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Please give a short answer.
Exercise 1. The Independent
t Test and the Mann-Whitney
U-Test
Independent
t Test
1. The independent samples
t-test is a
parametric test in which the following assumptions must be met
if the test is to be valid:
a. The variable grouping the data must be dichotomous
b. The data in the two samples from the two groups must be independent
c. The characteristic variable under study must be distributed normally and should be
continuous.
2. Develop a (real or fictitious) research question that could be addressed by an independent
t test:To see whether the students in section A of a school score is no different in History than those is
section B of the same school in a given year.
3. State the (real or fictitious) Null and Alternative Hypotheses:The null hypothesis (H0) and the alternative hypothesis (H1) can be written as follows:
H0: mean history scoreof section A = mean history scoreof section B
Against
H1: mean history scoreof section A ≠ mean history scoreof section B
4. Make up a set of data to analyze in SPSS that would qualify for the use of the independent samples
t
test. Over-all have at least 30 sets of data for this exercise. Make sure that the data meet all of the
necessary assumptions of the independent samples
t test.A dataset of 30 observations from each section recording history assessment scores were stored.
The dataset had two columns, one indicating the section of the students and the other the
corresponding history scores. The section variable is the grouping variable in this case. He dataset
is attached.
5. Use the
a-level of .05The level of significance used was 0.05 or 5%.
6. Enter the data into SPSS.
 Compute the mean, standard deviation and variance of each group.
 Follow the Step-by-Step Procedure for Using SPSS to compute the Independent Samples
t Test
(page 106 of K & K). Computing the test statistic, an independent samples
t test, that produces
SPSS Output similar to page 103 (K & K). Save the data entry file and send your data entry file
(input) from SPSS to the course instructor via email. Cut and paste your SPSS output results
below.
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The SPSS output showing the necessary computations are given hence:
Group Statistics
Section N Mean Std. Deviation Std. Error Mean
Marks Section A 30 46.4667 28.36673 5.17903
Section B 30 57.8667 30.14875 5.50438
Independent Samples Test
Levene's Test
for Equality of
Variances
t-test for Equality of Means
F Sig. t df Sig. (2-
tailed)
Mean
Difference
Std. Error
Difference
95% Confidence
Interval of the
Difference
Lower Upper
Marks
Equal
variances
assumed
.114 .736 -
1.508 58 .137 -11.40000 7.55782 -26.52862 3.72862
Equal
variances
not assumed
-
1.508 57.786 .137 -11.40000 7.55782 -26.52981 3.72981
7. Determine the Statistical Significance and State a Conclusion. (Use the example “Putting it All
Together” on page 107 (K & K)The statistical level of significance is 0.137 which is greater than the a-level which is 0.05.
Therefore the test is not significant and the test fails to reject null hypothesis. Hence there is not
enough evidence to support the conjecture that section A has better or worse scores than section
B.
Mann-Whitney
U-Test
1. The Mann-Whitney
U-Test is a
non-parametrictest in which the following assumptions must be met if
the test is to be valid:a.The grouping variable must be dichotomous
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b. The two samples from each group must be independent
c. The distribution of the data from the two groups must be the same
d. The data set must be either ordinal, interval or ratio scaled.
2. Develop a (real or fictitious) research question that could be addressed by a Mann-Whitney
U-Test:To test whether preference of shopping for women differ from that of men. The two variables
of interest are gender of respondent and how they rate their interest for shopping.
3. State the Null and Alternative Hypotheses:H0: Average preference rating of men = Average preference rating of women (null)
Against
H1: Average preference rating of men ≠ Average preference rating of women (alternate)
4. Make up a set of data to analyze in SPSS that would qualify for the use of the Mann-Whitney
U-Test.
Over-all have at least 20 sets of data for this exercise. Make sure that the data meet all of the necessary
assumptions of the Mann-Whitney
U-Test.
The first is a binary variable and the second an ordinal scaled variable with 1 being least liked and 5
being strongly likes. The size of the dataset is 30 observations.
5. Use the
a-level of .05.
The level of significance was taken to be 0.05.
6. Enter the data into SPSS. Follow the Step-by-Step Procedure for Using SPSS to Compute the Mann-
Whitney
U-Test (pages 114 – 116 K & K).
 Compute the Median and Interquartile Range for each group.
 Follow the Step-by-Step Procedure for Using SPSS to compute the Mann-Whitney
U-Test (page
117 of K & K). Compute the test statistic, the Mann-Whitney
U-Test that produces SPSS Output
similar to page 113 (K & K). Save the data entry file (input) and send via email to the course
instructor. Cut and paste your SPSS output results below.
Ranks
Gender N Mean Rank Sum of Ranks
Shopping
Female 16 13.66 218.50
Male 14 17.61 246.50
Total 30
3

