252x0821 3/30/08
ECO252 QBA2
SECOND EXAM
March 28 2008
Name
Class Hour___________________
Show your work! Make Diagrams! Exam is normed on 50 points. Answers without reasons are not
usually acceptable.
I. (8 points) Do all the following. Make diagrams!
x ~ N 11, 13  - If you are not using the supplement table, make sure that I know it.
1. P0  x  54 
2. Px  16 
3. P12  x  41
4.
x.055 (Do not try to use the t table to get this.)
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II. (5+ points) Do all the following. Look them over first – There is a section III in the in-class exam and
the computer problem is at the end. Show your work where appropriate. There is a penalty for not
doing Problem 1. Page 11 is left blank if you need more space for calculations.
Note the following:
1. This test is normed on 50 points, but there are more points possible including the take-home. You are unlikely to finish
the exam and might want to skip some questions.
2. A table identifying methods for comparing 2 samples is at the end of the exam.
3. If you answer ‘None of the above’ in any question, you should provide an alternative answer and explain why. You may
receive credit for this even if you are wrong.
4. Use a 5% significance level unless the question says otherwise.
5. Read problems carefully. A problem that looks like a problem on another exam may be quite different.
6. Make sure that you state your null and alternative hypothesis, that I know what method you are using and what the
conclusion is when you do a statistical test. Use a significance level of .05 unless you are told otherwise.
1. You wish to assess the stability of the price of a stock and you find closing prices for the last year.
Rather than computing a variance of the entire population you take a sample of seven randomly picked
closing prices and compute a sample standard deviation. The sample is below – compute the sample
standard deviation. Show your work! (3)
Row x
1 89
2 124
3 56
4 94
5 75
6 82
7 63
6
For your convenience the sum of the first six numbers in x is
x
 520 and the sum of the first six numbers
i 1
6
squared is
x
2
 47618 .
i 1
2) You wish to compare this stock against a second stock that your friend recommends. Your friend has taken a random
sample of 10 closing prices and assures us that the sample mean price of this stock is 117.699 and the sample standard
deviation is 55.2764. You don’t like your friend’s stock because 1) it has a larger variance, indicating that it is riskier
and it costs more per share. The values you get are in the y column with z y being the y values with the 117.699
subtracted and the result divided by 55.2764. Compare the variances using a statistical test of the equality of variances.
(2)
Row
y
zy
1
2
3
4
5
6
7
8
9
10
78.48
130.93
93.17
105.37
69.50
85.43
102.84
259.27
151.17
100.83
-0.71
0.24
-0.44
-0.22
-0.87
-0.58
-0.27
2.56
0.61
-0.31
3) Are you sure that stock y has a higher average price than stock x ? Using the results of 2) compare the
mean prices. If you do not assume equality of variances assume that you can use 14 degrees of freedom for
the test. (3)
4) Test stock y to see if it has a Normal distribution. How do your results from the test of Normality affect
your assessment of the results in 2) and 3)? (4)
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5) Using the sample means and standard deviations you found in 2) and 3) but assuming that both samples
are of size 100 and come from a Normal distribution, do an 11% confidence interval for the difference
between the means. (2)
[14]
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III. (18+ points) Do as many of the following as you can. (2 points each unless noted otherwise). Look
them over first – the computer problem is at the beginning. Show your work where appropriate.
1. Computer question.
a) Turn in your first computer output. Only do b, c and d if you did. (3)
b) (Meyer and Krueger) A corporation rents apartments within the city of Phoenix and in the
surrounding suburbs. It wishes to verify that the mean rent in the city is lower than in the suburbs. Two
independent random samples are taken. These appear below.
City
401.84
666.95
804.01
611.09
Suburb
458.98
994.09
810.44
764.69
815.86
755.37
715.30
314.14
584.52
650.46
904.77
587.72
870.44
970.26
639.96
657.92
403.64
617.37
444.47
567.60
574.94
538.26
506.58
695.45
752.60
735.26
313.08
752.66
398.33
732.83
667.61
762.35
670.29
458.07
396.20
656.04
676.23
364.37
953.06
728.25
187.23
878.82
720.20
745.79
793.68
764.80
879.91
737.99
566.75
279.74
918.40
654.05
841.70
648.31
1106.17
919.93
What are your null and alternative hypotheses? Three tests appear below – Which is correct for your null
hypotheses? (3)   .01
Test 1.
