Question

In a marketing survey, a random sample of 984 supermarket shoppers revealed that 276 always stock up on an item when they find that item at a real bargain price. (a) Let p represent the proportion of all supermarket shoppers who always stock up on an item when they find a real bargain.

Find a point estimate for p. (Use 3 decimal places.)

(b) Find a 90% confidence interval for p. (Use 3 decimal places.)

lower limit:

upper limit:

(c) What is the margin of error based on a 90% confidence interval? (Use 3 decimal places.)

Answer 1

Solution :

Given that,

n = 984

x = 276

Point estimate = sample proportion = = x / n = 276/984=0.280

1 -   = 1- 0.280 =0.720

At 90% confidence level

= 1 - 90%  

= 1 - 0.90 =0.10

/2 = 0.05

Z/2 = Z0.05 = 1.645 ( Using z table )

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

= 1.645 *((0.280*0.720) /984 )

E = 0.024

A 90% confidence interval for population proportion p is ,

- E < p < + E

0.280-0.024 < p < 0.280+ 0.024

0.256< p <0.304

The 90% confidence interval for the population proportion p is : 0.256 ,0.304