Question

A school bus driver records the time (in minutes) it takes to commute to school for six days. Those results are: 25, 22, 17, 20, 15, and 10. Assuming the population is normally distributed, develop a 99% confidence interval for the population mean. The confidence interval is:

• A.

  18.17 +/- 8.80

• B.

  18.17 +/- 4.40

• C.

  18.17 +/- 5.62

• D.

  None of these answers are correct.

Answer 1

For given sample mean is $\overline{x}=\frac{\sum x}{n}=18.17$

And standard deviation is

Create the following table.

data	data-mean	(data - mean)2
25	6.8333	46.69398889
22	3.8333	14.69418889
17	-1.1667	1.36118889
20	1.8333	3.36098889
15	-3.1667	10.02798889
10	-8.1667	66.69498889

Find the sum of numbers in the last column to get. $\sum (xi-\overline{x})^2=142.83$

So $s=\sqrt{\frac{\sum (xi-\overline{x})^2}{n-1}}=5.34$

t value for 99% CI for 5 df is TINV(0.01,5)=4.032

So Margin of Error $E=t*\frac{s}{\sqrt{n}}=8.80$

Hence answer here is A. 18.17 +/- 8.80