spss & minitab
TRANSCRIPT
Akram Ullah M.Sc Statistics UOP
Page 1
Practical # 1 One Sample t-Test (SPSS)
Ten individuals are chosen at random from a normal population and the heights
are found to be in inches 63, 63, 66, 67, 68, 69, 70, 70, 71 and 71. In the light of these
data, discuss the suggestion that mean height in the population is 66 inches.
Solution:
Step1: First we create a variable “Height” in the variable view and then data is entered in
the Data view.
Step 2: In order to apply “one sample t-test” first we click on the “Analyze” menu
and select “Compare means” then another menu will appear and click on “one sample t-
test”.
Step 3: A window will appear now shift variable “Height” to the test variable and
enter “66” for test value.
Step 4: Then click on “OK” and the results will appear in output window.
One-Sample Statistics
N Mean Std. Deviation Std. Error Mean
Height 10 67.8000 3.01109 .95219
One-Sample Test
Test Value = 66
T df Sig. (2-tailed) Mean Difference
95% Confidence Interval of the
Difference
Lower Upper
Height 1.890 9 .091 1.80000 -.3540 3.9540
Decision:
Since the p value is greater then “0.05” , hence we do not reject “H0” and
concluded that the mean height in the population is “66” inches.
Akram Ullah M.Sc Statistics UOP
Page 2
Practical # 2 Paired Samples t-Test (SPSS) The following data gives paired yields of two varieties of wheat. Each pair was
planted in a different locality. Test the hypothesis that the mean yields are equal. Variety 1: 45 32 58 57 60 38 47 51 42 38 Variety 2: 47 34 60 59 63 44 49 53 46 41
Solution:
Step 1: First we create two variables in the variable view. The “v1” variable is
create for Variety 1 and “v2” variable is create for Variety 2. Then enter the given data in
the “Data view”. Step 2: To apply “Paired samples t-test” we click on “Analyze” menu and select
“compare mean”, a menu will appear and then we click on “Paired samples t-test”. Step 3: A new dialog box will appear. We shift variable “v1” to variable 1 and
variable “v2” to variable 2 and click on “OK”. Step 4: The results will appear in the output window.
Paired Samples Statistics
Mean N Std. Deviation Std. Error Mean
Pair 1 v1 46.80 10 9.555 3.021
v2 49.60 10 9.168 2.899
Paired Samples Correlations
N Correlation Sig.
Pair 1 v1 & v2 10 .991 .000
Paired Samples Test
Paired Differences
t df
Sig. (2-
tailed)
Mean
Std.
Deviation
Std. Error
Mean
95% Confidence Interval
of the Difference
Lower Upper
Pair 1 v1 -
v2
-2.800 1.317 .416 -3.742 -1.858 -6.725 9 .000
Decision: Since the p value is less then “0.05”. Hence we reject “H0” and concluded
that the mean yields of Variety 1 and Variety 2 are not equal.
Akram Ullah M.Sc Statistics UOP
Page 3
Practical # 3 One Way ANOVA (SPSS) Given the data below test the hypothesis that the means of the three
populations are equal. Let α=0.05 Sample 1: 40 50 60 65 Sample 2: 70 65 66 50
Sample 3: 45 38 60 42
Solution:
Step 1: First we create two variables in the “variable view”. The variable
“sample” is created for the given data of the all three samples and “Factor” variable is
created for the subscripts of the samples. Assigned values “1” for sample 1, “2” for
sample 2 and “3” for sample 3. The data is entered in the “Data view”. Step 2: To apply “One Way ANOVA” first we click on “Analyze” menu and select
“compare mean”, another menu will appear and then we click on “One Way ANOVA”.
Step 3: A window will appear now we shift variable “s_data” to the “dependent
list” dialog box and variable “Factor” to “Factor” dialog box and click on “OK”. Step 4: The results will appear in the output window. Which are shown below.
One way
ANOVA
Sample data
Sum of Squares df Mean Square F Sig.
