252solngr1-071 2/13/07 (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 random sample of 11 young adult men was asked how many minutes of sports they
watched daily. Results are below.
50
48
65
74
66
30
40
60
60
60
50
Personalize the data as follows: add the digits of your student number to the last six numbers. Example: Ima
Badrisk has the student number 123456; so the last six numbers become {31, 42, 63, 64, 65, 66}.
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 99% confidence interval for the
mean. Does the mean from your interval differ significantly from 50 minutes? Why? What about 40
minutes?
2. If there are 0.9 million young adult men in the metropolitan area , do a confidence interval for the total
amount of time they watch in a day. (See problem 8.50.) It might help to convert your result to hours.
3. Show how would these results change if the 11 individuals were a random sample taken from a dormitory
with 100 male residents?
4. Assume that the population standard deviation is 10 (and that the sample of 11 is taken from a very large
population). Find z .0075 using the Normal table (If you have several values of z that you can use, pick the
average of the extreme ones.) and use it to compute a 98.5% confidence interval. Does the mean differ
significantly from 50 now? From 40? Why?
Extra Credit Problems
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. 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 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, t .20 is 20%. You can do the same for the F and chisquared tables in your book of tables. A good answer will explain what you did and contain the command
dialog and graphs.
252solngr1-071 2/8/07
1. (Keller & Warrack) A random sample of 11 young adult men was asked how many minutes of sports they
watched daily. Results are below.
50
48
65
74
66
30
40
60
60
60
50
Personalize the data as follows: add the digits of your student number to the last six numbers. Example: Ima
Badrisk has the student number 123456; so the last six numbers become {31, 42, 63, 64, 65, 56}.
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 99% confidence interval for the
mean. Does the mean from your interval differ significantly from 50 minutes? Why? What about 40
minutes?
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.
x12
x1
index
1
2
3
4
5
6
7
8
9
10
11
50
48
65
74
66
30
40
60
60
60
50
603
x 22
x2
2500
2304
4225
5476
4356
900
1600
3600
3600
3600
2500
34661
50
48
65
74
66
39
49
69
69
69
59
657
2500
2304
4225
5476
4356
1521
2401
4761
4761
4761
3481
40547
From Table 3   x  tn 1 s x is the formula for a two sided confidence interval when the population
2
standard deviation is unknown.
x
n1  n 2  15 ,
1
The means are x1 
s12 
s x1 
s 22 
s x2 

x12
s1

 nx12
n 1
2
2
 nx 22
n 1
s2
x
1
n

x
2
1
 34661 ,
x
2
 657 and
603
 54 .8182 and x 2 
11
34661  1154 .8182 
10
2

x
n
2
x

2
2
 40547 .
657
 59 .7273 .
11
1605 .61
 160 .56 s1  160 .56  12 .671
10
s12
160 .56

 14 .5965  3.821
n
11
n
x

 603 ,

40547  1159 .7273 2 1306 .15

 130 .61 s 2  130 .61  11 .429
10
10
s 22
130 .61

 11 .874  3.446
n
11

n
1    .99   .01

2
 .005
10
t n1  t.005
 3.169
2
1  x1  t n1 s x1  54.8182  3.1693.821  54.818  12.107 or 42.711 to 66.925.
2
 2  x 2  t n1 s x2  59.7173 3.1693.446  59.717  10.920 or 48.797 to 70.637.
2
Since 50 minutes is included in these intervals and 40 is not, our results are significantly different from 50
but not 40.
2
252solngr1-071 2/8/07
2. If there are 0.9 million young adult men in the metropolitan area , do a confidence interval for the total
amount of time they watch in a day. (See problem 8.50.) It might help to convert your result to hours.
Solution 1: 42.711 to 66.925 when multiplied by 0.9 million gives us 38.440 to 60.233 million minutes or
0.641 to 1.004 million hours.
Solution 2: 48.797 to 70.637 when multiplied by 0.9 million gives us 43.917 to 63.573 million minutes or
0.732 to 1.060 million hours.
3. Show how would these results change if the 11 individuals were a random sample taken from a dormitory
with 100 male residents?
If you interpreted this as a re-do of Problem 2, multiply solution below by 0.9 million.
Solution 1: If N  100 , the sample of 11 is more than 5% of the population, so use a finite population
correction. Recall that x1  54 .8182 and s x1 
s1
n
 3.821 . Now s x1 
s1
n
N n
100  11
 3.821
N 1
100  1
 3.821 0.8990  3.8210.94815   3.623 .
10
Recall that   .01 ,  2  .005 , t n1  t .005
 3.169 and that   x  tn 1 s x is the formula for a two
2
2
sided interval. 1  x1  t n1 s x1  54.8182  3.1693.623  54.818  11.481 or 43.337 to 66.299. The
2
interval is smaller, but it doesn’t change anything – the mean is still significantly different from 40 (but not
50).
Solution 2: If N  100 , the sample of 11 is more than 5% of the population, so use a finite population
correction. Recall that x 2  59 .7273 and s x2 
s2
 3.446 . Now
n
N n
100  11
 3.446 .
 3.446 0.8990  3.446 0.94815   3.267 .
N

