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 994 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.

5% 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.

(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 along with the margin of error.

Report the margin of error.

Report the confidence interval.

Report p̂.

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

I understand that I need to find the critical value, but I'm not sure how to find the critical values given the information provided.

Answer #1

(a) point estimate for p=x/n=268/994=0.2696

(b)confidence interval (0.02420, 0.2972)

lower limit=0.2420,

upper limit=0.2972

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

(1-alpha)*100% confidence interval for population proportion (p)=sample proportion ±z(1-alpha/2)*SE(p)

90% confidence interval for P=0.2696±z(1-0.0.05/2)*0.0141=0.2696±1.96*0.0141=0.2696±0.0276=(0.2420, 0.2972)

SE(p)=sqrt(p(1-p)/n)=sqrt(0.2696*(1-0.2696)/994)=0.0141

(c)

Report p̂ along with the margin of error 26.96% with margin of error 2.76%

Report the margin of error=2.76%

Report the confidence interval.

(24.20%, 29.72%)

Report p̂.=26.96%

(d) here 95% critical value=z(1-0.05/2)=1.96 ( using ms-excel=normsinv(1-0.05/2))

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