252solngr1 1/24/08 (Open this document in 'Page Layout' view!)
Name:
Class days and time:
Student number:
Please include this on what you hand in!
Graded Assignment 1
Please show your work! Write on one side of your paper. Neatness and whether the papers are stapled may affect your grade.
1. (Duane and Seward) A random sample of rents (to the nearest dollar amount) paid by students living off campus gives the data
below. We assume that rents are approximately Normally distributed.
910
820
780
870
860
800
820
810
900
730
920
650
Personalize the data as follows: add the third-to-last digit of your student number to the third-to-last rent above; add the
second to last digit to the second-to-last number above, change the last digit the same way. Make sure that your student number is
clearly visible on your paper.
Example: Ima Badrisk has student number 123456; so the sample of rents that she uses is below.
910
820
780
870
860
800
820
810
900
734
925
656
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 rent. The Off-Campus Housing Office says
that the average amount spent on rents is $870/month. Using the concept of significant difference, does it appear that the office is
stating the wrong rent? Why? (This was not an opinion question! If you don’t know what a significant difference is, find out!) Is the
mean significantly different from $880?
2. The university wants to publish a document showing its economic impact on the community. If university students rent 965 units,
do a confidence interval for the total amount of rent being paid. (See problem 8.50 in the text.)
3. Show how your results in 1) would change if the 12 rents were a random sample chosen from a population of only 100 rentals.
4. Assume that the population standard deviation is 77 (and that the sample of 12 is taken from a very large population). Find z .04
(the 96th percentile of z ) using the Normal table and use it and the mean that you found in 1) to compute a 92% confidence interval
for the average rent. Does the mean differ significantly from $870 ounces now? From $860? Why?
Extra Credit: (You have to do extra credit now – not after you receive an unacceptable final grade.)
5. a. Use the data above to compute a 90% confidence interval for the population standard deviation.
b. Assume that you got the sample standard deviation that you got above from a sample of 50, repeat a.
c. Fool around with the method for getting a confidence interval for a median and try to come close to a
90% 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.
1. (Duane and Seward) A random sample of rents (to the nearest dollar amount) paid by students living off
campus gives the data below. We assume that rents are approximately Normally distributed.
910
820
780
870
860
800
820
810
900
730
920
650
Personalize the data as follows: add the third-to-last digit of your student number to the third-tolast rent above; add the second to last digit to the second-to-last number above, change the last digit the
same way. Make sure that your student number is clearly visible on your paper.
Example: Ima Badrisk has student number 123456; so the sample of rents that she uses is below.
910
820
780
870
860
800
820
810
900
734
925
656
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
252solngr1 1/24/08 (Open this document in 'Page Layout' view!)
mean rent. The Off-Campus Housing Office says that the average amount spent on rents is $870/month.
Using the concept of significant difference, does it appear that the office is stating the wrong rent? Why?
(This was not an opinion question! If you don’t know what a significant difference is, find out!) Is the mean
significantly different from $860?
Solution: Three of the many possible solutions are shown.
Row
1
2
3
4
5
6
7
8
9
10
11
12
x12
x1
x 22
x2
910 828100
820 672400
780 608400
870 756900
860 739600
800 640000
820 672400
810 656100
900 810000
730 532900
920 846400
650 422500
9870 8185700
910 828100
820 672400
780 608400
870 756900
860 739600
800 640000
820 672400
810 656100
900 810000
734 538756
925 855625
656 430336
9885 8208617
x 32
x3
910 828100
820 672400
780 608400
870 756900
860 739600
800 640000
820 672400
810 656100
900 810000
739 546121
929 863041
659 434281
9897 8227343
Computation of Sample Variances using the computational formula.
n1  n 2  n3  12 ,
and
x
x
x
2
1
 8185700 ,
x
2
 9885 ,
x
2
2
 8208617 ,
x
3
 9897
 8227343 .
2
3
The means are x1 
x3 
 9870 ,
1
x