Test Statisticsa
Shopping
Mann-Whitney U 82.500
Wilcoxon W 218.500
Z -1.262
Asymp. Sig. (2-tailed) .207
Exact Sig. [2*(1-tailed Sig.)] .224b
a. Grouping Variable: Gender
b. Not corrected for ties.
7. Determine the Statistical Significance and State A Conclusion. (Use the example “Putting it All
Together”on page 117 (K & K).
The p-value was found to be 0.207, implying that it is greater than the level of significance and
so the test failed to reject the null hypothesis. Thus not enough evidence is present to suggest that
women and men regard shopping with differing levels of enthusiasm.
Exercise 2. The Paired
t Test and the Wilcoxon Matched-Pairs Signed Rank Test
Paired
t Test
1. The Paired
t Test is a
parametric test in which the following assumptions must be met if the test is to
be valid:
a. Pre-test and post-test observations are recorded on each controlled subject so data consists
of bivariate data points.
b. The two measures are normally distributed
c. The measurement scale is interval or ratio.
2. Develop a (real or fictitious) research question that could be addressed by the paired
t test:
To test that medicine A has been successful in improving the condition of diabetes among subjects in
a clinical trial of 30 subjects.
3. State the (real or fictitious) Null and Alternative Hypotheses:H0: Blood sugar(fasting) before treatment = Blood sugar(fasting) level after treatment (null)
Against
H1: Blood sugar before treatment > Blood sugar level after treatment (alternate)
4

4. Make up a set of data toanalyze in SPSS that would qualify for the use of the paired
t test. Over-all
have at least 30 pairs of data and a distribution that is not too badly skewed. Make sure that the data
meet all of the necessary assumptions of the paired t test.
The data consists of blood sugar readings before meal of 30 diabetic patients in two stages. First
before the treatment was applied and then after the treatment was applied. The dataset consists of
two columns with the blood sugar readings before and after the test.
5. Use the a-level of .05.
The level of significance used was 0.05 or 5%.
6. Enter the data into SPSS.
 Compute the mean and standard deviation of each group.
 Follow the Step-by-Step Procedure for Using SPSS to compute the Paired
t Test (pages 136 – 137
of K & K). Computing the test statistic, paired
t test, that produces SPSS Output similar to page
138 (K & K). Save the data entry file and send your data entry file (input) from SPSS to the
course instructor via email. Cut and paste your SPSS output results below.
Paired Samples Statistics
Mean N Std. Deviation Std. Error Mean
Pair 1 Before 128.8667 30 16.21139 2.95978
After 109.3667 30 16.21724 2.96085
Paired Samples Correlations
N Correlation Sig.
Pair 1 Before & After 30 .256 .171
Paired Samples Test
Paired Differences t df Sig. (2-
tailed)Mean Std.
Deviation
Std. Error
Mean
95% Confidence Interval
of the Difference
Lower Upper
5