Between Groups 546.000 2 273.000 2.804 .113
Within Groups 876.250 9 97.361
Total 1422.250 11
Decision:
Since the p value is greater then 0.05. Hence we do not reject “H0” and
concluded that all the three means are equal.
Akram Ullah M.Sc Statistics UOP
Page 4
Practical # 4 Correlation Coefficient (SPSS) Calculate co-efficient of correlation between X and Y from the following data. X: 1 2 3 4 5 Y: 2 5 3 8 7 Solution:
Step 1: First we create two variables “X” and “Y” in the “variable view”. And then
the data is entered in the “Data view”. Step 2: To apply “correlation” first we click on “Analyze” menu and select
“correlate”, another menu will appear and then we click on “bivariate”. Step 3: A window will appear now we shift variables “X” and “Y” to the
“variables” dialog box and click on “option”. Another window will appear we mark
means and standard deviations and then click on “continue” button. Step 4: Then click on “OK” and the results will appear in the output window.
Correlations
Descriptive Statistics
Mean Std. Deviation N
X 3.00 1.581 5
Y 5.00 2.550 5
Correlations
X Y
X Pearson Correlation 1 .806
Sig. (2-tailed) .099
N 5 5
Y Pearson Correlation .806 1
Sig. (2-tailed) .099
N 5 5
Decision:
Since the p value is greater then 0.05. Hence we do not reject our null hypothesis
and concluded that the correlation is significant.
Akram Ullah M.Sc Statistics UOP
Page 5
Practical # 5 Sign Test (SPSS) Apply Sign test Weight before: 125 195 171 140 201 170 176 195 139 Weight after: 136 201 158 145 195 175 190 190 145
Solution:
Step 1: First we create two variables “before” and “after” in “Variable View”. And
then data is entered in the “Data view”. Step 2: To apply “sign test” first we click on “Analyze” menu and select “Non-
paramedics test” and click on “2-related samples”.
Step-3: A new window will appear where we have to shift the variable “after” to
the dialog box “variable 1” and the variable “before” to dialog box “variable 2”. And click
on “sign”. Step-4: Then we click on “OK” and the output will appear in out put window.
Sign Test
Frequencies
N
Weight before - Weight after Negative Differencesa 7
Positive Differencesb 3
Tiesc 0
Total 10
a. Weight before < Weight after
b. Weight before > Weight after
c. Weight before = Weight after
Test Statisticsb
Weight before - Weight after
Exact Sig. (2-tailed) .344a
a. Binomial distribution used.
b. Sign Test
Decision:
The p value is greater then 0.05. Hence we do not reject H0 and concluded that the
programme does not affect on the average weight of recruits.
Akram Ullah M.Sc Statistics UOP
Page 6
Practical # 6 Willcoxon Sign rankTest (SPSS) Apply
Willcoxon Sign rank test Weight before: 125 195 171 140 201 170 176 195 139 Weight after: 136 201 158 145 195 175 190 190 145
Solution:
Step 1: First we create two variables “before” and “after” in “Variable View”. And
then data is entered in the “Data view”. Step 2: To apply “sign test” first we click on “Analyze” menu and select “Non-
parametic test” and click on “2-related samples”.
Step-3: A new window will appear where we have to shift the variable “after” to
the dialog box “variable 1” and the variable “before” to dialog box “variable 2”. And click
on “willcoxon”. Step-4: Then we click on “OK” and the output will appear in out put window.
Sign Test
Frequencies
N
Weight before - Weight after Negative Differencesa 7
Positive Differencesb 3
Tiesc 0
Total 10
a. Weight before < Weight after
b. Weight before > Weight after
c. Weight before = Weight after
Test Statisticsb
Weight before - Weight after
Exact Sig. (2-tailed) .344a
a. Binomial distribution used.
b. Sign Test
Decision:
The p value is greater then 0.05. Hence we do not reject H0 and concluded that the
programme does not affect on the average weight of recruits.
