252solngr1 2/05/04 (Open this document in 'Page Layout' view!)
Please show your work! Neatness and whether the papers are stapled may affect your grade.
1. A manufacturer is concerned that a machine is not adequately filling a 200 oz container. 15 measurements are taken. Results are below.
198 204 189 182 205 195 188 201 200 203 201 193 194 196 190
Compute the sample standard deviation using the computational formula. Use this sample standard deviation to compute a 90% confidence interval for the mean. Does the mean differ significantly from 200 oz? Why?
2. How would your results change if the sample of 15 had been taken from a population of 100?
3. Assume that the population standard deviation is 7.00 (and that the sample of 15 is taken from a very large population). Find z and use it to compute a 92% confidence interval. Does the mean differ
.
04 significantly from 200 oz? Why?
Solution:
1) index x
2 x
1 198 39204
2 204 41616
3 189 35721
4 182 33124
5 205 42025
6 195 38025
7 188 35344
8 201 40401
9 200 40000
10 203 41209
11 201 40401
12 193 37249
13 194 37636
14 196 38416
15 190 36100 sum 2939 576471 s x

 x s x
2 x



 n

2939 x x n

2

2939

,
1
15 n x
2


 x
2 
576471
195 .
9333
576471
623 .
12928


44 .
5093
14
44 .
5093

6 .
67152 .

, n

15
15

195 .
9333

2
14 s x
 s x  s n
2 x 
44 .
5993

2 .
9673 n 15
The formula is in table 20 of the Supplement.

1 .
7226
.
10 x
 t


Confidence interval:




2
 s
2 x

.
05
  x
 t
195 .
9333

 n

1
 t
2
 

2

1
 s
 x t
.
05
is the formula for a two sided interval.

1 .
761


1 .
761
1 .
7227


195 .
9333

3 .
0337 or 192.900 to 198.967. If we ask if the mean is significantly different from 200, our null hypothesis is H
0
:
 
200 and since 200 is not between the top and the bottom of the confidence interval, reject H
0
and say that the mean is significantly different from 200.
If we ask if the mean is significantly different from 198, the null hypothesis is H
0
:
 
198 and 198 is between the top and the bottom of the confidence interval. So we reject H
0
or we can say that the mean is significantly different from 198.

252solngr1 2/05/04 (Open this document in 'Page Layout' view!)
2) If N s
Recall that
Confidence interval:
 x

 x s
 x n x
N
N
 t

 
2

1 s

 n
1

1 .
7226
195 .
9333 x

  x
,


195 .
9333
100
.
10


15


1 .
7226
.
05 and t
0 .
8586
 t
.
05
195 .
9333

1 .
7226

0 .
92660


1 .
5962

1 .
761 .
.

100 t
 

2

1
 s x
,

1
2

1 .
761

1 .
5962



 n

1
2

is the formula for a two sided interval.

2 .
8109 or 193.122 to 198.744. The interval is smaller, but it doesn’t change anything – the mean is still significantly different from 200 but not
198.

100 , the sample of 15 is more than 5% of the population, so use
3) a) Find z
.
04
. Please don’t tell me that because P

0
 z

0 .
04


.
0160 , that z
.
04
is .0160.
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
so that
.
04
 z z .
.
04
From the diagram, we want one point P
 z
.
04


.
04 or
From the interior of the Normal table the closest we can come to .4600 is P

P
0

0
 z
 z
.
04
 z

1 .
75




.
.
4600
4599
.
. This means that
Check: P
 z z
.
04


1 .
75
1 .
75


.
P
 z

0
 
0
 z

1 .
75


.
5

.
4599

.
0401

.
04 .
Normal Curve with Mean 0 and Standard Deviation 1
The Area to the Right of 1.75 is 0.0401
0.4
0.3
0.2
0.1
0.0
-5 -4 -3 -2 -1 0 1 2 3 4
Data Axis b) We know that x

195 .
9333 , n

15 and
 
7 . So
 x

 n

7
15

7
15
2

3 .
2667
=1.8074. The 92% confidence interval has 1
   confidence interval is
  x
 z

2
 x
199.09. If we test the null hypothesis

H
195 .
9333
0
:
 

.
92 or
 
.
08
1 .
75

1 .
8074

200
, so
 z

2
 z
.
04
195 .
93


3 .
16
against the alternative hypothesis
1 .
75
. The
or 192.77 to
H
0
:
 
200 , since 200 is not on the confidence interval, we reject the null hypothesis. The result is not significantly different from 198 because 198 is within the interval.
Extra Credit:
4) a. Use the data above to compute a 90% confidence interval for the population standard deviation.

