252solngr1-072 10/02/07 (Open in ‘Print Layout’ format.)
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 z .02 (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,
t .20 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.
252solngr1-072 10/02/07
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?
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.
1
2
3
4
5
6
7
8
9
10
11
12
13
x12
x1
index
x 22
x2
7.81 60.9961
7.92 62.7264
7.94 63.0436
8.00 64.0000
7.95 63.2025
7.91 62.5681
7.98 63.6804
8.05 64.8025
8.17 66.7489
7.80 60.8400
7.80 60.8400
7.80 60.8400
7.90 62.4100
103.03 816.6985
7.81 60.9961
7.92 62.7264
7.94 63.0436
8.00 64.0000
7.95 63.2025
7.91 62.5681
7.98 63.6804
8.05 64.8025
8.17 66.7489
7.80 60.8400
7.89 62.2521
7.89 62.2521
7.99 63.8401
103.30 820.9528
Computation of Sample Variances using the computational formula.
x12  816 .6985 ,
n1  n 2  13 ,
x1  103.03 ,
x 2  103.30 and

The means are x1 
s12 
s 22 

x12
 nx12
n 1
x
2
2
 nx 22
n 1
x
1
n



103 .03
 7.92538 and x 2 
13
x
n
2

x
2
2
 820 .9528 .
103 .30
 7.94615
13

0.146123
816 .6985  137.92538 

 0.0121769 s1  0.0121769  0.1103
12
12

820 .9528  137.94615 2 0.115108

 0.00959231 s 2  0.00959231  0.09794
12
12
2
Computation of Standard Errors
s x1 
s1
s x2 
s2

s12
0.0121769

 0.000936685  0.03060
n
13

s 22
0.00959231

 0.00073787  0.02716
n
13
n
n
Finding t
The significance level is given as 90%. n1  n 2  13
1    .90   .10   .05
t n1  t 12  1.782
2

2
.05
2
252solngr1-072 10/02/07
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 x1  7.92538 and x 2  7.94615
1  x1  t n1 s x1  7.92538 1.7820.03060  7.925  0.055 or 7.870 to 7.980.
2
 2  x 2  t n1 s x2  7.94615 1.7820.02716  7.946  0.048 or 7.898 to 7.994.
2
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 n1  n 2  13 , x1  7.92538, x 2  7.94615, s12  0.0121769 ,
s1  0.0121769  0.1103 , s 22  0.00959231 and s 2  0.00959231  0.09794
Solution 1: If N  100 , the sample of 13 is more than 5% of the population, so use a finite population
correction. Recall that x1  54 .8182 . Now s x1 
.
s1
n
N n

N 1
s12  N  n 


n  N 1 
0.0121769  100  13 

  .0082315  0.02869 . You may have found that the s x1  0.03060 in
13
 100  1 
 100  13 

  .8788  0.9374 and that the size of the confidence interval is
 100  1 
12
reduced by the same factor. t n1  t .05
 1.782 . The interval becomes
problem 1 is multiplied by
2
1  x1  t n1 s x1  7.92538 1.7820.02869  7.925  0.051 or 7.874 to 7.976. If you did not find that
2
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 N  100 , the sample of 13 is more than 5% of the population, so use a finite population
s 22
0.00959231

 0.00073787  0.02716 .
n
13
n
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  2  x 2  t n1 s x2  7.94615 1.7820.02546  7.946  0.045 or 7.901 to 7.991. The interval is
correction. Recall that previously we had s x2 
s2

2
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 z .02 (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?
3
252solngr1-072 10/02/07
Solution: Make a diagram! The diagram for z will be a Normal curve centered at zero and will show one
point, z .02 , 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 z .02 .
From the diagram, we want one point z .02 so that Pz  z.02   .0200 or P0  z  z.02   .4800 . On the
interior of the Normal table we cannot find .4800 exactly. If you look at the 2.0 row you will find
z
2.0
0.00
0.4772
0.01
0.4778
0.02
0.4783
0.03
0.4788
0.04
0.4793
0.05
0.4798
0.06
0.4803
0.07
0.4808
0.08
0.4812
0.09
0.4817
0.0
0.0000
0.0040
0.0080
0.0120
0.0160
0.0199
0.0239
0.0279
0.0319
0.0359
This means that the best we could do are P0  z  2.05   .4798 or P0  z  2.06   .4803 . 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
z.02  2.054 . 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
z
This means that P0  z  0.02   .0080 . It does not give us any information about z .02 .

10 2
 7.6923  2.7735 . This is, of course, gigantic compared to the standard
13
n
13
deviation that we got before. Our results are now not significantly different from almost anything. We had
x 

10

x1  7.92538 and x 2  7.94615
Solution 1: 1  x1  z x1  7.92538 2.0542.7735  7.925  5.697 or 2.22 to 13.62
Solution 2:  2  x2  z x2  7.94615 2.0542.7735  7.946  5.697 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. n1  n 2  13 . The problem says that   .02 and
(the supplement pg 1 or Table 3),
n1
 21
2
n  1s

2
2
2 
n  1s

2

 .29912
 3.5706
2
1
2


2
 .01 . From the outline
n1
. We use  2 
  .20112  26.2170 and
2
2
.
Solution 1: s12  0.0121769 and s1  0.0121769  0.1103 .The formula becomes
12 0.0121769
 2 
26 .2170
0.07463    0.6397 .
12 0.0121769
3.5706
or 0.005570   2  0.4092 . If we take square roots, we get
b. We will repeat a) with n  45 . Now DF  n  1  44 . From the supplement pg 2 (or Table 3), the
formula for large samples is
s 2 DF
z   2 DF
2
 
s 2 DF
 z   2 DF
. Since the  2 table has no values for 44
2
degrees of freedom, we must use the large sample formula. We use z.01  2.327 and
2 DF  2(44 )  88  9.3808 .
4
252solngr1-072 10/02/07
Solution 1: s1  0.0121769  0.1103 . The formula becomes
0.1103  9.3808
2.327  9.3808
 