2
1
 nx12
s12
x

s 22
x

2
2
x
2
3
1
n

9870
 822 .50 , x 2 
12
x
n
2

9885
 823 .75 and
12
9897
 824 .75
12
3
n
x
n 1
 nx 22
n 1

67625 .00
8185700  12 822 .50 2

 6147 .73 s1  6147 .73  78 .407
11
11

65848 .25
8208617  12 823 .75 2

 5986 .20 s 2  5986 .20  77 .371
11
11
64792 .25
8227343  12 824 .75 2

 5890 .20 s 3  5890 .20  76 .748
11
n 1
11
If you used another formula, you should lose some credit but get the same answer as I did.
s 32 
 nx32

Computation of Standard Errors. We will use the formula   x  tn1s x , where s x  s
2
s x1 
s1
s x3 
s3

s
s12
6147 .73

 512 .31  22 .634 s x2  2 
n
12
n

s 32
5890 .20

 490 .85  22 .155
n
12
n
n
.
s 22
5986 .20

 498 .85  22 .335
n
12
n
Finding t
The significance level is given as 90%. n1  n 2  n3  12 . 1    .90   .10  2  .05
t n1  t 11  1.796 {ttable}. Please repeat this rhyme after me “Sigma   means

2
.05
z and s
means use t (unless you have a very high freedom degree).”
2
252solngr1 1/24/08 (Open this document in 'Page Layout' view!)
Putting it all together
Remember x1  822 .50 , x 2  823 .75 and x 3  824 .75 . The formula for a 2-sided interval is
  x  t n1 s  822.50  1.79622.634  822.50  40.65 or 781.85 to 863.15.
1
1

2
x1
 2  x 2  t n1 s x2  823.75  1.79622.335  823.75  40.11 or 783.64 to 863.86.
2
 3  x3  tn1 s x3  824.75  1.79622.155  824.75  39.79 or 784.96 to 864.54.
2
The Off-Campus Housing Office says that the average amount spent on rents is $870/month. Using the
concept of significant difference it appears that the office is stating the wrong rent? In all of these cases it
appears that the mean is significantly different from 870 since 870 is not on any of the confidence intervals.
The same is not true of 860, since it falls within all of these intervals. Minitab output follows. The
instruction is ‘onet’ followed by the appropriate column name. The instruction is modified by the
subcommand ‘conf 90.’ This indicates a 90% confidence level.
MTB > onet c1;
SUBC> conf 90.
One-Sample T: Rents1
Variable
N
Mean
Rents1
12 822.5
StDev
78.4
SE Mean
22.6
90% CI
(781.9, 863.1)
MTB > onet c2;
SUBC> conf 90.
One-Sample T: Rents2
Variable
N
Mean
Rents2
12 823.8
StDev
77.4
SE Mean
22.3
90% CI
(783.6, 863.9)
MTB > onet c3;
SUBC> conf 90.
One-Sample T: Rents3
Variable
N
Mean
Rents3
12 824.8
StDev
76.7
SE Mean
22.2
90% CI
(785.0, 864.5)
2. The university wants to publish a document showing its economic impact on the community. If university
students rent 965 units, do a confidence interval for the total amount of rent being paid. (See problem 8.50
in the text.)
Solution 1: 781.85 to 863.15 when multiplied by 965 gives us $754,485 to $832,939.
Solution 2: 783.64 to 863.86 when multiplied by 965 gives us $756,212 to $833,625.
Solution 3: 784.96 to 864.54 when multiplied by 965 gives us $757,486 to $834,281.
3. Show how your results in 1) would change if the 12 rents were a random sample chosen from a
population of only 100 rentals.
Solution: Recall that n  n  n  12 t n1  t 11  1.796 , x  822 .50 , x  823 .75 , x  824 .75 ,
1
2

