Question

A local health care company wants to estimate the mean weekly elder day-care cost. A sample of 10 facilities shows a mean of $250 per week, with a standard deviation of $25. What is the 90% confidence interval for the population mean?

Answer 1

Answer:

Step-by-step explanation:

Mean "X"=250

s = 25

n=10

Because Standard deviation is unknown and sample size is small, we must employ the T-distribution

df=n-1 = 10-1

df=9

For 90% confidence, t = 1.833

E=t*s/$\sqrt{n$ = 1.833*(25/$\sqrt{10}$)

E=14.49

The 90% confidence interval is X±E=250±14.49 or 235.51, 264.49