Akram Ullah M.Sc Statistics UOP
Page 7
Practical # 7 Median Test (SPSS) The following data represent the operating times in hours for three types of
scientific pocket calculators before a recharge is required: Calculator A: 4.9 6.1 4.3 4.6 5.3 Calculator B: 5.5 5.4 6.2 5.8 5.5 5.2 4.8 Calculator C: 6.4 6.8 5.6 6.5 6.3 6.6 Use the Median Test to test
the hypothesis that the operating times for all three calculators are equal. Solution: Step 1: First we create two variables in the “Variable View”. Step 2: To apply “Median Test” first we click on “Analyze” menu and select “Non-
parametric test” and click on “K-independent”. Step 3: Now we shift the variable “factor” to the factor dialog box and “response” to the
response dialog box. and click on “Median”. Step 4: Then we click
on “OK” and the output will appear in out put window. Median Test
Frequencies
factor
1 2 3
response > Median 1 2 6
<= Median 4 5 0
Test Statisticsb
response
N 18
Median 5.55
Chi-Square 9.086a
df 2
Asymp. Sig. .011
a. 6 cells (100.0%) have expected frequencies less than 5. The minimum expected cell
frequency is 2.5.
Decision: Since p value is greater then 0.01. Hence we do not reject (H0) and
concluded that the operating times for all three calculators are equal.
Akram Ullah M.Sc Statistics UOP
Page 8
Practical # 8 Run Test (SPSS) Each day a sample of 10 production items was taken and the mean weight computed.
Following are the first 20 daily means:13.0 12.8 12.9 13.0 12.9 12.6 12.6
12.7 12.9 13.1 13.1 13.2 13.3 13.2 13.1 12.9 13.2 13.3 13.2 Are the number of runs below and above the median significant at the 5 percent level? Solution: Step 1: First we create a variable “mean” in the “Variable View”. And data is
entered in the “Data view”.
Step 2: To apply the “run test” first we click on “Analyze” menu and select “Non-
parametric test” and click on “runs”.
Step 3: A new window will appear where we have to shift the variable “mean” to
the “test variable” dialog box and click on the “option” button. A new window will
appear we click on the “descriptive” dialog box and then click on the “continue” button.
Step 4: We click on “OK” and the output will appear in the output window.
NPar Tests
Descriptive Statistics
N Mean Std. Deviation Minimum Maximum
mean 20 13.005 .2139 12.6 13.3
Runs Test
mean
Test Valuea 13.1
Cases < Test Value 10
Cases >= Test Value 10
Total Cases 20
Number of Runs 6
Z -2.068
Asymp. Sig. (2-tailed) .039
a. Median
Decision: Since p value is less then 0.05. Hence we reject (H0) and concluded that the
number of runs below and above the median is significant.
Akram Ullah M.Sc Statistics UOP
Page 9
Practical # 9 Kalmogorov – Smirnov Test (SPSS) The following scores were obtained by rolling a six-sided die 10 times: Scores: 3 4 4 2 6 6 3 4 2 5 Use
the Kalmogorov – Smirnov Test to test the hypothesis that at is a sample from a uniform
distribution at α = “0.05”. Solution: Step 1: First of all we have to create a variable “score” in the “Variable View”.
And the given data is entered in the “Data view”. Step 2: To apply the “Kalmogorov – Smirnov Test” first we click on “Analyze”
menu and select “Non-parametric test” and click on “1- sample K-S…”. Step 3: A window will appear where we have to shift the variable “score” to the
“test variable list” and click on the “option” button. A new window will appear we click
on the “descriptive” dialog box and then click on the “continue” button. Then we click
on “uniform” dialog box. Step 4: Then we click on “OK” and the output will appear in the output window.
NPar Tests
Descriptive Statistics
N Mean Std. Deviation Minimum Maximum
score 10 3.90 1.449 2 6
One-Sample Kolmogorov-Smirnov Test
score
N 10
Uniform Parametersa,,b Minimum 2
Maximum 6
Most Extreme Differences Absolute .200
Positive .200
Negative -.200
Kolmogorov-Smirnov Z .632
Asymp. Sig. (2-tailed) .819
a. Test distribution is Uniform.
b. Calculated from data.