Pair
1
Before -
After 19.50000 19.77241 3.60993 12.11686 26.88314 5.402 29 .000
7. Determine the Statistical Significance and State a Conclusion. (Use the example “Putting it All
Together” on page 138 (K & K)).The p-value was found to be less than 0.0001 and hence much lower than the level of
significance assumed which is 0.05. Thus the test rejected the null hypothesis at 5% level of
significance.
Wilcoxon Matched-Pairs Test
Review pages 137 - 144 in Kellar & Kelvin (K & K).
1. The Wilcoxon Matched-Pairs Test is a test in which the following assumptions must be met if the test
is to be valid:a. Pre-test and post-test observations are recorded on each controlled subject so data consists
of bivariate data points.
b. The two measures follow the same distribution.
c. The measurement scale is ordered, interval or ratio.
2. Develop a (real or fictitious) research question that could be addressed by the Wilcoxon Matched-
Pairs Test:
To check whether a treatment plan for treating joint pains in athletes was successful in reducing pain
or not.
3. State the (real or fictitious) Null and Alternative Hypotheses:
H0: Pain level before treatment = Pain level after treatment (null)
Against
H1: Pain level before treatment > Pain level after treatment (alternate)
4. Make up a set of data to analyze in SPSS that would qualify for the use of the Wilcoxon Matched-
Pairs Test. Over-all have at least 20 pairs of data. Make sure that the data meet all of the necessary
assumptions of the Wilcoxon Matched-Pairs Test.
The pain levels of 30 athletes suffering from acute joint pain who were undergoing the treatment was
recorded before and after the treatment in two separate columns. The pain level was recorded in a
scale of 1 to 10 with 1 being least and 10 being strong pain.
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5. Use the a-level of .05.
The level of significance used was 0.05 or 5%.
6. Enter the data into SPSS.
 Compute the median and interquartile range for each group.
 Follow the Step-by-Step Procedure for Using SPSS to compute the Wilcoxon Matched-Pairs
Statistics (page 144 of K & K). Computing the test statistic, Wilcoxon Matched-Pairs, that
produces SPSS Output similar to page 143 (K & K). Save the data entry file and send your data
entry file (input) from SPSS to the course instructor via email. Cut and paste your SPSS output
results below.
Ranks
N Mean Rank Sum of Ranks
PainAfter - PainBefore
Negative Ranks 20a 15.68 313.50
Positive Ranks 6b 6.25 37.50
Ties 4c
Total 30
a. PainAfter < PainBefore
b. PainAfter > PainBefore
c. PainAfter = PainBefore
Test Statisticsa
PainAfter -
PainBefore
Z -3.531b
Asymp. Sig. (2-tailed) .000
a. Wilcoxon Signed Ranks Test
b. Based on positive ranks.
7. Determine the Statistical Significance and State a Conclusion. (Use the example “Putting it All
Together” on page 144 (K & K)).The p-vale was found to be less than 0.0001 and hence much lower than the value of
significance 0.05. Therefore the null is to be rejected at 5% level and it suggests that the treatment did
indeed reduce pain in the patients.
7

Exercise 3. The One-Way ANOVA and the Kruskal-Wallis
H-Test
The One-Way ANOVA
Review pages 151 – 171in Kellar & Kelvin (K & K).
1. The One-Way ANOVA is a
parametric test in which the following assumptions must be met if the test
is to be valid:a. The sample of data of interest is an independent random sample.
b. The grouping variable has more than 2 categories.
c. The measure of the data of interest is distributed normally.
d. The variable is measured in either interval or ratio scale.
e. The variance is homogeneous among all of the groups.
2. List and define each of the commonly used post hoc tests for analysis of variance models.a.Least significance difference test is equivalent to multiple t-tests where the pooled estimate
of the variance is employed instead of comparing the variance which are common to all
groups.
b.Scheffe Test is based on the formula for t-test or f-ratio test. The critical value determines
whether the resulting F statistic is significant or not.
c.Tukey’s Honestly Significant Difference is the least powerful test of all the others defined
here. The critical values remain the same for all comparisons irrespective of the number of
mean comparisons.
d.Student Newman-Keuls test is similar to Tukey’s HSD but the critical values do not stay
equal and reflect the variable which are being compared.
e.Duncan Test is compared in the same way as Student Newman Keuls however in this case
the critical value is not at stringent as the latter.
f. Bonferroni Correction divides the alpha level by the number of comparisons of groups that
are to be made and the test for significance is carried out accordingly.
g.Tukey’s wholly significant difference makes use of critical values which are the average of
that in Tukey’s HSD and Student Newman-keuls. It is a moderated measure between the two.
3. Develop a (real or fictitious) research question that could be addressed by a One-Way ANOVA:To test whether the scores in Math of students in the tenth grade differs among schools from
three districts A, B, C.
4. State the (real or fictitious) Null and Alternative Hypotheses:
H0: Average math score is equal for all districts A, B, C (null)
Against
H1: At least one inequality among the math scores exists over the districts (alternate)
5. Make up a set of data to analyze in SPSS that would qualify for the use of the One-Way ANOVA.
Over-all have at least 30 data points in each of the three groups for this exercise. Make sure that data
8

