Graded Assignment 1

252solngr1-072 10/02/07

Graded Assignment 1

Please show your work! Neatness and whether the papers are stapled may affect your grade.

1. (Keller & Warrack) A Federal agency is checking weights of an ‘8 ounce’ container of a product. A random sample of 13 bottles is taken. Results are below.

7.81 7.92 7.94 8.00 7.95 7.91 7.98 8.05 8.17 7.80 7.80 7.80 7.90

Personalize the data as follows: change the last digit of the weights of the last three bottles to the last three digits of your student number. Example: Ima Badrisk has the student number 123456; so the last three numbers become {7.84, 7.85, 7.96}.

Compute the sample standard deviation using the computational formula (if you don’t know what that means, find out!). Use this sample standard deviation to compute a 90% confidence interval for the mean. Using the concept of significant difference, does it appear that these 6 ounce containers are mislabeled? Why? Is the mean significantly different from 7.96 ounces?

2. If a store has 1000 bottles of the product, do a confidence interval for the total weight of this stock. (See problem 8.50 in the text.) Convert your result to pounds.

3. Show how these results would change if the 13 bottles were taken from a trial batch of only 100 bottles.

4. Assume that the population standard deviation is 10 (and that the sample of 13 is taken from a very large population). Find (the 98th percentile) using the Normal table and use it to compute a 96% confidence interval. Does the mean differ significantly from 8 ounces now? From 7.96? Why?

Extra Credit:

5. a. Use the data above to compute a 98% confidence interval for the population standard deviation.

b. Assume that you got the sample standard deviation that you got above from a sample of 45, repeat a.

c. Fool around with the method for getting a confidence interval for a median and try to come close to a 99% confidence interval for the median.

6. Use the computer labs to access Minitab. Click on the command part of the display (the blank upper sheet). Use editor and ‘enable commands’to start. Check some numbers in the Normal, t, Chi-Squared or F tables using the set of Minitab routines that I have prepared. To use the new set of routines, follow the instructions inAreadoc1. There are several things that you can do. For the Normal distribution use the computer to check the answers to Examples 6.1-6.4 on pp 198-200 in the text. For the t-table pick a number of degrees of freedom and show that for that number of degrees of freedom, the probability above, say, on the t-table is 20%. You can do the same for the F and chi-squared tables in your book of tables. A good answer will explain what you did and contain the command dialog and graphs.

1. (Keller & Warrack) A Federal agency is checking weights of an ‘8 ounce’ container of a product. A random sample of 13 bottles is taken. Results are below.

7.81 7.92 7.94 8.00 7.95 7.91 7.98 8.05 8.17 7.80 7.80 7.80 7.90

Solution: Two data sets for computations of the variance are shown here. They will be referred to as solution 1 and solution 2. The first represents a student number of 000000 and the second 999999.

index

1 7.81 60.9961 7.81 60.9961

2 7.92 62.7264 7.92 62.7264

3 7.94 63.0436 7.94 63.0436

4 8.00 64.0000 8.00 64.0000

5 7.95 63.2025 7.95 63.2025

6 7.91 62.5681 7.91 62.5681

7 7.98 63.6804 7.98 63.6804

8 8.05 64.8025 8.05 64.8025

9 8.17 66.7489 8.17 66.7489

10 7.80 60.8400 7.80 60.8400

11 7.80 60.8400 7.89 62.2521

12 7.80 60.8400 7.89 62.2521

13 7.90 62.4100 7.99 63.8401

103.03 816.6985 103.30 820.9528

Computation of Sample Variances using the computational formula.

, , , and .

The means are 7.92538 and 7.94615

Computation of Standard Errors

Finding t

The significance level is given as 90%.

Please, please repeat after me

“Sigma means z and s means use t

(unless you have a very high freedom degree).”

Putting it all together

Remember 7.92538 and 7.94615

or 7.870 to 7.980. or 7.898 to 7.994.

Yes! In a statistical sense, these sample means are significantly different from 8 ounces, but they are not significantly different from 7.96 ounces, since 8 ounces is not included in the interval, but 7.96 is. Note that your interval could include 8 if you got a standard error of 0.04209 or larger.

2. If a store has 1000 bottles of the product, do a confidence interval for the total weight of this stock. (See problem 8.50 in the text.) Convert your result to pounds.

Solution 1: 7.870 to 7.980 when multiplied by 100 gives us 7878 to 7980 ounces. If we divide these by 16 and round to the nearest pound we get 492 to 499 pounds. A finite population correction is not needed here

Solution 2:7.898 to 7.994 when multiplied by 100 gives us 7898 to 7994 ounces. If we divide these by 16 and round to the nearest pound we get 494 to 500 pounds.

3. Show how these results would change if the 13 bottles were taken from a trial batch of only 100 bottles.

Solution: Recall that , 7.92538, 7.94615, , , and

Solution 1: If , the sample of 13 is more than 5% of the population, so use a finite population correction. Recall that . Now .. You may have found that the in problem 1 is multiplied by and that the size of the confidence interval is reduced by the same factor. . The interval becomes or 7.874 to 7.976. If you did not find that the mean was significantly different from 8 before, it may be now. You may even find that it is different from 7.96.

