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

utsa1981 · 08-21-2008, 09:59 PM

directory.? 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.

cidyah · 08-21-2008, 09:59 PM

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.

08-21-2008, 09:59 PM	#1 (permalink)
utsa1981 Junior Member Join Date: Aug 2008 Posts: 1	A sample of 20 pages was taken without replacement from the 1,591-page phone directory.? 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.

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08-21-2008, 09:59 PM	#2 (permalink)
cidyah Junior Member Join Date: May 2008 Posts: 6	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.