Decision: Since the p value is greater then 0.05. Hence we do not reject (H0) and
concluded that at is a sample from a uniform distribution.
Akram Ullah M.Sc Statistics UOP
Page 10
Practical # 10 Mann-Whitney U test (SPSS) Two independent samples given below are from two populations. Sample 1: 38 49 45 29 31 35 Sample 2: 31 42 22 26 43 37 25 30 47 Apply the “Mann-Whitney U test” to test the hypothesis that the two samples come
from population having identical distribution. Use a “0.05”level of significance. Solution: Step 1: First of all we have to create two variables in the “Variable View”. The
variable “response” is created for the given data of the two samples and the variable
“Factor” is created for the subscripts of the samples. Assigned values “1” for sample 1
and “2” for sample 2. And the data is entered in the “Data view”. Step 2: To apply “Kalmogorov – Smirnov Test and Mann-Whitney U test” first of
all we have to click on the “Analyze” menu and select “Non-parametric test” and click on
“2-independent samples”. Step 3: A new window will appear where we have to shift the variable “factor” to
the “grouping variable” dialog box and “response” to the “test variable list” dialog box
and click on “Define Groups” a new window will appear and we enter “1” in the “Group
1” dialog box and “2” in the “Group 2”dialog box and click on continue then click on
“Kalmogorov – Smirnov” and “Mann-Whitney U”. Step 4: Then we click on “OK” and the output will appear in out put window.
NPar Tests
Descriptive Statistics
N Mean Std. Deviation Minimum Maximum
respons 15 35.33 8.482 22 49
factor 15 1.60 .507 1 2
Mann-Whitney Test
Ranks
factor N Mean Rank Sum of Ranks
respons sample 1 6 9.42 56.50
sample 2 9 7.06 63.50
Total 15
Akram Ullah M.Sc Statistics UOP
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Test Statisticsb
respons
Mann-Whitney U 18.500
Wilcoxon W 63.500
Z -1.003
Asymp. Sig. (2-tailed) .316
Exact Sig. [2*(1-tailed Sig.)] .328a
a. Not corrected for ties.
b. Grouping Variable: factor
Decision:
Since the p value is greater then 0.05. Hence we do not reject (H0) and concluded
that the two samples come from population having identical distribution.
Akram Ullah M.Sc Statistics UOP
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MINITAB
PRACTICAL
Practical # 11 One Way ANOVA (MINITAB)
Akram Ullah M.Sc Statistics UOP
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Given the data below, test the hypothesis that the means of the three
populations are equal.
Sample 1: 45 60 38 57
Sample 2: 73 58 55 75
Sample 3: 47 42 65 41
Solution:
Step 1: First of all we enter the data of all the three samples separately in column
C1,C2 and C3 in the “Data window”. And assigned labels “sample 1” to C1, “sample 2”
to C2 an sample 3” to C3.
Step 2: In order to apply “One Way ANOVA” first we click on “stat” menu and
select “ANOVA”, another menu will appear and then we click on “oneway (unstacked)”.
Step 3: A new window will appear now we shift all the three variables in to the
“responses” dialog box.
Step 4: Then click on “OK” and the results will appear in the session window.
One-Way Analysis of Variance
Analysis of Variance
Source DF SS MS F P
Factor 2 675 338 3.03 0.099
Error 9 1004 112
Total 11 1679
Individual 95% CIs For Mean
Based on Pooled StDev
Level N Mean StDev ----------+---------+---------+-----
-
C1 4 50.00 10.30 (---------*---------)
C2 4 65.25 10.21 (---------*---------)
C3 4 48.75 11.15 (---------*---------)
----------+---------+---------+-----
-
Pooled StDev = 10.56 48 60 72
Decision:
Since the p value is greater then “0.05” therefore we do not reject our null
hypothesis.
Practical # 12 One Sample t-Test (MINITAB)
Apply the one sample t-test in the following data.