meet all of the necessary assumptions of the One-Way ANOVA. If necessary for your final results, pick
and use the appropriate post hoc test.The data consists of average school math scores from 45 schools from over 3 districts A, B and
C. The two variables district and score shows the district from which the data point has been recorded
and the average score of the school in math.
6. State your a-level:
The level of significance used was 0.05 or 5%.
7. Enter the data into SPSS.
 Follow the Step-by-Step Procedures for using SPSS pages 164 – 166 of K & K computing the test
statistics similar to pages 169 & 170 of K & K. Save the data entry file and send your data entry
file (input) from SPSS to the course instructor via email. Cut and paste your SPSS output results
below.
ANOVA
SCore
Sum of Squares df Mean Square F Sig.
Between Groups 5006.944 2 2503.472 2.750 .075
Within Groups 38234.033 42 910.334
Total 43240.978 44
8. Determine the results of your analysis and state a conclusion. (Use the example “Putting it All
Together: on page 171 (K & K).
The p-value was found to be 0.075 which is less than level of significance 5% and hence the alternative
was not accepted since not enough evidence is found to reject the null hypothesis. The scores are thus
the same over the three districts.
Kruskal-Wallis
H-Test
Review pages 171 – 177 in K & K.
1. The Kruskal-Wallis
H-Test is a
non-parametric test in which the following assumptions must be met if
the test is to be valid:a. Measures of the variables of interest constitute of mutually independent samples.
b. The grouping variable has more than two categories.
9

c. The variables of interest are at least ordinal in scale of measurement.
2. Develop a (real or fictitious) research question that could be addressed by the Kruskal-Wallis
H-Test:To test whether the salary of different employee groups, namely minorities, men and women
are different in a company.
3. State the (real or fictitious) Null and Alternative Hypotheses:
H0: mean salary of women= mean salary of men= mean salary of minorities
Against
H1: At least one inequality among the salaries among the groups (alternate)
4. Make up a set of data to analyze in SPSS that would qualify for the use of the Kruskal-Wallis
H-Test.
Over-all have at least 10 – 15 data pints in each of the three groups for this exercise. Make sure that
data meet all of the necessary assumptions of the Kruskal-Wallis
H-Test. If necessary for your final
results, pick and use the appropriate post hoc test.
5. State you’re a-level:
The level of significance used was 0.05 or 5%.
6. Enter the data into SPSS.
 Follow the Step-by-Step Procedures for using SPSS pages 174 – 175 of K & K. Save the data
entry file and send your data entry file (input) from SPSS to the course instructor via email. Cut
and paste your SPSS output results below.
Ranks
Group N Mean Rank
Salary
Minorities 9 15.44
Females 10 16.05
Males 11 15.05
Total 30
Test Statisticsa,b
Salary
Chi-Square .069
df 2
Asymp. Sig. .966
10

a. Kruskal Wallis Test
b. Grouping Variable:
Group
7. Determine the results of your analysis and state a conclusion. (Use the example “Putting it all
together: on page 177 in K & K.
The p-value was found to be 0.966 which is greater than the level of significance 0.05 and hence the
test failed to reject the null hypothesis based on given data. The existing evidence fails to support the
validity of the alternate conjecture that the salary level of the groups in the company are not same.
Exercise 4. Comparing the Means of Three or More Related Groups: Repeated Measures ANOVA and
Friedman’s ANOVA by Rank
Comparing the Means of Three or More Related Groups – Repeated Measures ANOVA
Review pages 215 – 232.
1. What type of data is required for the Repeated Measures ANOVA?
Interval or ratio scaled data is necessary for repeated measures ANOVA. There must be at least three
related measures of the characteristic under scrutiny. These can be of the same variable at different
time or places or of three related variables.
2. Develop a (real or fictitious) research question that could be addressed by a Repeated Measures
ANOVA:
11

3. State the Null and Alternative Hypotheses:
H0: Blood sugar at baseline= after 3 months= after 6 months (End of trial)
Against
H1: Not H0
4. Make up a set of data to analyze in SPSS that would qualify for the use of the Repeated Measures
ANOVA. Over-all have at least 30 data points for each of the three groups. Make sure that the data
meet all of the necessary assumptions of the Repeated Measures ANOVA test.
The dataset consists of three columns recording blood sugar before meal of 30 participants at
baseline, after 3 months of treatment and after 6 months of treatment.
5. What is your alpha level?
The alpha level used in 0.05 or 5% level of significance.
6. Enter the data into SPSS. Follow the Step-by-Step Procedure for Using SPSS to Compute the
Repeated Measures ANOVA (pages 226 & 227 K & K)
 Compute the Repeated Measures ANOVA that produces SPSS output similar to pages 230-231 of
K & K. Save the data entry file (input) and send via email to the course instructor. Cut and paste
your SPSS output results below.
Descriptive Statistics
Mean Std. Deviation N
Baseline 127.4667 24.48044 30
Mid 95.1667 8.55039 30
End 78.9333 6.03972 30
Multivariate Testsa
Effect Value F Hypothesis df Error df Sig. Partial Eta
Squared
Time
Pillai's Trace .849 78.703b 2.000 28.000 .000 .849
Wilks' Lambda .151 78.703b 2.000 28.000 .000 .849
Hotelling's Trace 5.622 78.703b 2.000 28.000 .000 .849
Roy's Largest Root 5.622 78.703b 2.000 28.000 .000 .849
a. Design: Intercept
Within Subjects Design: Time
12

