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 1000 supermarket shoppers revealed that 268 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:

-5% of the confidence intervals created using this method would include the true proportion of shoppers who stock up on bargains.

-95% of all confidence intervals would include the true proportion of shoppers who stock up on bargains.

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

-5% of all confidence intervals would include the true proportion of shoppers who stock up on bargains.

(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 along with the margin of error.

Report p̂.

Report the confidence interval.

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 = 1000

x = 268

a) Point estimate = sample proportion = = x / n = 268 / 1000 = 0.2680

1 - = 1 - 0.2680 = 0.7320

b) At 95% confidence level

= 1 - 95%

=1 - 0.95 =0.05

/2
= 0.025

Z/2
= Z_{0.025 = 1.960}

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

= 1.96 (((0.2680 * 0.7320) / 1000)

= 0.027

A 95% confidence interval for population proportion p is ,

± E

= 0.2680 ± 0.027

= ( 0.241, 0.295 )

lower limit = 0.241

upper limit = 0.295

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

c) Report p̂ along with the margin of error

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

E = 1.96 (((0.2680 * 0.7320) / 1000)

E = 0.027

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