3
2
.05
1
2
3
 6147 .73 ,
 5986 .20 and
 5890 .20 .
Solution 1: If N  100 , the sample of 12 is more than 5% of the population, so use a finite population
s12
s 22
s32
s2  N  n 
6147 .73  100  12 
N n
 1 

  455 .387  21 .34 .. The
 .
12  100  1 
n  N 1 
n N 1
interval becomes 1  x1  t n1 s x1  822.50  1.79621.34  822.50  38.33 or 784.17 to 860.83.
correction. Now s x1 
s1
2
Solution 2: If N  100 , the sample of 12 is more than 5% of the population, so use a finite population
s2  N  n 
5986 .20  100  12 
N n
 1 

  443 .422  21 .06 .. The
 .
12  100  1 
n  N 1 
n N 1
interval becomes  2  x 2  t n1 s x2  823.75  1.79621.06  823.75  37.82 or 785.93 to 861.57.
correction. Now s x1 
s1
2
3
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Solution 3: If N  100 , the sample of 12 is more than 5% of the population, so use a finite population
s2  N  n 
5890 .20  100  12 
N n
 1 

  436 .311  20 .89 .. The
 .
12  100  1 
N

1
n
N

1


n
interval becomes  3  x3  t n1 s x3  824.75  1.79620.89  824.75  37.51 or 787.24 to 862.26.
correction. Now s x1 
s1
2
It would have been easier for me to multiply the original standard errors ( s x1  22.634 , s x2  22.335 and
 100  12 

  .8888  0.9428 , but the results would be the same. It is still true that
 100  1 
the results are significantly different from 870, but not 860.
s x3  22.155 ) by
4. Assume that the population standard deviation is 77 (and that the sample of 12 is taken from a very large
population). Find z .04 (the 96th percentile of z ) using the Normal table and use it and the mean that you
found in 1) to compute a 92% confidence interval for the average rent. Does the mean differ significantly
from $870 ounces now? From $880? Why?
Solution: Recall that n1  n 2  n3  12 , x1  822 .50 , x 2  823 .75 and x 3  824 .75 . Our new data is
  77 . We will use the formula   x  z  x , where  x  
2
n

77

12
77 2
 494 .0833
12
= 22.228.
We must now find z .04 . Make a diagram! The diagram for z will be a Normal curve centered
at zero and will show one point, z .04 , which has 4% above it (and 96% below it!) and is above zero
because zero has 50% below it. Since zero has 50% above it, the diagram will show 46% between zero and
z .04 .
From the diagram, we want one point z .04 so that Pz  z.04   .0400 or P0  z  z.04   .4600 . On the
interior of the Normal table{norm} we cannot find .4600 exactly. If you look at the 1.7 row you will find
0.00
0.01
0.02
0.03
0.04
0.05
0.06
0.07
0.08
0.09
z
1.7
0.4554
0.4564
0.4573
0.4582
0.4591
0.4599
0.4608
0.4616
0.4625
0.4633
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  1.75   .4599 or P0  z  1.76   .4608 . 1.75 is
definitely better than 1.76 and something like 1.751 might be even better. Any of these are acceptable. I will
use z.04  1.75 . 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  0.04   .0160 . It does not give us any information about z .04 .
1  x1  z  2  x  822 .50  1.75 22 .228   822 .50  38.90 or 783.60 to 861.40.
 2  x 2  z  2  x  823 .75  1.75 22.228   823 .75  38.90 or 784.85 to 862.65.
 3  x3  z  2  x  824 .75  1.75 22.227   824 .75  38 .90 or 785.85 to 863.65.
It is still true that the results are significantly different from 870, but not 860.
We really ought to check our value of z .04 . We could try Pz  1.75   Pz  0  P0  z  1.75 
 .5  .4599  .0401  .04 , or we could show that 1.75 is the 96th percentile. Pz  1.75 
 Pz  0  P0  z  1.75   .5  .4599  .9599  .96 .
4
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Extra Credit Problem
(You have to do extra credit now – not after you receive an unacceptable final grade.)
5. a. Use the data above to compute a 90% confidence interval for the population standard deviation.
b. Assume that you got the sample standard deviation that you got above from a sample of 50, repeat a.
c. Fool around with the method for getting a confidence interval for a median and try to come close to a
90% confidence interval for the median.
Solution: a) n1  n 2  n3  12 , s12  6147 .73 , s 22  5986 .20 and s32  5890 .20 . The problem says that
  .10 and