Akram Ullah M.Sc Statistics UOP
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12 20 31 35 38 40 17 22 27
Solution:
Step 1: First we entered the given data in C1 in the “Data window”.
Step 2: In order to apply “one sample t-test” first we click on the “stat” menu and
select “Basic Statistics” then another menu will appear and click on “1-Sample t…”.
Step 3: A window will appear now we shift variable “C1” to the test variables and
click on “test mean” .
Step 4: We click on “OK” and the result will appear in the session window.
T-Test of the Mean
Test of mu = 0.00 vs mu not = 0.00
Variable N Mean StDev SE Mean T P
C1 9 26.89 9.80 3.27 8.23 0.0000
Decision:
Since the p value is less then “0.05”. Hence we reject our null hypothesis.
Practical # 13 Two sample t-Test (MINITAB)
Akram Ullah M.Sc Statistics UOP
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The following data gives paired yields of two varieties of wheat. Each pair was
planted in a different locality. Test the hypothesis that the mean yields are equal.
Variety 1: 32 58 57 38 47 51 38
Variety 2: 47 60 63 49 53 46 41
Solution:
Step 1: First we have enter the values of variety 1 in “C1” and variety 2 in “C2” in
the “Data window”.
Step 2: Then we click on the “stat” menu and select “Basic Statistics” then
another menu will appear and click on “2-Sample t…”.
Step 3: A window will appear now we click on “samples in different columns” the
the two dilogs boxes will appear then we have to shift shift variable “C1” to the “First”
dilog box and shioft “C2” to the “second” dilog box.
Step 4: We click on “OK” and the result will appear in the session window.
Two Sample T-Test and Confidence Interval
Two sample T for C1 vs C2
N Mean StDev SE Mean
C1 7 45.9 10.1 3.8
C2 7 51.29 7.89 3.0
95% CI for mu C1 - mu C2: ( -16.1, 5.2)
T-Test mu C1 = mu C2 (vs not =): T= -1.12 P=0.29 DF= 11
Decision:
Since the p value is greater then “0.05”. Hence we do not reject “H0” and
concluded that the mean yields of Variety 1 and Variety 2 are equal.
Practical # 14 One sample sign test (MINITAB)
Akram Ullah M.Sc Statistics UOP
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A sample of size 8 was chosen from a population. The sample observations are
given below: 3.55, 5.62, 3.93, 3.46, 2.95, 5.55, 3.11 and 1.90.
Use sign test to test the hypothesis that the median of the population equal to 3.
Solution:
Step 1: First we enter the values of the given sample in the column “C1” in the
“Data window”.
Step: 2 An apply “1-sample sign test” we click on “stat” menu and select
“Nonparametrics” a new menu will appear and then we click on “1-sample sign”.
Step 3: A new window will appear then we shift “C1” to the “variables” dialog
box and click on “test mean” a dialog box will appear then we enter “2” in the “test
median” dialog box.
Step 4: We click on “OK” and the result will appear in the “session window”.
Sign Test for Median
Sign test of median = 3.000 versus not = 3.000
N Below Equal Above P Median
C1 7 1 0 6 0.1250 3.550
Decision:
Since the p value is greater then 0.05. Hence we do not reject our null hypothesis
and concluded that the median of the population is equal to 3.
Practical # 15 Regression (MINITAB) Use the following data to test the hypothesis that the regression is linear at the 0.05
Akram Ullah M.Sc Statistics UOP
Page 17
level of significance.
X: 1 2 3 4 5 6 7 8
Y: 3 5 8 11 12 17 19 20
Solution:
Step 1: First we enter the values of “X” in to “C1” and the values of “Y” in to “C2”
in the “Data window”. And assigned labels “X” to “C1” and “Y” to “C2”.
Step 2: To apply “Regression” first we click on “stat” menu and select
“regression”, a new menu will appear and click on “regression”.
Step 3: A new window will appear then we shift variable “X” to “Response” dialog box
and variable “Y” to “Predictors” dialog box.
Step 4: Click on “OK” and the results will appear in the session window.
