In a marketing survey, a random sample of 1020 supermarket shoppers revealed that 268 always stock up on an item when they find that item at a real bargain price. a. Find a 95% confidence interval for p?
b. What is the margin of error based on a 95% confidence interval?

a) 95% of the confidence intervals created using this method would include the true proportion of shoppers who stock up on bargains.

b) Margin of error  = 0.0270.

Step-by-step explanation:

a)

= point estimate of p

From Table, critical values of ,

Lower limit

upper limit = 0.2627 + 0.0276

                 = 0.2903

Correct option:

95% of the confidence intervals created using this method would include the true proportion of shoppers who stock up on bargains.

b) Margin of error =Z SE

                              = 1.96 X 0.0141

                              = 0.0270.

b

step-by-step explanation: