# deviation

• Mar 14th 2008, 03:39 PM
Aala
deviation
8.48 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.

Add all numbers together: Mean 6943 sq mm
Divide total by 20: Standard deviation 347.15

(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. (Data are from a project by MBA student Daniel R. Dalach.) DisplayAds

8.64 Biting an unpopped kernel of popcorn hurts! As an experiment, a self-confessed connoisseur of
cheap popcorn carefully counted 773 kernels and put them in a popper. After popping, the unpopped
kernels were counted. There were 86. (a) Construct a 90 percent confidence interval for the proportion
of all kernels that would not pop. (b) Check the normality assumption. (c) Try the Very Quick
Rule. Does it work well here? Why, or why not? (d) Why might this sample not be typical?
• Mar 14th 2008, 05:38 PM
TKHunny
Your "mean" is really just the sum.

Your "standard deviation" is the mean.

You don't have a standard deviation.

Let's get those straight before we try to move on.
• Mar 14th 2008, 06:02 PM
Aala
I am lost Homework
I thought that once you added all numbers to get the mean, then you divide the sum by the number of numbers added to get the standard deviation. I am really confused(Crying)
• Mar 15th 2008, 05:06 AM
mr fantastic
Quote:

Originally Posted by Aala
8.48 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.

Add all numbers together: Mean 6943 sq mm
Divide total by 20: Standard deviation 347.15

(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. (Data are from a project by MBA student Daniel R. Dalach.) DisplayAds

8.64 Biting an unpopped kernel of popcorn hurts! As an experiment, a self-confessed connoisseur of
cheap popcorn carefully counted 773 kernels and put them in a popper. After popping, the unpopped
kernels were counted. There were 86. (a) Construct a 90 percent confidence interval for the proportion
of all kernels that would not pop. (b) Check the normality assumption. (c) Try the Very Quick
Rule. Does it work well here? Why, or why not? (d) Why might this sample not be typical?

You need to nail the basics first - read this so that you know how to calcuate mean and standard deviation.
• Mar 16th 2008, 06:00 PM
Aala
Hello I am still confused

I read the material but I am still lost. I know that there is a formula to enter in excel, but I am not grasping the concept. I really need help.(Crying)
• Mar 17th 2008, 05:36 AM
TKHunny
We can write the book for you, but why would that be different from what you already have available. You need a qualified local tutor.

Try these.
=sum()
=average()
=stdev()
=stdevp()
Read the help files associated with them.

Again, if you do not know how to enter these formulas in Excel, or to get at the help files, you need a qualified local tutor or you need to take a class in Excel.

There must be some basic, fundamental understanding or this format simply is not very beneficial.
• May 15th 2008, 08:50 PM
cesar75
[quote=Aala;117126]Hello I am still confused

I read the material but I am still lost. I know that there is a formula to enter in excel, but I am not grasping the concept. I really need help please with this assignment