2
 .05 . From the outline
(the supplement pg 1 or Table 3),
n 1
 21
2
n  1s 2

2
2 
n  1s 2

2

 .29511
2
1

n1
. We use  2 
  .20511  19.6751 and
2
2
 4.5748 {chiSq}.
11 6147 .73   2  11 6147 .73
or 3442 .34   2  14782 .07 . If we
19 .6451
4.5748
take square roots, we get 58.67    121 .58 . The other solutions are very similar.
Solution 1: The formula becomes
b. We will repeat a) with n  50 . Now DF  n  1  49 . From the supplement pg 2 (or Table 3), the formula
for large samples is
s 2 DF
z   2 DF
 
s 2 DF
 z   2 DF
2
. Since the  2 table has no values for 49 degrees
2
of freedom, we must use the large sample formula. We use z .05  1.645 {ttable} and
2 DF  2(49 )  98  9.8995 .
Solution 1: s x1  22.634 , s x2  22.335 and s x3  22.155 . The formula becomes
22 .634  9.8995
1.645  9.8995
 
22 .634  9.8995
 1.645  9.8995
or
224 .065
224 .065
 
or 19.4087    17.1446 .
11 .5452
8.2545
c. We are now trying to find an approximate 99% confidence interval for the median. This was one many
people blew because they didn’t read the question! Some people 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. We must put them in order as below.
Rank
1
2
3
4
5
6
7
8
9
10
11
12
x1
x2
650
730
780
800
810
820
820
860
870
900
910
920
656
734
780
800
810
820
820
860
870
900
910
925
x3
659
739
780
800
810
820
820
860
870
900
910
929
5
252solngr1 1/24/08 (Open this document in 'Page Layout' view!)
We could, instead, just list their ranks.
Row
r1
x1
x 2 r2
1
2
3
4
5
6
7
8
9
10
11
12
910
820
780
870
860
800
820
810
900
730
920
650
11.0
6.5
3.0
9.0
8.0
4.0
6.5
5.0
10.0
2.0
12.0
1.0
910
820
780
870
860
800
820
810
900
734
925
656
11.0
6.5
3.0
9.0
8.0
4.0
6.5
5.0
10.0
2.0
12.0
1.0
x3
r3
910
820
780
870
860
800
820
810
900
739
929
659
11.0
6.5
3.0
9.0
8.0
4.0
6.5
5.0
10.0
2.0
12.0
1.0
It says on the outline that, if we use the k th numbers from the end,   2Px  k  1 . We want  to be
10% or lower which means Px  k  1  .05 . There are two ways to do this. If we take the easy way out
and use a Normal approximation, z .05  1.645 so that
n  1  z .2 n
12  1  1.645 12 13  5.69845