MTB > TwoSample c1 c2;
SUBC>
Confidence 99.0.
Two-Sample T-Test and CI: City, Suburb
Two-sample T for City vs Suburb
SE
N Mean StDev Mean
City
30
590
169
31
Suburb 30
743
189
34
Difference = mu (City) - mu (Suburb)
Estimate for difference: -153.0
99% CI for difference: (-276.2, -29.7)
T-Test of difference = 0 (vs not =): T-Value = -3.31
P-Value = 0.002
DF = 57
Test 2.
MTB > TwoSample c1 c2;
SUBC>
Confidence 99.0;
SUBC>
Alternative -1.
Two-Sample T-Test and CI: City, Suburb
Two-sample T for City vs Suburb
SE
N Mean StDev Mean
City
30
590
169
31
Suburb 30
743
189
34
Difference = mu (City) - mu (Suburb)
Estimate for difference: -153.0
99% upper bound for difference: -42.3
T-Test of difference = 0 (vs <): T-Value = -3.31
P-Value = 0.001
DF = 57
Test 3.
MTB > TwoSample c1 c2;
SUBC>
Confidence 99.0;
SUBC>
Alternative 1.
Two-Sample T-Test and CI: City, Suburb
Two-sample T for City vs Suburb
SE
N Mean StDev Mean
City
30
590
169
31
Suburb 30
743
189
34
Difference = mu (City) - mu (Suburb)
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252x0821 3/30/08
Estimate for difference: -153.0
99% lower bound for difference: -263.7
T-Test of difference = 0 (vs >): T-Value = -3.31
P-Value = 0.999
DF = 57
3) From the output, but using the correct format for a confidence interval, what is an appropriate confidence
interval to correct your hypotheses? (2)
4) What is your conclusion? Why? (2)
5) What method was used by the computer? D1, D2, D3, D4, D5a, D5b, D6a, D6b, D7? (1)
6) The following tests were run after the original hypotheses tests. What do they tell us about the
appropriateness of the method? Why? (1)
[12]
MTB > NormTest c1;
SUBC>
KSTest.
Probability Plot of City
MTB > NormTest c2;
SUBC>
KSTest.
Probability Plot of Suburb
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2. A ‘robust’ test procedure is one that
a) Can only be done with a computer
b) Requires an underlying Normal distribution
c) Is sensitive to slight violations of its assumptions.
d) Is insensitive to slight violations of its assumptions.
3. (Ng -219, 18) Assume that you have the following information: s12  4 , s 22  6 , n1  16 and n 2  25
and you wish to do a pooled-variance t test, your ŝ p and degrees of freedom are (3) [17]
a) 2.45, 41
b) 2.24, 41
c) 2.29, 41
d) 2.00, 41
e) 2.45, 39
f) 2.24, 39
g) 2.29, 39
h) 2.00, 39
i) 2.45, 16
j) 2.24, 16
k) 2.29, 16
l) 2.00, 16
m) 2.45, 25
n) 2.24, 25
o) 2.29, 25
p) 2.00, 25
e) It’s more appropriate to add standard
errors and use z
4. If I want to test to see if the mean of x1 is smaller than the mean of x 2 my null hypothesis is:
(Note: D  1   2 ) Only check one answer!
(2)
a) 1   2 and D  0
b) 1   2 and D  0
e) 1   2 and D  0
f) 1   2 and D  0
c) 1   2 and D  0
d) 1   2 and D  0
g) 1   2 and D  0
h) 1   2 and D  0
5. Consumers are asked to take the Pepsi Challenge. They were asked they which cola they preferred and
the number that preferred Pepsi was recorded. Sample 1 was males and sample 2 was females. The
following was run on Minitab.
[19]
MTB > PTwo 109 46 52 13;
SUBC>
Pooled.