Regression Analysis
The regression equation is
C1 = - 0.023 + 0.381 C2
Predictor Coef StDev T P
Constant -0.0228 0.2642 -0.09 0.934
C2 0.38087 0.01988 19.16 0.000
S = 0.3356 R-Sq = 98.4% R-Sq(adj) = 98.1%
Analysis of Variance
Source DF SS MS F P
Regression 1 41.324 41.324 366.93 0.000
Error 6 0.676 0.113
Total 7 42.000
Decision: Since the p value is less then “0.05” . Hence we reject our null hypothesis and concluded that the variables “X” and “Y” are dependent to each other.
Practical # 16 Mann-Whitney U test (MINITAB)
Two independent samples given below are from two populations.
Akram Ullah M.Sc Statistics UOP
Page 18
Sample 1: 38 49 45 29 31 35
Sample 2: 31 42 22 26 43 37
Apply the “Mann-Whitney U test” to test the hypothesis that the two samples
come from population having identical distribution. Use a “0.05”level of significance.
Solution:
Step 1: First we have entered the data of sample 1 in “C1” and the data of
sample 2 in “C2” in the “Data window”.
Step: 2 To apply “Mann-Whitney U test” we click on “stat” menu and select
“Nonparametric” a new menu will appear and then we click on “Mann-Whitney”.
Step 3: A new window will appear then we shift “C1” to the “First Sample” dialog
box and also shift “C2” to the “Second Sample” dialog box.
Step 4: We click on “OK” and the result will appear in the “session window”.
Mann-Whitney Confidence Interval and Test C1 N = 6 Median = 36.50
C2 N = 6 Median = 34.00
Point estimate for ETA1-ETA2 is 4.50
95.5 Percent CI for ETA1-ETA2 is (-7.99,16.00)
W = 44.5
Test of ETA1 = ETA2 vs ETA1 not = ETA2 is significant at 0.4233
The test is significant at 0.4225 (adjusted for ties)
Cannot reject at alpha = 0.05
Practical # 17 Kruskal Wallis Test(MINITAB) The following data represent the operating times in hours for three types of
scientific pocket calculators before a recharge is required:
Akram Ullah M.Sc Statistics UOP
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Calculator A: 4.9 6.1 4.3 4.6 5.3 Calculator B: 5.5 5.4 6.2 5.8 5.5 5.2 4.8 Calculator C: 6.4 6.8 5.6 6.5 6.3 6.6 Use the Kruskal Willis Test to test the hypothesis that the operating times for all
three calculators are equal.
Solution:
Step 1: First of all we have entered the values of all the three samples in the
column “C1” and their corrosponding subscripts in “C2” in the “Data window”.
Step 2: An ordered to apply “Kruskal wallis test” we click on “stat” menu and
select “Nonparametrics” a new menu will appear and then we click on “kurskal -wallis”.
Step 3: A new window will appear then we shift “C1” to the “Response” dialog
box and also shift “C2” to the “Factor” dialog box.
Step 4: We click on “OK” and the result will appear in the “session window”.
Kruskal-Wallis Test Kruskal-Wallis Test on C1
C2 N Median Ave Rank Z
1 5 4.900 5.0 -2.22
2 7 5.500 8.0 -0.95
3 6 6.450 15.0 3.09
Overall 18 9.5
H = 10.47 DF = 2 P = 0.005
H = 10.48 DF = 2 P = 0.005 (adjusted for ties)
Decision:
Since the p value is less then “0.05”. Hence we reject our null hypothesis (H0) and
concluded that the operating times for all three calculators are not equal.
Practical # 18 Run Test (MINITAB)
Akram Ullah M.Sc Statistics UOP
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Each day a sample of 10 production items was taken and the mean weight
computed. Following are the first 20 daily means:
13.0 12.8 12.9 13.0 13.1 12.9 12.6 12.6 12.7 12.9 13.1 13.1
13.2 13.3 13.2 13.1 12.9 13.2 13.3 13.2
Are the number of runs below and above the median significant at the 5 percent level?
