Question

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?

Answer 1

Answer:

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)

$\hat{p}$ = point estimate of p $= 268/1020 =0.2627$

$\hat{q}= 1 - 0.2627 = 0.7373$

$SE = \sqrt{\hat{p}\hat{q}/n}\\\\=\sqrt{0.2627\times 0.7373/1020}=0.0138\\\alpha = 0.05$

From Table, critical values of $Z= \pm1.96$ ,

Lower limit

                  $= \hat{p} - Z SE\\ = 0.2627 - (1.96 X 0.0138) \\ = 0.2627 - 0.0276 \\ = 0.236$

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.