Question

For this problem, carry at least four digits after the decimal in your calculations. Answers may vary slightly due to rounding.

In a marketing survey, a random sample of 1002 supermarket shoppers revealed that 280 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. (Enter a number. Round your answer to four decimal places.)

(b)

Find a 95% confidence interval for p. (For each answer, enter a number. Round your answers to three decimal places.)
lower limit    
upper limit    

Give a brief explanation of the meaning of the interval.

We are 95% confident that the true proportion of shoppers who stock up on bargains falls outside this interval. We are 5% confident that the true proportion of shoppers who stock up on bargains falls above this interval.     We are 95% confident that the true proportion of shoppers who stock up on bargains falls within this interval. We are 5% confident that the true proportion of shoppers who stock up on bargains falls within this interval.

(c)

As a news writer, how would you report the survey results on the percentage of supermarket shoppers who stock up on items when they find the item is a real bargain?

Report the margin of error. Report p̂.     Report the confidence interval. Report p̂ along with the margin of error.

What is the margin of error based on a 95% confidence interval? (Enter a number. Round your answer to three decimal places.)

Answer 1

Solution :

Given that,

n = 1002

x = 280

(A)Point estimate = sample proportion = = x / n = 280/1002=0.2794

AnsB

1 - = 1 - 0.2794=0.7206

At 95% confidence level

= 1 - 95%  

= 1 - 0.95 =0.05

/2 = 0.025

Z/2 = Z _{0.025 = 1.96} ( Using z table ( see the 0.025 value in standard normal (z) table corresponding z value is 1.96 )   

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

= 1.96 (((0.2794*0.7206 /1002 )

E = 0.0278

A 95% confidence interval for population proportion p is ,

- E < p < + E

0.2794 -0.0278< p < 0.2794 +0.0278

0.2517< p < 0.3072

(lower limit =0.252 , upper limit =0.307)

(C)We are 95% confident that the true proportion of shoppers who stock up on bargains falls within this interval. (lower limit =0.252 , upper limit =0.307)