Solution:
Step 1: First we enter the values of the given sample in the column “C1” in the
“Data window”.
Step 2: To apply “runs test” we click on “stat” menu and select “Nonparametrics”
a new menu will appear and then we click on “Runs test”.
Step 3: A new window will appear then we shift “C1” to the “variables” dialog
box and click on “Above and below the median”.
Step 4: Then click on “OK” and the result will appear in the “session window”.
Runs Test
C1
K = 13.0050
The observed number of runs = 6
The expected number of runs = 11.0000
10 Observations above K 10 below
* N Small -- The following approximation may be invalid
The test is significant at 0.0218
Akram Ullah M.Sc Statistics UOP
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Practical # 19 Median test (MINITAB) The following data represent the operating times in hours for three types of
scientific pocket calculators before a recharge is required: Calculator A: 4.9 6.1 4.3 4.6 5.3 Calculator B: 5.5 5.4 6.2 5.8 5.5
Calculator C: 6.4 6.8 5.6 6.5 6.3 Use the Mood’s Median Test to test the hypothesis that the operating times for
all three calculators are equal.
Solution:
Step 1: First we have entered the values of all the three samples in the column
“C1” and their corresponding subscripts in “C2” in the “Data window”.
Step 2: To apply “Kruskal wallis test” we click on “stat” menu and select
“Nonparametric” a new menu will appear and then we click on “kurskal-wallis”.
Step 3: A window will appear then we shift “C1” to the “Response” dialog box
and also shift “C2” to the “Factor” dialog box.
Step 4: Then click on “OK” and the result will appear in the “session window”.
Mood Median Test Mood median test for C1
Chi-Square = 3.75 DF = 2 P = 0.153
Individual 95.0% CIs
C2 N<= N> Median Q3-Q1 ---------+---------+---------+-------
1 4 1 4.90 1.25 (--------+----------------)
2 3 2 5.50 0.55 (-+---------)
3 1 4 6.40 0.70 (----------+-----)
---------+---------+---------+-------
4.90 5.60 6.30
Overall median = 5.60
* NOTE * Levels with < 6 observations have confidence < 95.0%
Decision:
Since according p value is greater then “0.05”. Hence we do not reject our null
hypothesis and concluded that the operating times for all three calculators are equal.
Akram Ullah M.Sc Statistics UOP
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Practical # 20 Chi square test (MINITAB)
from the adult male population of seven large cities, random samples of sizes
indicated below were taken, and the numbers of married and single men recorded.
City A B C D E F G
Married
Single
133 164 155 106 153 123 116
36 57 40 37 55 33 36
Test the hypothesis at the “0.05” level of significance that the proportions of
married men are the same in all the seven cities.
Solution:
Step 1: First of all we enter the data of all the seven cities separately in column
C1, C2, C3, C4, C5, C6 and C7 respectively in the “Data window”.
Step 2: To apply “Chi-square test” first we click on “stat” menu and select “tables” then
another menu will appear and then we click on “Chi-square test…”.
Step 3: A window will appear now we shift all the seven variables in to the
“Column containing the table” dialog box.
Step 4: Then we click on “OK” and the results will appear in the session window.
Chi-Square Test
Expected counts are printed below observed counts
A B C D E F G
Total
1 133 164 155 106 153 123 146
980
130.00 170.00 150.00 110.00 160.00 120.00 140.00
2 36 57 40 37 55 33 36
294
39.00 51.00 45.00 33.00 48.00 36.00 42.00
Total 169 221 195 143 208 156 182
1274
Chi-Sq = 0.069 + 0.212 + 0.167 + 0.145 + 0.306 + 0.075 + 0.257 +
0.231 + 0.706 + 0.556 + 0.485 + 1.021 + 0.250 + 0.857 = 5.337
DF = 6, P-Value = 0.501
Decision: Since the p value is greater then “0.05” we do not reject our null
hypothesis and concluded that the proportions of married men are the same in all the seven cities.
Akram Ullah M.Sc Statistics UOP
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