Question

Sleeping outlier: A simple random sample of nine college freshmen were asked how many hours of sleep they typically got per night. The results were

8.5 24 9 8.5 8 6.5 6 7.5 9.5

Notice that one joker said that he sleeps 24 a day.

(a) The data contain an outlier that is clearly a mistake. Eliminate the outlier, then construct a 99.9% confidence interval for the mean amount of sleep from the remaining values. Round the answers to at least two decimal places.

(b) Leave the outlier in and construct the 99.9% confidence interval. Round the answers to at least two decimal places.

Answer 1

6, 6.5, 6.75, 8, 8.5, 8.5, 9, 9.5, 24

Median = 8.5

Q1 = (6.5 + 6.75)/2 = 6.625

Q3 = (9 + 9.5)/2 = 9.25

IQR = Q3 - Q1 = 9.25 - 6.625 = 2.625

Q1 - 1.5 * IQR = 6.625 - 1.5 * 2.625 = 2.6875

Q3 + 1.5 * IQR = 9.25 + 1.5 * 2.625 = 13.1875

So the outlier is 24.

= (6 + 6.5 + 6.75 + 8 + 8.5 + 8.5 + 9 + 9.5)/8 = 7.844

s = sqrt(((6 - 7.844)^2 + (6.5 - 7.844)^2 + (6.75 - 7.844)^2 + (8 - 7.844)^2 + (8.5 - 7.844)^2 + (8.5 - 7.844)^2 + (9 - 7.844)^2 + (9.5 - 7.844)^2)/7) = 1.274

At 99.9% confidence level the critical value is t^* = 5.409

The 99.9% confidence interval is

+/- t^* * s/

= 7.844 +/- 5.409 * 1.274/

= 7.844 +/- 2.436

= 5.408, 10.280

b) = (6 + 6.5 + 6.75 + 8 + 8.5 + 8.5 + 9 + 9.5 + 24)/9 = 9.639

s = sqrt(((6 - 9.639)^2 + (6.5 - 9.639)^2 + (6.75 - 9.639)^2 + (8 - 9.639)^2 + (8.5 - 9.639)^2 + (8.5 - 9.639)^2 + (9 - 9.639)^2 + (9.5 - 9.639)^2 + (24 - 9.639)^2)/8) = 5.516

At 99.9% confidence level the critical value is t^* = 5.042

The 99.9% confidence interval is

+/- t^* * s/

= 9.639 +/- 5.042 * 5.516/

= 9.639 +/- 9.271

= 0.368, 18.910