252solngr1 2/05/04 (Open this document in 'Page Layout' view!)
Solution: From the supplement pg 1, s x

6 .
67152 , n

15 ,
 
.
10 and
 n

1
 s

2
 2

2

2
.
05
 
 n

1
 s
2
 2
. We use


. We know
2
1


2
2
 n

1


2
 
.
05
 s
2 x

44 .
5093 ,
23 .
6848 and
 2

1
1
 
2

26 .
3083

.
95
 
2


6 .
5706 . The formula becomes
44 .
5093
23 .
6848
 
94 .
8361 . If we take square roots, we get 5 .
129
2
 

44 .
5093
6 .
5706

9 .
738
or
b. Assume that you got the sample standard deviation that you got above from a sample of 45, repeat a.
Solution: From the supplement pg 2, z
 s x

6 .
67152
, n

45 ,
 
2 s
.
10 and

2

2 DF
2 DF

.
05 . We use z
.
05 s 2 DF
 z


2 DF

2
1 .
645
and
. We now have
2 DF


6 .
67152

2 ( 44 )
9 .
3808
1 .
645

9 .
3808

 
88



9 .
3808
6 .
67152

1 .
645


. The formula becomes
9 .
3808
or
 
9 .
3808
5 .
681

8 .
090 .
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.
The numbers in order are x
1 x
2 x
3 x
4 x
5 x
6 x
7 x
8 x
9 x
10 x
11 x
12 x
13 x
14 x
15
182 188 189 190 193 194 195 196 198 200 201 201 203 204 205
It says on the outline that
 
2 P
 x
 k

1

. If we check a Binomial table with n find that the first quantity below 5% is
 
2

.
01758 dangerously use


.
03516 k

5 , k

and
1

P
P

190
4 ,
 
 x

3


.
01758

2



201

.
05923



.
1

. So if k

4 ,
.
03516
11846 and
 k

96 .
48 %
P


1

15
3
and and p

.
50 , we
. If you are willing to live
193
  
201


1

.
11846

88 .
15 % . If we use the formula k
 n

1
 z

.
2 n

15

1

1 .
645 15

4 .
81
2 2 it tells us to use the 4 th number from the end, since, if we want to be conservative we round the answer
, down.
How I got these results
‘MTB >’ is the Minitab prompt. The retrieval is done using the ‘file’ pull-down menu and the ‘open worksheet’ command followed by finding where I put the data. Other instructions were typed in the ‘session’ window.
I put the data in column 1 in Minitab and used the ‘Gsummary’ command to get the mean and standard deviation.
Welcome to Minitab, press F1 for help
MTB > GSummary c1;
SUBC> Confidence 90.0.

252solngr1 2/05/04 (Open this document in 'Page Layout' view!)
184 188 192 196
9 0 % Confidence Inter vals
200 204
A nderson-Darling Normality Test
A -Squared
P-V alue
0.23
0.763
M ean
StDev
V ariance
Skew ness
Kurtosis
N
195.93
6.67
44.50
-0.505497
-0.417569
15
M inimum
1st Q uartile
M edian
3rd Q uartile
M aximum
182.00
190.00
196.00
201.00
205.00
90% C onfidence Interv al for Mean
192.90
198.97
90% C onfidence Interv al for Median
192.72
201.00
90% C onfidence Interv al for StDev
5.13
9.74
Mean
Median
192 194 196 198 200 202
MTB > let c2=c1*c1
MTB > Print c1 c2
I computed the square of C2 in c1 and got the sums for computing the variance.
Row C1 C2
1 198 39204
2 204 41616
3 189 35721
4 182 33124
5 205 42025
6 195 38025
7 188 35344
8 201 40401
9 200 40000
10 203 41209
11 201 40401
12 193 37249
13 194 37636
14 196 38416
15 190 36100
MTB > sum c1
Sum of C1 = 2939
MTB > sum c2
Sum of C2 = 576471
MTB > describe c1
Variable N N* Mean SE Mean StDev Minimum Q1 Median Q3