Solution 2: If , the sample of 13 is more than 5% of the population, so use a finite population correction. Recall that previously we had . This will be cut by about 7% as in Solution 1. If we multiply 0.02716 by 0.we get about 0.02546. This means or 7.901 to 7.991. The interval is smaller, but it doesn’t change anything – the mean is still significantly different from 8 (but not 7.96).

4. Assume that the population standard deviation is 10 (and that the sample of 11 is taken from a very large population). Find (the 98th percentile) using the Normal table and use it to compute a 96% confidence interval. Does the mean differ significantly from 8 ounces now? From 7.96? Why?

Solution: Make a diagram! The diagram for will be a Normal curve centered at zero and will show one point, , which has 2% above it (and 98% below it!) and is above zero because zero has 50% below it. Since zero has 50% above it, the diagram will show 48% between zero and .

From the diagram, we want one point so that or . On the interior of the Normal table we cannot find .4800 exactly. If you look at the 2.0row you will find

0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09

2.00.47720.47780.47830.47880.47930.47980.48030.48080.48120.4817

This means that the best we could do are or . 2.05 is a little better than 2.06 and something like 2.054 might be even better. Any of these are acceptable. I will use .Note that the first line of the table reads as below.

0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09

0.00.00000.00400.00800.01200.01600.01990.02390.02790.03190.0359

This means that . It does not give us any information about .

This is, of course, gigantic compared to the standard deviation that we got before. Our results are now not significantly different from almost anything. We had 7.92538 and7.94615

Solution 1: or 2.22 to 13.62

Solution 2: or 2.25 to 13.64

5. a. Use the data above to compute a 98% confidence interval for the population standard deviation.

b. Assume that you got the sample standard deviation that you got above from a sample of 45, repeat a.

c. Fool around with the method for getting a confidence interval for a median and try to come close to a 99% confidence interval for the median. This was one everyone blew because they didn’t read the question! However, one person tried to use the formula for the confidence interval for the mean on every other parameter. Though that can work in some unusual cases, it is a bad idea.

Solution: a. . The problem says that and . From the outline
(the supplement pg 1 or Table 3), . We use 26.2170 and 3.5706 .

Solution 1:and .The formula becomes or . If we take square roots, we get .

b. We will repeat a) with . Now From the supplement pg 2 (or Table 3), the formula for large samples is . Since the table has no values for 44 degrees of freedom, we must use the large sample formula. We use and .

Solution 1:. The formula becomes or or .

c. We are now trying to find an an approximate 99% confidence interval for the median.

The numbers in order are

Solution 2:

7.80 7.81 7.89 7.89 7.91 7.92 7.94 7.95 7.98 7.99 8.008.058.17

It says on the outline that, if we use the numbers from the end,. We want to be 1% or lower which means . There are two ways to do this. If we take the easy way out and use a Normal approximation This seems to be telling us to use the numbers that are 2nd from each end or 7.81 and 8.05. (To be conservative, round the result down.) This really looks sloppy, so the next solution is preferred.

To be more precise, use the Binomial table with . Possible intervals are to , to etc.The part of the table for is shown below. We need only look at the last column.

13 0 0.87752 0.51334 0.25419 0.12091 0.05498 0.02376 0.00969 0.00370 0.00131 0.00042 0.00012

1 0.99275 0.86458 0.62134 0.39828 0.23365 0.12671 0.06367 0.02958 0.01263 0.00490 0.00171

2 0.99973 0.97549 0.86612 0.69196 0.50165 0.33260 0.20248 0.11319 0.05790 0.02691 0.01123

3 0.99999 0.99690 0.96584 0.88200 0.74732 0.58425 0.42061 0.27827 0.16858 0.09292 0.04614

4 1.00000 0.99971 0.99354 0.96584 0.90087 0.79396 0.65431 0.50050 0.35304 0.22795 0.13342

5 1.00000 0.99998 0.99908 0.99247 0.96996 0.91979 0.83460 0.71589 0.57440 0.42681 0.29053

6 1.00000 1.00000 0.99990 0.99873 0.99300 0.97571 0.93762 0.87053 0.77116 0.64374 0.50000

7 1.00000 1.00000 0.99999 0.99984 0.99875 0.99435 0.98178 0.95380 0.90233 0.82123 0.70947

8 1.00000 1.00000 1.00000 0.99998 0.99983 0.99901 0.99597 0.98743 0.96792 0.93015 0.86658

9 1.00000 1.00000 1.00000 1.00000 0.99998 0.99987 0.99935 0.99749 0.99221 0.97966 0.95386

10 1.00000 1.00000 1.00000 1.00000 1.00000 0.99999 0.99993 0.99965 0.99868 0.99586 0.98877

11 1.00000 1.00000 1.00000 1.00000 1.00000 1.00000 0.99999 0.99997 0.99986 0.99948 0.99829

