# Thread: Sampling Question - Help!!!

1. ## Sampling Question - Help!!!

A sample of 20 pages was taken without replacement from the 1,591-page phone directory Ameritech Pages Plus Yellow Pages. One each page the mean area devoted to display ads was measured ( a display ad is a large block of multicolered 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 +/- square millimeters with 99 percent confidence? (d)If this is not a reasonable requirement suggest one that is.

2. Originally Posted by chocolat771
A sample of 20 pages was taken without replacement from the 1,591-page phone directory Ameritech Pages Plus Yellow Pages. One each page the mean area devoted to display ads was measured ( a display ad is a large block of multicolered 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 +/- square millimeters with 99 percent confidence? (d)If this is not a reasonable requirement suggest one that is.

First let's take this one step at a time. You need a sample mean and standard deviation. Your number set is:

0,260,356,403,536,0,268,369,428,536,268,396,469,53 6,162,338,403,536,536,130

Enter that here --> Mean, Variance, and Standard Deviation

I get $\bar{X} = 346.5$
$\sigma = 166.0643$

Now, since your sample size is less than 30, we should not do the normal test, but rather, the Student t-test. To see how that works, go here:

Student t-Test Confidence Interval for Mean

Using our mean and standard deviation that we calculated, n=20, and 95 for our confidence interval, I get:

$268.7804 < \mu < 424.2196$

PS - Scroll down in the math for a message on how to calculate this using Excel.

For your questions c and d, you can calculate a 99% confidence interval for your current sample size, but I'm not sure I understand what you mean by what sample size needed to ensure 99% confidence.

3. Originally Posted by chocolat771
A sample of 20 pages was taken without replacement from the 1,591-page phone directory Ameritech Pages Plus Yellow Pages. One each page the mean area devoted to display ads was measured ( a display ad is a large block of multicolered 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 +/- ...... square millimeters with 99 percent confidence? (d)If this is not a reasonable requirement suggest one that is.

I need help understanding what the questions are asking and how to input info into excel or minitab. Mr F says: I'd suggest that you go to a minitab or excel reference book or manual.

There is a value missing where I've inserted the red dots. Part (c) cannot be done without the missing value.

(a) Use the t-distribution because population variance is unknown and sample size is small (n < 50).

df = 20 - 1 = 19. $t_{\alpha/2} = t_{0.025} = 2.093$ (using http://www.anu.edu.au/nceph/surfstat...e/tables/t.php, for example).

$\bar{x} - t_{\alpha/2} \frac{s}{\sqrt{n}} < \mu < \bar{x} + t_{\alpha/2} \frac{s}{\sqrt{n}}$

where $\bar{x}$ is the sample mean and s is the sample sd.

(b) For sample sizes less than ~15: The t-distribution can be used if the population is close to normal. The t-distribution should not be used if the population is non-normal or if outliers are present in the sample data set.

For sample sizes ~15 or greater and less than ~40: t-distribution can be safely used whether or not the population is normal, except in the presence of outliers or strong skewness in the data set.

Your data set has outliers .....