0.1103  9.3808
 2.327  9.3808
or
1.0347
1.0347
 
or 0.0884    0.1470 .
11 .7078
7.0538
c. We are now trying to find an an approximate 99% confidence interval for the median.
The numbers in order are
Solution 2: x1
x6
x 7 x8 x 9 x10 x11 x12
x 2 x3 x 4 x5
x13
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
It says on the outline that, if we use the k th numbers from the end,   2Px  k  1 . We want  to be 1%
or lower which means Px  k  1  .005 . There are two ways to do this. If we take the easy way out and
n  1  z .2 n
13  1  2.576 13 14  9.2879

 2.3560 . This seems to
2
2
2
be telling us to use the numbers that are 2nd from each end or x 2  7.81 and x12  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 n  13 . Possible intervals are x1 to x13 , x 2 to
x12 etc. The part of the table for n  13 is shown below. We need only look at the last  p  .5 column.
use a Normal approximation k 
13
0
1
2
3
4
5
6
7
8
9
10
11
12
13
0.87752
0.99275
0.99973
0.99999
1.00000
1.00000
1.00000
1.00000
1.00000
1.00000
1.00000
1.00000
1.00000
1.00000
Solution 2:
x1
0.51334
0.86458
0.97549
0.99690
0.99971
0.99998
1.00000
1.00000
1.00000
1.00000
1.00000
1.00000
1.00000
1.00000
x2
0.25419
0.62134
0.86612
0.96584
0.99354
0.99908
0.99990
0.99999
1.00000
1.00000
1.00000
1.00000
1.00000
1.00000
x3
0.12091
0.39828
0.69196
0.88200
0.96584
0.99247
0.99873
0.99984
0.99998
1.00000
1.00000
1.00000
1.00000
1.00000
x4

0.05498
0.23365
0.50165
0.74732
0.90087
0.96996
0.99300
0.99875
0.99983
0.99998
1.00000
1.00000
1.00000
1.00000
x5
x6
0.02376
0.12671
0.33260
0.58425
0.79396
0.91979
0.97571
0.99435
0.99901
0.99987
0.99999
1.00000
1.00000
1.00000
x7
0.00969
0.06367
0.20248
0.42061
0.65431
0.83460
0.93762
0.98178
0.99597
0.99935
0.99993
0.99999
1.00000
1.00000
x8
0.00370
0.02958
0.11319
0.27827
0.50050
0.71589
0.87053
0.95380
0.98743
0.99749
0.99965
0.99997
1.00000
1.00000
x9
x10
0.00131
0.01263
0.05790
0.16858
0.35304
0.57440
0.77116
0.90233
0.96792
0.99221
0.99868
0.99986
0.99999
1.00000
0.00042
0.00490
0.02691
0.09292
0.22795
0.42681
0.64374
0.82123
0.93015
0.97966
0.99586
0.99948
0.99997
1.00000
x11 x12
0.00012
0.00171
0.01123
0.04614
0.13342
0.29053
0.50000
0.70947
0.86658
0.95386
0.98877
0.99829
0.99988
1.00000
x13
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
k
x1 to x13 or 7.80 to 8.17 1
x 2 to x12 or 7.81 to 8.05 2
  2Px  k  1
Significance  1  
2Px  0  2.00012   .00024
.9998
2Px  1  2.00171   .00342
.9966
x3 to x11 or 7.89 to 8.00 3
.9775
2Px  2  2.01123   .02246
.9077
2Px  3  2.04614   .09228
x 4 to x10 or 7.89 to 7.99 4
If we are being conservative we will take the smallest interval with a significance level above 99%. This
will be as before x 2  7.81 to x12  8.05.
5
252solngr1-072 10/02/07
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 6.16.4 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, t .20 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.
10
10
10
Results: I looked at the tables and found t.10
 1.372 , z.10  1.282 ,  2 .10  15.9872 ,  2.90  4.8650 ,
10,10  2.32 and F 10,10  1
F.10
 0.431 . For the numbers with .10 as a subscript, I checked that the
.90
2.32
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)
n
Do you want the area to the right of a value? (Y or N)
y
Enter the value for which you want the area to the right.
DATA> 1.372
...working...
t Curve Area
6
252solngr1-072 10/02/07
Data Display
mode
median
0
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)
n
Do you want the area to the right of a value? (Y or N)
y
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)
n
Do you want the area to the right of a value? (Y or N)
y
Enter the value for which you want the area to the right.
DATA> 15.9872
7
252solngr1-072 10/02/07
...working...
ChiSquare Curve Area
Data Display
std_dev
mode
median
4.47214
8.00000
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)
l
Please answer Yes or No.
y
Enter the value for which you want the area to the left.
DATA> 4.8650
...working...
Chi Squared Curve Area
Data Display
std_dev
mode
median
4.47214
8.00000
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
8
252solngr1-072 10/02/07
or between two values. (This macro uses C100-C116 and K100-K120)
Enter the degrees of freedom.DF2 must be above 4.
DATA> 10
DATA> 10
Do you want the area to the left of a value? (Y or N)
n
Do you want the area to the right of a value? (Y or N)
y
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
DATA> 10
Do you want the area to the left of a value? (Y or N)
y
Enter the value for which you want the area to the left.
DATA> .431
9
252solngr1-072 10/02/07
...working...
F Curve Area
Data Display
mode
0.818182
std dev
0.968246
10