample Size
a = 1 - (Confidence Level/100)
Za/2 = Z-table value
CI = Confidence Interval
Mean(x)=1498
Sample Size(n)=1800
Sample proportion = x/n =0.83222
Confidence Interval = [ 0.83222 ±Z a/2 ( Sqrt ( 0.83222*0.16778) /1800)]
= [ 0.83222 - 1.96* Sqrt(0.00008) , 0.8322 + 1.96* Sqrt(0.00008) ]
= [ 0.81496,0.84948]