Question

At a large grocery store, based on a sample of size 19. the cashler error has a sample mean of 29.1 dollars per day and a standard deviation of 5.9 dollars. Find a 87% confidence interval for the population cashier error. You may assume that the cashier errors are normally distributed. Answer: We are 87% confident that the true population mean of cashier error lies somewhere between E and Round to the nearest whole dollar, do not include dollar symbol in your answer.

Answer 1

dfs given n=19, F = 2901, se 5.9, x = 0.13 te 19-1 =18, tc = 1.587 for a=0.13 Required & 7% Col for mean is- 29.1 1.58 X 5.9 7 un ng ( 29.1- 2.149, 29,1 + 2.148) = ( 26. 952) 31.248) I

We are 87% confident that the true population mean of cashier error lies somewhere between 26.952 and 31.248

I.e  $26.952 < \mu < 31.248$

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