Question

In a survery of 134 households, a Food Marketing Institute found that 68 households spend more than $125 a week on groceries. Please find the 95% confidence interval for the true proportion of the households that spend more than $125 a week on groceries.

A. Enter your answer as an open-interval (i.e., parentheses) using decimals (not percents) accurate to three decimal places.

Confidence interval =

B. Express the same answer as a tri-linear inequality using decimals (not percents) accurate to three decimal places.

< p <

C. Express the same answer using the point estimate and margin of error. Give your answers as decimals, to three places.

p =    ±

Answer 1

Solution :

Given that,

a) n = 134

x = 68

Point estimate = sample proportion = = x / n = 68 / 134 = 0.507

1 - = 1- 0.507 = 0.493

At 95% confidence level

= 1 - 95%

=1 - 0.95 =0.05

/2 = 0.025

Z/2 = Z_{0.025 = 1.960}

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

= 1.96 (((0. 507 * 0.493) / 134)

= 0.043

b) A 95% confidence interval for population proportion p is ,

- E < p < + E

0.507 - 0.043 < p < 0.507 + 0.043

0.464< p < 0.550

c) ± E  

0.507 ± 0.043

0.464 , 0.550