252solngr1 2/05/04 (Open this document in 'Page Layout' view!)
C1 15 0 195.93 1.72 6.67 182.00 190.00 196.00 201.00
Variable Maximum
C1 205.00
MTB > Onet c1; This does a 90% confidence interval and a test for a mean of 200 using s.
SUBC> Test 200;
SUBC> Confidence 90.
Test of mu = 200 vs not = 200
Variable N Mean StDev SE Mean 90% CI T P
C1 15 195.933 6.670 1.722 (192.900, 198.967) -2.36 0.033
MTB > OneZ c1;
SUBC> Sigma 7;
This does a 92% confidence interval and a test for a mean of 200 using sigma.
SUBC> Test 200;
SUBC> Confidence 92.
Test of mu = 200 vs not = 200
The assumed standard deviation = 7
Variable N Mean StDev SE Mean 92% CI Z P
C1 15 195.933 6.670 1.807 (192.769, 199.098) -2.25 0.024
MTB > %normarea5a This does the graph shown above.
Executing from file: normarea5a.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> 1.75
...working...
MTB > let c5=c1
MTB > Sort c5 c5;
SUBC> By c5.
MTB > print c5
This sorts c1, which I moved to c5.
C5
182 188 189 190 193 194 195 196 198 200 201 201 203
204 205
Extra Credit:
5. Check some numbers in the t, Chi-Squared or F tables using the new set of Minitab routines that I have prepared. To use the new set of routines, set up a file to hold your work. Then go to http://courses.wcupa.edu/rbove , open the Minitab folder and download any of the following:
Normal Distribution Area programs:
NormArea5A.txt

252solngr1 2/05/04 (Open this document in 'Page Layout' view!) or NormArea5C.txt and NormArea5.txt. t Distribution Area programs: tAreaA.txt or tAreaC.txt and tArea.txt
Chi-squared Distribution Area programs:
ChiAreaA.txt or ChiAreaC.txt and ChiArea.txt
F Distribution area programs:
FAreaA.txt or FAreaC.txt and FArea.txt
Use Notepad (under ‘tools’ in Minitab’) to convert their extensions from .txt back to .mac. To see how they are used, look at http://courses.wcupa.edu/rbove/Minitab/Area.doc
.
Routines like tAreaA are self prompting. To use routines like tAreaC, you need to set up your data in advance. If you want to use one of the worksheets that are mentioned in http://courses.wcupa.edu/rbove/Minitab/Area.doc
, click on ‘File’ and then ‘Open Worksheet.’ Copy a URL like the ones below into File Name.’ http://courses.wcupa.edu/rbove/Minitab/252PrA1d-f.MTW
http://courses.wcupa.edu/rbove/Minitab/tEx1.MTW
http://courses.wcupa.edu/rbove/Minitab/ChiEx1.MTW
http://courses.wcupa.edu/rbove/Minitab/FEx1.MTW
Results: I looked at the tables and found t
.
10

1 .
372 , z
.
10

1 .
282 ,
 2
.
10

23 .
2093 ,
 2
.
90

4 .
8650 , F

10 , 10

.
10

2 .
32 and F

10 , 10

.
90
 1
2 .
32

0 .
431 . 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.
MTB > %tareaA
Executing from file: tareaA.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...

252solngr1 2/05/04 (Open this document in 'Page Layout' view!) t Curve with 10 Degrees of Freedom and Standard Deviation 1.11803
The Area to the Right of 1.372 is 0.1000
0.4
0.3
0.2
0.1
0.0
-5.0
-2.5
0.0
Data A xis
2.5
mode 0 median 0
MTB > %normarea5a
Executing from file: normarea5a.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 with Mean 0 and Standard Deviation 1
The Area to the Right of 1.282 is 0.0999
5.0
0.4
0.3
0.2
0.1
0.0
-5 -4 -3 -2 -1 0
Data A xis
1 2 3 4

252solngr1 2/05/04 (Open this document in 'Page Layout' view!)
MTB > %ChiareaA
Executing from file: ChiareaA.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> 1.282
...working...
ChiSquare Curve with 10 Degrees of Freedom and Standard Deviation 4.47214
The Area to the Right of 1.282 is 0.9995
0.10
0.08
0.06
0.04
0.02
0.00
0 10 20
Data A xis
30
mode 8.00000 median 9.33333
MTB > %chiareaA
Executing from file: chiareaA.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
...working...
40

252solngr1 2/05/04 (Open this document in 'Page Layout' view!)
ChiSquare Curve with 10 Degrees of Freedom and Standard Deviation 4.47214
The Area to the Left of 4.865 is 0.1000
0.10
0.08
0.06
0.04
0.02
0.00
0 10 20
Data A xis
30
mode 8.00000 median 9.33333
MTB > %fareaA
Executing from file: fareaA.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
...working...
F Curve with numerator DF of 10 and Denominator DF of 10
The Area to the Right of 2.32 is 0.1003
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0.0
0 2 4 6 8
Data A xis
10 12 14
mode 0.818182
40
16

252solngr1 2/05/04 (Open this document in 'Page Layout' view!)
std dev 0.968246
MTB > %fareaA
Executing from file: fareaA.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
...working...
F Curve with Numerator DF of 10 Denominator DF of 10
The Area to the Left of 0.431 is 0.1003
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0.0
0 2 4 6 8
Data A xis
10 12 14
mode 0.818182
std dev 0.968246
16