12 1.00000 1.00000 1.00000 1.00000 1.00000 1.00000 1.00000 1.00000 0.99999 0.99997 0.99988

13 1.00000 1.00000 1.00000 1.00000 1.00000 1.00000 1.00000 1.00000 1.00000 1.00000 1.00000

Solution 2:

7.80 7.81 7.89 7.89 7.91 7.92 7.94 7.95 7.98 7.99 8.00 8.05 8.17

Interval Significance

to or 7.80 to 8.171 .9998

to or 7.81 to 8.052 .9966

to or 7.89 to 8.003 .9775

to or 7.89 to 7.994 .9077

If we are being conservative we will take the smallest interval with a significance level above 99%. This will be as before 7.81 to 8.05.

Extra Credit Minitab Problem

5. Check some numbers in the Normal, t, Chi-Squared or F tables using the new set of Minitab routines that I have prepared. To use the new set of routines, follow the instructions in Areadoc1. There are several things that you can do. For the Normal distribution use the computer to check the answers to Examples 6.1-6.4 on pp 198-200 in the text. For the t-table pick a number of degrees of freedom and show that for that number of degrees of freedom, the probability above, say, is 20%. You can do the same for the F and chi-squared tables in your book of tables. A good answer will explain what you did and contain the command dialog and graphs.

Results: I looked at the tables and found , , , , and. For the numbers with .10 as a subscript, I checked that the probability above them was .10, for the numbers with .90 as a subscript, I checked that the probability below them was .10.

————— 9/19/2005 5:33:43 PM ————————————————————

Welcome to Minitab, press F1 for help.

MTB > WOpen "C:\Documents and Settings\rbove\My Documents\Minitab\notmuch.MTW".

Retrieving worksheet from file: 'C:\Documents and Settings\rbove\My

Documents\Minitab\notmuch.MTW'

Worksheet was saved on Thu Apr 14 2005

Results for: notmuch.MTW

MTB > %tarea6a

Executing from file: tarea6a.MAC

Graphic display of t curve areas

Finds and displays areas to the left or right of a given value

or between two values. (This macro uses C100-C116 and K100-K120)

Enter the degrees of freedom.

DATA> 10

Do you want the area to the left of a value? (Y or N)

Do you want the area to the right of a value? (Y or N)

Enter the value for which you want the area to the right.

DATA> 1.372

...working...

t Curve Area

Data Display

mode 0

median 0

MTB > %normarea6a

Executing from file: normarea6a.MAC

Graphic display of normal curve areas

Finds and displays areas to the left or right of a given value

or between two values. (This macro uses C100-C116 and K100-K116)

Enter the mean and standard deviation of the normal curve.

DATA> 0

DATA> 1

Do you want the area to the left of a value? (Y or N)

Do you want the area to the right of a value? (Y or N)

Enter the value for which you want the area to the right.

DATA> 1.282

...working...

Normal Curve Area

MTB > %chiarea6a

Executing from file: chiarea6a.MAC

Graphic display of chi square curve areas

Finds and displays areas to the left or right of a given value

or between two values. (This macro uses C100-C116 and K100-K120)

Enter the degrees of freedom.

DATA> 10

Do you want the area to the left of a value? (Y or N)

Do you want the area to the right of a value? (Y or N)

Enter the value for which you want the area to the right.

DATA> 15.9872

...working...

ChiSquare Curve Area

Data Display

std_dev 4.47214

mode 8.00000

median 9.33333

MTB > %chiarea6a

Executing from file: chiarea6a.MAC

Graphic display of chi square curve areas

Finds and displays areas to the left or right of a given value

or between two values. (This macro uses C100-C116 and K100-K120)

Enter the degrees of freedom.

DATA> 10

Do you want the area to the left of a value? (Y or N)

Please answer Yes or No.

Enter the value for which you want the area to the left.

DATA> 4.8650

...working...

Chi Squared Curve Area

Data Display

std_dev 4.47214

mode 8.00000

median 9.33333

MTB > %farea6a

Executing from file: farea6a.MAC

Graphic display of F curve areas

Finds and displays areas to the left or right of a given value

or between two values. (This macro uses C100-C116 and K100-K120)

Enter the degrees of freedom.DF2 must be above 4.

DATA> 10

Do you want the area to the left of a value? (Y or N)

Do you want the area to the right of a value? (Y or N)

Enter the value for which you want the area to the right.

DATA> 2.32

...working...

F Curve Area

Data Display

mode 0.818182

std dev 0.968246

MTB > %farea6a

Executing from file: farea6a.MAC

Graphic display of F curve areas

Finds and displays areas to the left or right of a given value

or between two values. (This macro uses C100-C116 and K100-K120)

Enter the degrees of freedom.DF2 must be above 4.

DATA> 10

Do you want the area to the left of a value? (Y or N)

Enter the value for which you want the area to the left.

DATA> .431

...working...

F Curve Area

Data Display

mode 0.818182

std dev 0.968246