Question

A new cheese spread is tested in ve local supermarkets. A sample of 600 shoppers try the product,

and 348 indicate that they like it. Construct a 95% confidence interval estimate for the percentage of

shoppers who like the new cheese spread. What are the upper and lower confidence intervals?

Answer 1

Solution :

Given that,

n = 600

x = 348

Point estimate = sample proportion = = x / n = 348/600=0.58

1 -   = 1- 0.58 =0.42

At 95% confidence level the z is

= 1 - 95% = 1 - 0.95= 0.05

/ 2 = 0.025

Z/2 = Z0.025 = 1.96 ( Using z table )

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

= 1.96 *((0.58*0.42) / 600)

= 0.0395

A 95% confidence interval for population proportion p is ,

- E < p < + E

0.58-0.0395< p < 0.58+ 0.0395

0.5405< p < 0.6195

The 95% confidence interval for the population proportion p is : 0.5405, 0.6195