Question

A random sample of 815 adults surveyed resulted in 20% preferred honey to jam/jelly on their peanut butter sandwich. Find the 99% confidence interval for the proportion of adults that prefer honey.

What is the UL, Upper limit of the confidence interval?

Answer 1

Solution :

Given that,

n = 815

Point estimate = sample proportion = =20%=0.20

1 - = 1-0.20=0.80

At 99% confidence level the z is ,

  = 1 - 99% = 1 - 0.99 = 0.01

Z = Z0.01 = 2.326( Using z table )

Margin of error = E = Z * (( * (1 - )) / n)

= 2.326 (((0.20*0.80) /815 )

= 0.033

A 99% upper confidence interval is ,

+ E

0.20+0.033

0.233

Upper limit of the confidence interval is 0.233