Test and CI for Two Proportions
Sample
X
N Sample p
1
46 109 0.422018
2
13
52 0.250000
Difference = p (1) - p (2)
Estimate for difference: 0.172018
95% CI for difference: (0.0221925, 0.321844)
Test for difference = 0 (vs not = 0): Z = 2.12
P-Value = 0.034
On the basis of the printout above we can say one of the following.
a) At a 99% confidence level we can say that we have enough evidence to state that the proportion
of men that prefer Pepsi differs from the proportion of women that prefer Pepsi
b) At a 95% confidence level we can say that we have enough evidence to state that the proportion
of men that prefer Pepsi differs from the proportion of women that prefer Pepsi
c) At a 99% confidence level we can say that we have enough evidence to state that the proportion
of men that prefer Pepsi equals the proportion of women that prefer Pepsi.
d) At a 96% confidence level there is insufficient evidence to indicate that the proportion of men
that prefer Pepsi differs from the proportion of women that prefer Pepsi
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6. (Lenzi) A group of runners run a 100 meter dash before and after running a marathon. Their times are
shown below. Pat – how long did they wait before running the second dash?
Row
1
2
3
4
5
6
7
Before
12.4
11.8
12.5
12.0
11.5
11.2
12.9
After
12.6
12.2
12.4
12.7
12.0
11.8
12.7
d
-0.2
-0.4
0.1
-0.7
-0.5
-0.6
0.2
Minitab printed out the following statistics.
Variable
Before
After
d
N
7
7
7
N* Mean
0 12.043
0 12.343
0 -0.300
SE Mean
0.226
0.134
0.131
StDev
0.597
0.355
0.346
Minimum
11.200
11.800
-0.700
Q1
11.500
12.000
-0.600
Median
12.000
12.400
-0.400
Q3
Maximum
12.500 12.900
12.700 12.700
0.100 0.200
Can we show that they were slower after the marathon?
a) How many degrees of freedom do we have in this problem? (1)
b) What are our null and alternative hypotheses? (1)
c) What is the approximate p-value for our result? (3) Show your work! (2 points if you do not do
a p-value)
d) On the basis of your p-value, what is our conclusion if the confidence level is 95%? Why? (1)
e) What if the confidence level is 99%? Why? (You do not need a p-value to answer this part of
the question, though is would help.) (1)
[26]
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7. A researcher takes independent random samples salaries of 18 women (Sample 1) and 18 men (Sample
2) who are fairly recent Business graduates with the following results.
Row Women
Men Difference
Minitab gives the following statistics.
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
64709
47105
28972
31449
42574
59051
26838
56651
64929
57497
38290
67106
67280
40826
60826
46207
58976
45809
40824
54465
68433
54941
54050
53043
45680
40399
57584
78224
53722
34915
59636
50499
53502
77186
60208
48381
23885
-7360
-39461
-23492
-11476
6008
-18842
16252
7345
-20727
-15432
32191
7644
-9673
7324
-30979
-1232
-2572
Descriptive Statistics: Women, Men, d
Variable
Women
Men
d
N
18
18
18
N* Mean SE Mean
0 50283 3156
0 54761 2708
0 -4478 4462
StDev
13392
11488
18929
Test the statement that women have a significantly lower salary than men.
a) What are your null and alternative hypotheses?
b) (Extra Credit) If you do not assume that variances of the two samples are equal
(i) How many degrees of freedom do you have? (4)
(ii) If you use the formula t 
d  D0
, what is the value of s d ? (2)
sd
(iii) Compute the t ratio and test the hypothesis, clearly stating your conclusions
  .05  (2)
c) If you assume that variances of the two samples are equal
(i) How many degrees of freedom do you have? (1)
d  D0
(ii) If you use the formula t 
, what is the value of s d ? (3)
sd
(iii) Compute the t ratio and test the hypothesis, clearly stating your conclusions
  .05  (2)
[32]
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8. (Meyer and Krueger again) Back to the Phoenix problem. The people in the problem on page 3 are still
obsessing over the relationship of rents to whether an apartment is urban or suburban. The computer output
from a Chi-Squared test is below.
Results for: 251x0821-06.MTW
MTB > WSave "C:\Documents and Settings\RBOVE\My
Documents\Minitab\251x0821-06.MTW";
SUBC>
Replace.
Saving file as: 'C:\Documents and Settings\RBOVE\My
Documents\Minitab\251x0821-06.MTW'
MTB > print c1-c3
Data Display
Row Rent
City
1 <500
48
2 500-599
51
3 600-699
30
4 700 up
22
Suburb
2
11
17
19
MTB > ChiSquare c2 c3.
