Question

2. The manager of the dairy section of a large supermarket took a random sample of 250 egg cartons and found that 40 cartons had at least one broken egg.

a) Find a 95% confidence interval for p, the proportion of egg cartons that have at least one broken egg.

b) Express your confidence interval in words.

c) What is the margin of error?

d) What sample size would be needed for a margin of error of 3%?

Answer 1

Solution :

Given that,

a) Point estimate = sample proportion = = x / n = 40 / 250 = 0.16

1 - = 1 - 0.16 = 0.84

Z/2 = Z_0.025 = 1.96

Margin of error = E = Z _{/ 2} * (( * (1 - )) / n)

= 1.96 (((0.16 * 0.84) / 250)

= 0.045

A 95% confidence interval for population proportion p is ,

± E  

= 0.16  ± 0.045

= ( 0.115, 0.205 )

b) We are 95% confident that the true proportion of egg cartons that have at least one broken egg between 0.115 and 0.205.

c) Margin of error = E = Z _{/ 2} * (( * (1 - )) / n)

= 1.96 (((0.16 * 0.84) / 250)

= 0.045

d) sample size = n = (Z _{/ 2} / E )² * * (1 - )

= (1.96 / 0.03)² * 0.16 * 0.84

= 573.67

sample size = n = 574