A sample of 20 pages was taken without replacement from the 1,591-page phone directory.?

Question

A sample of 20 pages was taken without replacement from the 1,591-page phone directory Ameritech Pages Plus Yellow Pages. On each page, the mean area devoted to display ads was measured (a display ad is a large block of multicolored illustrations, maps, and text). The data (in square millimeters) are shown below:

0 260 356 403 536 0 268 369 428 536

268 396 469 536 162 338 403 536 536 130

a) Construct a 95 percent confidence interval for the true mean.

(b) Why might normality be an issue here? (c) What sample size would be needed to obtain an error of ±10 square millimeters with 99 percent confidence? (d) If this is not a reasonable requirement, suggest one that is.

Answer 1

Number of cases 20

To find the mean, add all of the observations and divide by 20

Mean 346.5

Squared deviations

(0-346.5)^2 = (-346.5)^2 = 120062.25

(260-346.5)^2 = (-86.5)^2 = 7482.25

(356-346.5)^2 = (9.5)^2 = 90.25

(403-346.5)^2 = (56.5)^2 = 3192.25

(536-346.5)^2 = (189.5)^2 = 35910.25

(0-346.5)^2 = (-346.5)^2 = 120062.25

(268-346.5)^2 = (-78.5)^2 = 6162.25

(369-346.5)^2 = (22.5)^2 = 506.25

(428-346.5)^2 = (81.5)^2 = 6642.25

(536-346.5)^2 = (189.5)^2 = 35910.25

(268-346.5)^2 = (-78.5)^2 = 6162.25

(396-346.5)^2 = (49.5)^2 = 2450.25

(469-346.5)^2 = (122.5)^2 = 15006.25

(536-346.5)^2 = (189.5)^2 = 35910.25

(162-346.5)^2 = (-184.5)^2 = 34040.25

(338-346.5)^2 = (-8.5)^2 = 72.25

(403-346.5)^2 = (56.5)^2 = 3192.25

(536-346.5)^2 = (189.5)^2 = 35910.25

(536-346.5)^2 = (189.5)^2 = 35910.25

(130-346.5)^2 = (-216.5)^2 = 46872.25

Add the squared deviations and divide by 19

Variance = 551547/19

Variance 29028.7895

Standard deviation = sqrt(variance) = 170.3784

a) 95 % limits

Since we do not know the population standard deviation and estimate it by the sample, we use the t-distribution with 20-1=19 degrees of freedom assuming normality.

Sample mean 346.5

Standard deviation = 170.3784

Standard error of mean = sd / sqrt(n)

SE = 170.3784/4.4721

Standard error of mean 38.0978

Confidence limits 346.5-(38.0978)(2.093)

and 346.5+(38.0978)(2.093)

95 % limits for mean is (266.7614, 426.2386)

b) Construct a histogram. Group the values into intervals and see if the data represent a bell-shaped curve. If it doesn't, normality would be an issue.

c) sample size n = (2.57 * 170.3784 / 10) ^2

n=1917.32 = 1917

d) The sample size is reasonable.