1
100  1
n
10
Recall that   .01 ,  2  .005 , t n1  t .005
 3.169 and that   x  tn 1 s x is the formula for a two
s x2 
s2
2
2
sided interval.  2  x 2  t n1 s x2  59.7273 3.1693.267  59.7273 10.348 or 43.3390 to 70.075. The
2
interval is smaller, but it doesn’t change anything – the mean is still significantly different from 40 (but not
50).
4. Assume that the population standard deviation is 10 (and that the sample of 11 is taken from a very large
population). Find z .0075 using the Normal table (If you have several values of z that you can use, pick the
average of the extreme ones.) and use it to compute a 98.5% confidence interval. Does the mean differ
significantly from 50 now? From 40? Why?
Solution: Make a diagram! The diagram for z will be a Normal curve centered at zero and will show one
point, z .0075 , which has 0.75% above it (and 99.25% below it!) and is above zero because zero has 50%
below it. Since zero has 50% above it, the diagram will show 49.25% between zero and z .0005 .
From the diagram, we want one point z .0025 so that Pz  z .0075   .0075 or P0  z  z .0025   .4925 .
On the interior of the Normal table we can find to .4925 exactly. In fact, it says P0  z  z 0   .4925 for
2.43. This means that we will say z .0075  2.43 .
If the confidence level is 1    .9850   .0150 and
x 


10


2
 .0075 .   10 , so
10 2
 9.0909  3.0151 .
11
n
11
Solution 1: 1  x1  z x1  54.8182  2.433.0151  54.82  7.33 or 47.50 to 62.15
Solution 2:  2  x 2  z x2  59.7273 2.433.0151  59.73  7.33 or 52.40 to 67.06
3
252solngr1-071 2/8/07
First Extra Credit Problem
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.
Solution: a. The problem says that   .02 and
n  1s

2
2
2
2 
n  1s

2
1

2

2
 .01 . From the supplement pg 1 (or Table 3),
n1
n 1
10
10
. We use  2 
  .01
 23.2093 and  21   .99
 2.5582 .
2
2
2
Solution 1: s12  160 .56 s1  160 .56  12 .671 The formula becomes
10 160 .56   2  10 160 .56
23 .2093
2.5582
or
69 .1792   2  627 .6288 . If we take square roots, we get 8.3174    25.0525 .
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 .
Solution 1: s1  160 .56  12 .671 . The formula becomes
12 .671  9.3808
2.327  9.3808
 
12 .671  9.3808
or
 2.327  9.3808
118 .8641
118 .8641
 
or 10.15    16.8511 .
11 .7078
7.0538
c. We 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.
The numbers in order are
x3
x5
x6
x7
x8
x9
x10
Solution 1:
x1
x11
x2
x4
30
40
48
50
50
60
60
60
65
66
74
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
11  1  2.576 11 12  8.5436


 1.7282 . This seems to
2
2
2
be telling us to use the numbers that are 1st from each end or x1  52 and x11  74. (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  15 . Possible intervals are x1 to x11 , x 2 to
x10 etc. Let’s try a few intervals.
use a Normal approximation k 
4
252solngr1-071 2/8/07
The binomial table for n  11 is copied here. We need only look at the last  p  .5 column.
11
0
1
2
3
4
5
6
7
8
9
10
11
0.89534
0.99482
0.99984
1.00000
1.00000
1.00000
1.00000
1.00000
1.00000
1.00000
1.00000
1.00000
0.56880
0.89811
0.98476
0.99845
0.99989
0.99999
1.00000
1.00000
1.00000
1.00000
1.00000
1.00000
0.31381
0.69736
0.91044
0.98147
0.99725
0.99970
0.99998
1.00000
1.00000
1.00000
1.00000
1.00000
0.16734
0.49219
0.77881
0.93056
0.98411
0.99734
0.99968
0.99997
1.00000
1.00000
1.00000
1.00000
0.08590
0.32212
0.61740
0.83886
0.94959
0.98835
0.99803
0.99976
0.99998
1.00000
1.00000
1.00000
0.04224
0.19710
0.45520
0.71330
0.88537
0.96567
0.99244
0.99881
0.99987
0.99999
1.00000
1.00000
0.01977
0.11299
0.31274
0.56956
0.78970
0.92178
0.97838
0.99571
0.99942
0.99995
1.00000
1.00000
0.00875
0.06058
0.20013
0.42555
0.66831
0.85132
0.94986
0.98776
0.99796
0.99979
0.99999
1.00000
0.00363
0.03023
0.11892
0.29628
0.53277
0.75350
0.90065
0.97072
0.99408
0.99927
0.99996
1.00000
0.00139
0.01393
0.06522
0.19112
0.39714
0.63312
0.82620
0.93904
0.98520
0.99779
0.99985
1.00000
0.00049
0.00586
0.03271
0.11328
0.27441
0.50000
0.72559
0.88672
0.96729
0.99414
0.99951
1.00000
  2Px  k  1
Interval
k
x1 to x11 or 30 to 74
x 2 to x10 or 40 to 66
1
2Px  0  2.00049   .00098
2
2Px  1  2.00586   .01172
x3 to x9 or 48 to 65
3
2Px  2  2.03271   .06542
4
2Px  3  2.11328   .22656
x 4 to x8 or 50 to 60
Notice that we could have answered the question by finding the largest value of k with Px  k  1  .005 .
Since the smallest interval with a significance level below 1% is 30 to 74, this is the best that we can do.
We can check our results using the Normal distribution. The outline says, using a continuity correction,


k  1  1 2  np 

k  .5  .5n 

 P x  k  1  1 2   P z 
 P z 



2
npq
.
5
n




.

1  1  .5  11.5  
0.5  5.5 

k  1 Px  0.5  P  z  
 Pz  3.02   .5  .4987  .0013
  P z 
1.65831 






11
.
5
.
5







 2  1  .5  11.5  
1.5  5.5 

k  2 Px  1.5  P  z  
 Pz  2.41  .5  .4920  .0080
  Pz 
1.65831 


11.5.5  


Since we need Px  k  1  .005 , k  1 was correct.
5
252solngr1-071 2/8/07
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
252solngr1-071 2/8/07
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
252solngr1-071 2/8/07
...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
252solngr1-071 2/8/07
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
252solngr1-071 2/8/07
...working...
F Curve Area
Data Display
mode
0.818182
std dev
0.968246
10