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 1008 supermarket shoppers revealed that 270 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 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.

95% 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 p̂.

Report p̂ along with the margin of error.    

Report the margin of error.

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

given data are:-

sample size (n) = 1008

number of shoppers,who revealed that they always stock up on an item when they find that item at a real bargain price(x)= 270

a).a point estimate for p = 0.2679

[ interpretation:-

]

b). 95% confidence interval for p be:-

lower limit	0.241
upper limit	0.295

[ interpretation:-

z critical value for 95% confidence interval (both tailed) be:-

= 1.960

the 95 % confidence interval be:-

]

explanation:-

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

c). as a reporter i will :-

Report p̂ along with the margin of error.

[ because if i report the margin of error, then anyone will easily understand how much the proportions deviates from the estimated proportion.if he wishes, then he will also be able to calculate the confidence interval easily,if needed.]

the margin of error based on a 95% confidence interval be = 0.027

[ interpretation:-

the margin of error be :-

= 0.027 ]

***If you have any doubt regarding the problem please write it in the comment box..if you are satisfied please give me a LIKE if possible...