Chi-Square Test: City, Suburb
Expected counts are printed below observed counts
Chi-Square contributions are printed below expected counts
City
48
37.75
2.783
Suburb
2
12.25
8.577
Total
50
2
51
46.81
0.375
11
15.19
1.156
62
3
30
35.48
0.848
17
11.52
2.613
47
4
22
30.95
2.591
19
10.05
7.983
41
Total
151
49
200
1
Chi-Sq = 26.925, DF = 3, P-Value = 0.000
a) The above is a Chi-squared test of (1)
i) independence
ii) homogeneity
iii) goodness-of-fit
iv) none of the above
b) What is the null hypothesis of this test and, assuming a 95% confidence level, what is the
conclusion? (2) [35]
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9. Back to Phoenix again. The people in the previous Phoenix problems are now sure that the distribution
of rents is not Normal but skewed to the right. They select a random sample of 10 rents in the city and
another random sample of 10 rents in the suburbs (1-City, 2-Suburb). The researchers now believe that
rentals in the city are lower than rentals in the country. The researchers will do the following test.
a) T-test of paired data
b) Wilcoxon signed rank test
c) T-test of means of independent samples
d) Wilcoxon-Mann-Whitney test
e) None of the above
10. Assuming that a rank test of some sort is done in Problem 9, what will be our null hypothesis, and,
assuming that the smaller of the two sums of ranks is 44 and that we are working with a 95% confidence
level, what will be our conclusion and why? (3) [40]
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Blank page for calculations
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ECO252 QBA2
SECOND EXAM
March 28, 2008
TAKE HOME SECTION
Name: _________________________
Student Number: _________________________
Class hours registered and attended (if different):_________________________
IV. Neatness Counts! Show your work! Always state your hypotheses and conclusions clearly. (19+
points). In each section state clearly what number you are using to personalize data. There is a penalty for
failing to include your student number on this page, not clarifying version number in each section and not
including class hour somewhere. Please write on only one side of the paper. Be prepared to turn in your
Minitab output for the first computer problem and to answer the questions on the problem sheet
about it or a similar problem.
1. (Moore, McCabe et. al.) A large public university took a survey of 865 students to find out if there was a
relationship between the chosen major and whether the students had student loans. The students’ majors
were categorized as Agriculture, Child Development, Engineering, Liberal Arts, Business, Science and
Technology. Before you start personalize the data as follows. Let a be the second-to-last digit of your
student number. Change the number of Science majors with loans to 31  a and the number of business
majors who have loans to 24  a for every part of this problem. The total number of students in the survey
will not change. Put your version of the table below on top of the first page of your solution. Use a 99%
confidence level in this problem.
Loan
None
Ag
32
35
Ch
37
50
Engg
98
137
Lib
89
124
Bus
24
51
Sci
31
29
Tech
57
71
a) Compute the proportion of non-science majors that have loans in order to test the hypothesis that science
majors are more likely to have loans than other majors. Tell which group you consider sample 1. State H 0
and H 1 in terms of the proportions involved and also in terms of the difference between the proportions,
explaining whether this difference is a statistic from sample 1 minus a statistic from sample 2 or the
reverse. (1)
b) Use a test ratio to test your hypotheses from a) (2)
c) Use a critical value for the difference between proportions to test your hypotheses from a) (2)
d) Use an appropriate confidence interval to test your hypotheses from a) (2)
e) Treat each major separately and test the hypothesis that the proportion of students that have loans is
independent of major (4)
f) If you did section 1e, follow your analysis with a Marascuilo procedure to compare the proportion of
business students that have loans with the proportions for the other 6 majors. Tell which differences are
significant. (3) [14]
g) (Extra credit) Check your results using Minitab.
(i) To do a chi-squared test on an O table that is in Columns c22-c28, simply put the row labels in
Column c21 and print out your data. Then type in
ChiSquare c22 – c28.
The computer will print back the columns with their names, but below each number from the O table you
O  E 2
, the contribution of the value of O to the chi-square
E
total. Use the p-value to find out if we reject the hypothesis of equal proportions at the 1% significance
level.
will find the corresponding values of E and
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252x0821 3/30/08
(ii) To do a test of the alternative hypothesis H 1 : p1  p 2 , where p1 
x1
x
and p 2  2 , use the
n1
n2
command below, substituting your numbers for x1 , n1 , x 2 and n 2 .
