Whole Foods is an all-natural grocery chain that has 50,000square foot stores, u

Question

Whole Foods is an all-natural grocery chain that has 50,000square foot stores, up from the industry average of 34,000 squarefeet. Sales per square foot of supermarkets average just under $400per square foot, as reported by USA Today in an aritcle on"A whole new ballgame ingrocery shopping." Suppose that salesper square foot in the most recent fiscal year are recorded for arandom sample of 10 Whole Foods supermarkets. The data (salesdollars per square foot) are as follows: 854, 858, 801, 892, 849,807, 894, 863, 829, 815. Using the fact that the sample mean =846.2 and s = 32.866, and find a 95% confidence interval for thetrue mean sales dollars per square foot of all Whole Foodssupermarkets during th emost recent fiscal year. Are we 95%confident that this mean is greater than $800, the historicalaverage for Whole Foods? Ugh...I don't know where to begin with this...! Whole Foods is an all-natural grocery chain that has 50,000square foot stores, up from the industry average of 34,000 squarefeet. Sales per square foot of supermarkets average just under $400per square foot, as reported by USA Today in an aritcle on"A whole new ballgame ingrocery shopping." Suppose that salesper square foot in the most recent fiscal year are recorded for arandom sample of 10 Whole Foods supermarkets. The data (salesdollars per square foot) are as follows: 854, 858, 801, 892, 849,807, 894, 863, 829, 815. Using the fact that the sample mean =846.2 and s = 32.866, and find a 95% confidence interval for thetrue mean sales dollars per square foot of all Whole Foodssupermarkets during th emost recent fiscal year. Are we 95%confident that this mean is greater than $800, the historicalaverage for Whole Foods? Ugh...I don't know where to begin with this...!

Explanation / Answer

Is s standard error? the "formula" for confidence interval is sample mean +- (standard error)*(critical value) using t-distribution because I assumed I was given the standarderror. In my book t(0.97,9) = 2.262, two sided, degrees of freedom =9, observations minus one so the 95% confidence interval is: 846.2 +- 32.866*2.262 (771.86, 920.54) second part hypothesis testing H0 the mean is greater than 800, x > 800 HA the mean is less or equal to 800, x tc so you don't reject H0, we are 95%confident that this mean is greater than $800

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