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 980 supermarket shoppers revealed that 264 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. 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.5% 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 p̂. 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.)
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 980 supermarket shoppers revealed that 264 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. 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.5% 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 p̂. 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.)
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
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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 980 supermarket shoppers revealed that 264 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.
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.5% 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 p̂. 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.)
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