MTB > PTwo x1 n1 x 2 n 2 ;
SUBC>
Confidence 99.0;
SUBC>
Alternative 1;
SUBC>
Pooled.
x1
x
, x 2 , n 2 and p 2  2 a p-value for a z-test and Fisher’s
n1
n2
exact test (results should be somewhat similar to the z-test) and a 1-sided 99% confidence interval.
The computer will print back x1 , n1 , p1 
2. (Moore, McCabe et. al) An absolutely tactless psychology professor has divided faculty members into
categories the professor labels ‘Fat’ and ‘Fit’. A random sample of scores on a test of ‘ego strength’ of the
‘Fat’ faculty is labeled x1 . A sample of ‘ego strength’ of the ‘Fit’ faculty is labeled x 2 . d  x1  x 2 .
Use a 95% confidence level in this problem.
The professor has computed
Fat scores = 64.96,
x
2
1
x
1
 Sum of
 Sum of squares of Fat scores
= 307.607,
x
x
2
 Sum of scores of Fit = 90.02,
2
2
 Sum of squares of Fit scores
= 581.239,
 d  Sum of diff = -25.06 and
 d  Sum of squares of diff =
2
51.8198.
Row
1
2
3
4
5
6
7
8
9
10
11
12
13
14
Fat
Fit
Diff
x1
x2
d  x1  x 2
4.99
4.24
4.74
4.93
4.16
5.53
4.12
5.10
4.47
5.30
3.12
3.77
5.09
5.40
6.68
6.42
7.32
6.38
6.16
5.93
7.08
6.37
6.53
6.68
5.71
6.20
6.04
6.52
-1.69
-2.18
-2.58
-1.45
-2.00
-0.40
-2.96
-1.27
-2.06
-1.38
-2.59
-2.43
-0.95
-1.12
To personalize the data remove row b , where b is the last digit of your student number. Please state
clearly what row you removed. At this point you will have n1  n 2  13 rows of data. You will need the
mean and variance of all three columns of data if you do all sections of this problem. You can save yourself
considerable effort by using the computational formula for the variance with the sums and sums of squares
that the professor computed with the value or value squared of the numbers you removed subtracted.
The professor got the following results.
Variable
Fat
Fit
diff
n
14
14
14
Mean
4.640
6.430
-1.790
SE Mean
0.184
0.115
0.196
StDev
0.690
0.431
0.732
Median
4.835
6.400
-1.845
Your results should be relatively similar. Credit for computing the sample statistics needed is included in
the relevant parts of this problem. State hypotheses and conclusions clearly in each segment of the problem.
a) Assume that x1 and x 2 are independent random samples and test the hypothesis that the population
mean of the ego strength of the ‘fit’ faculty is above the population mean of the ‘fat’ faculty. Assume that
the data comes from the Normal distribution and that the variances for the ‘fit’ and ‘fat’ populations are
similar. (3)
b) (Extra credit) Assume that x1 and x 2 are independent random samples and test the hypothesis that the
population mean of the ego strength of the ‘fit’ faculty is above the population mean of the ‘fat’ faculty.
Assume that the data comes from the Normal distribution and that the variances for the ‘fit’ and ‘fat’
populations are not similar. (3)
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252x0721 3/19/07
c) Assume that x1 and x 2 are independent random samples. How would we decide whether the method in
a) of b) is correct? Do the appropriate test. Assume that the data comes from the Normal distribution.
Should we have used a) or b)? (2) [22]
d) Compute the mean and variance of the column of differences and test the column to see if the Normal
distribution works for these data. (4)
e) Assume that we had rejected the hypothesis that the distributions in the populations that the columns
come from is Normal, do a one-sided test to see whether the ego strength of the ‘Fat and ‘Fit’ people
differs. (2)
f) In the remainder of this problem assume that the x1 and x 2 columns are not independent random
samples but instead represent the ego strength of the same 14 or 13 faculty members before and after a
fitness program. Assuming that the Normal distribution applies, can we say that the ego strength of the
faculty has increased? (2)
g) Repeat f) under the assumption that the Normal distribution does not apply. (1)
h) Use the Wilcoxon signed rank test, to test to see if the median of the d column is -2. (2) [35]
i) Extra credit. Use Minitab to check your work.
