Question

# For this problem, carry at least four digits after the decimal in your calculations. Answers may...

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.

(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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