Question

Reming At a large grocery store, based on a sample of size 15, the cashier error has a sample mean of 29.3 dollars per day and a standard deviation of 5.4 dollars. Find a 84% confidence interval for the population cashier error. You may assume that the cashier errors are normally distributed. Answer: We are 84% confident that the true population mean of cashier error lies somewhere between Band Round to the nearest whole dollar, do not include dollar symbol in your answer

Answer 1

Now , df=degrees of freedom=n-1=15-1=14

The critical value is , $t_{df,\alpha/2}=t_{14,0.16/2}=1.484$ ; The excel function is , =TINV(0.15,14)

Therefore , the 84% confidence interval is ,

S X = tdf, 0/2 n

$29.3\pm 1.484*\frac{5.4}{\sqrt{15}}$

$29.3\pm 2.0691$

Therefore ,

We are 84% confident that the true population mean of cashier error lies somewhere between 27 and 31.