The commands that you might need are as follows – remember that the subcommand ’Alternative -1’
gives a left-sided test and ’Alternative +1’ gives a right sided test. If this subcommand is not used a
2-sided test will appear.
The basic command to compare two means for data in c2 and c3 is
MTB > TwoSample c2 c3.
This will produce a 2-sided test using Method D3. A semicolon followed by the Alterative subcommand
will produce a 1-sided test. Adding the subcommand ’Pooled’ switches the method to D2. Remember
that a semicolon tells Minitab that a subcommand is coming and a period tells Minitab that the command is
complete. To use Method C4 on the same two columns use the command
MTB > Paired c2 c3.
This also can be modified with the Alternative command.
To test C2 for Normality using a Lilliefors test use
MTB > NormTest c4;
SUBC>
KSTest.
There are two other tests for Normality baked into Minitab. These are the Anderson-Darling test and the
Ryan-Joiner test. The graph produced by any of these can be analyzed by the Fat Pencil Test. To get a basic
explanation of these tests use the Stat pull-down menu hit basic statistics and then Normality Test. Finally
hit ‘help’ and investigate the topics available. There will be a small bonus for those of you who mention
Minitab’s problems with English grammar. To use the Anderson-Darling test, use the NormTest command
without a subcommand. To use the Ryan-Joiner test use
MTB > NormTest c4;
SUBC>
RJTest.
A really impressive paper might compare the results of the 3 tests and then show the results of an internet
search on the differences between them.
The other two tests that are relevant here can be accessed by using the Stat pull-down menu and
the Nonparametrics option. The instruction for a left-sided (Wilcoxon)-Mann-Whitney test would be
MTB > Mann-Whitney 95.0 c2 c3;
SUBC>
Alternative -1.
Minitab’s instructions for a 2-sided Wilcoxon signed rank test of a median of -2 from one sample in C4
would be
MTB > WTest -2 c4.
To do a one-sided test comparing samples in two columns take d  x1  x 2 and do a test that the median of
d is zero. Again Alternative can be used to get a 1-sided test.
Also there is some advice from last term’s Take-home.
To fake computation of a sample variance or standard deviation of the data in column c1 using
column c2 for the squares,
MTB
MTB
MTB
MTB
MTB
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let C2 = C1*C1
name k1 'sum'
name k2 'sumsq'
let k1 = sum(c1)
let k2 = sum(c2)
* performs multiplication
** would do a power, but multiplication
is more accurate.
This is equivalent to let k2 = ssq(c1)
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252x0721 3/19/07
MTB > print k1 k2
Data Display
sum
sumsq
MTB
MTB
MTB
MTB
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3047.24
468657
This is a progress report for my data
set.
name k1 'meanx'
let k1 = k1/count(c1)
/means division. Count gives n.
let k2 = k2 - (count(c1))*k1*k1
print k1 k2
Data Display
meanx
sumsq
152.362
4372.53
MTB > name k2 'varx'
MTB > let k2 = k2/((count(c1))-1)
MTB > print k1 k2
Data Display
meanx
varx
152.362
230.133
MTB > name k2 'stdevx'
MTB > let k2 = sqrt(k2)
MTB > print k1 k2
Sqrt gives a square root.
Data Display
meanx
stdevx
152.362
15.1701
Print C1, C2
To check for equal variances for data in C1 and C2, use
MTB > VarTest c1 c2;
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Unstacked.
Both an F test and a Levine test will be run. The Levine test is for non-Normal data so you want the F test
results.
To check your mean and standard deviation, use
`
MTB > describe C1
To put a items in column C1 in order in column C2, use
MTB > Sort c1 c2;
SUBC>
By c1.
3. Sorry. This is all I’ve got.
The methods were listed in the outline in the following table.
Paired Samples
Location - Normal distribution.
Method D4
Compare means.
Independent Samples
Methods D1- D3
Location - Distribution not
Normal. Compare medians.
Method D5b
Method D5a
Proportions
Method D6b
Method D6a
Variability - Normal distribution.
Compare variances.
Method D7
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