 3.6507 . This seems to be telling us to use the
2
2
2
numbers that are 3rd from each end or x1,3  780 and x1,10  900. Actually, these are the same for all three
k
columns shown here. (To be conservative, round the value of k down.) This looks sloppy for such a low
sample size, so the next solution is preferred.
To be more precise, use the Binomial table with n  12 . Possible intervals are x i ,1 to x i ,12 , x i , 2
to x i ,11 etc. (Here i is the column number and the second subscript is the rank of the number in the
column.) The part of the Binomial table for n  12 {bin} is shown below. We need only look at the last
 p  .5 column.
n x
12 0
1
2
3
4
5
6
7
8
9
10
11
12
.01
0.88638
0.99383
0.99979
1.00000
1.00000
1.00000
1.00000
1.00000
1.00000
1.00000
1.00000
1.00000
1.00000
.05
0.54036
0.88164
0.98043
0.99776
0.99982
0.99999
1.00000
1.00000
1.00000
1.00000
1.00000
1.00000
1.00000
.10
0.28243
0.65900
0.88913
0.97436
0.99567
0.99946
0.99995
1.00000
1.00000
1.00000
1.00000
1.00000
1.00000
.15
0.14224
0.44346
0.73582
0.90779
0.97608
0.99536
0.99933
0.99993
0.99999
1.00000
1.00000
1.00000
1.00000
.20
0.06872
0.27488
0.55835
0.79457
0.92744
0.98059
0.99610
0.99942
0.99994
1.00000
1.00000
1.00000
1.00000
.25
0.03168
0.15838
0.39068
0.64878
0.84236
0.94560
0.98575
0.99722
0.99961
0.99996
1.00000
1.00000
1.00000
Let us list the possible intervals for
k
Interval
  2Px  k  1
x i ,1 to x i ,12
1
2Px  0  2.00024   .00048
x i , 2 to x i ,11
2
x i ,3 to x i ,10
3
x i , 4 to x i ,9
4
2Px  1  2.00317   .00634
2Px  2  2.01929   .03858
2Px  3  2.07300   .14600
.30
0.01384
0.08503
0.25282
0.49252
0.72366
0.88215
0.96140
0.99051
0.99831
0.99979
0.99998
1.00000
1.00000
.35
0.00569
0.04244
0.15129
0.34665
0.58335
0.78726
0.91537
0.97449
0.99439
0.99915
0.99992
1.00000
1.00000
.40
0.00218
0.01959
0.08344
0.22534
0.43818
0.66521
0.84179
0.94269
0.98473
0.99719
0.99968
0.99998
1.00000
.45
0.00077
0.00829
0.04214
0.13447
0.30443
0.52693
0.73931
0.88826
0.96443
0.99212
0.99892
0.99993
1.00000
.50
0.00024
0.00317
0.01929
0.07300
0.19385
0.38721
0.61279
0.80615
0.92700
0.98071
0.99683
0.99976
1.00000
Significance  1  
.9995
.9937
.9614
.8540
If we are being conservative we will take the smallest interval with a significance level above 99%. This
will be as before x1,3  780 and x1,10  900 and, again, will be the same for the other two columns.
6
252solngr1 1/24/08 (Open this document in 'Page Layout' view!)
Extra Credit Minitab Problem
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 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 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.
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. This was run in 2005, but there is no reason to believe that the new version of Minitab
will give different results.
————— 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
#This is a dummy worksheet.
MTB > %tarea6a
# It is used to locate %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
#Verify that 1.372 is a 10% value of t.
...working...
t Curve Area
7
252solngr1 1/24/08 (Open this document in 'Page Layout' view!)
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
#Verify that 1.282 is the 10% value of z.
...working...
Normal Curve Area
8
252solngr1 1/24/08 (Open this document in 'Page Layout' view!)
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
#Verify that 15.9872 is a 10% value of chi-squared.
...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)
y
Enter the value for which you want the area to the left.
DATA> 4.8650
#Verify that 4.865 is a 90% value of chi-squared.
...working...
Chi Squared Curve Area
9
252solngr1 1/24/08 (Open this document in 'Page Layout' view!)
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
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
#Verify that 2.32 is a 10% value of F.
...working...
F Curve Area
Data Display
mode
0.818182
std dev
0.968246
10
252solngr1 1/24/08 (Open this document in 'Page Layout' view!)
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
# Verify that 0.431 is a 90% value of F.
# 0.431 is 1 divided by 2.31.
# It does not appear on the F table.
# See material on F distribution for explanation.
...working...
F Curve Area
Data Display
mode
0.818182
std dev
0.968246
11