Question

In a marketing survey, a random sample of 1055 supermarket shoppers revealed that 328 always stock up on an item when they find that item at a bargain price. Construct a 90% confidence interval for the true percentage of supermarket shoppers that stock up on an item when they find it at a bargain price. Interpret your result. [

Answer 1

Solution :

Given that,

n = 1055

x = 328

Point estimate = sample proportion = = x / n = 328 / 1055 = 0.311

1 - = 1 - 0.311 = 0.689

At 90% confidence level

= 1 - 90%

=1 - 0.90 =0.10

/2 = 0.05

Z/2 = Z0.05  = 1.645

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

= 1.645 (((0.311 * 0.689) / 1055 )

= 0.023

A 90% confidence interval for population proportion p is ,

± E

= 0.311  ± 0.023

= ( 0.288, 0.334 )

We are 90% confident that the true percentage of supermarket shoppers that stock up on an item when they find it at a bargain price between 0.288 and 0.334.