b. Exact statistic
Tests of Within-Subjects Effects
Measure: Fasting
Source Type III Sum of
Squares
df Mean
Square
F Sig. Partial Eta
Squared
Time
Sphericity Assumed 36622.956 2 18311.478 76.692 .000 .726
Greenhouse-
Geisser 36622.956 1.207 30350.983 76.692 .000 .726
Huynh-Feldt 36622.956 1.230 29762.903 76.692 .000 .726
Lower-bound 36622.956 1.000 36622.956 76.692 .000 .726
Error(Time)
Sphericity Assumed 13848.378 58 238.765
Greenhouse-
Geisser 13848.378 34.993 395.749
Huynh-Feldt 13848.378 35.684 388.081
Lower-bound 13848.378 29.000 477.530
7. Determine the statistical significance and state a conclusion. (Use the example “Putting it all
together” page 232 K & K.
The p-value was found to be less than 0.0001 and hence the null was rejected at 5% level of
significance, in favor of the alternative. This suggests that the treatment had positive impact in
reducing blood sugar.
13

Friedman’s ANOVA by Rank
Review pages 233-239 K & K.
1. What type of data is required for the Friedman’s ANOVA by Rank?
There must be at least three related measures of the characteristic under scrutiny. These can be of the
same variable at different time or places or of three related variables. The data must be measured on
at least ordinal scale where it could be ordinal, interval or ratio.
2. Develop a (real or fictitious) research question that could be addressed by a Friedman’s ANOVA by
Rank:
Determine whether listening to a particular type of music, namely, Rock, classical or pop/RnB music
has any impact on effort to improve athletic performance of a student or not.
2. State the Null and Alternative Hypotheses:H0: No significant difference in efforts among the music types
Against
H1: There is significant difference between two or more groups in terms of music being played
3. Make up a set of data to analyze in SPSS that would qualify for the use of the Friedman’s ANOVA
by Rank. Over-all have at least 10 data points for each of the groups. Make sure that the data
meet all of the necessary assumptions of the Friedman’s ANOVA by Rank test.The data consists of three variables, or groups, namely, Classical, Pop and Rock, with measure
of effort recorded when each music type was being played. The effort measure is an ordinal
measure on scale of 1-5 with 1 being minimum effort and 5 being maximum.
5. What is your alpha level?
The level of significance used was 0.05 or 5%.
6. Enter the data into SPSS. Follow the Step-by-Step Procedure for Using SPSS to Compute the
Friedman’s ANOVA by Rank (pages 236 & 237 K & K)
 Compute the Friedman’s ANOVA by Rank that produces SPSS output similar to pages 235 of K &
K. Save the data entry file (input) and send via email to the course instructor. Cut and paste
your SPSS output results below.
Descriptive Statistics
N Mean Std. Deviation Minimum Maximum
Classical 30 3.2333 1.40647 1.00 5.00
Pop_RnB 30 2.7333 1.28475 1.00 5.00
14

Rock 30 3.1000 1.56139 1.00 5.00
Ranks
Mean Rank
Classical 2.12
Pop_RnB 1.82
Rock 2.07
Test Statisticsa
N 30
Chi-Square 1.755
df 2
Asymp. Sig. .416
a. Friedman Test
7. Determine the statistical significance and state a conclusion. (Use the example “Putting it all
together“page 234 K & K.
The p-vale was found to be 0.416 which is greater than the assumed level of significance 0.05 and
hence the null could not be rejected at 5% level of significance. Thus no music type that the student
was listening to was found to have any significant impact on the athletic effort of the students.
15

Exercise 5. Correlation Coefficients
Correlation Coefficients
Review pages 263 to 282 (K & K)
1. What is the difference between the Pearson and the Spearman Correlation Coefficients?
The Spearman’s correlation coefficient is a non-parametric test whereas the Pearson correlation is
parametric. For Pearson’s correlation it is assumed that the data is normally distributed. The former is
primarily used to test for monotonic relationships between the data variables when they are ratio,
interval or even ordinal scale which cannot be tackled using Pearson’s correlation coefficient which
requires only interval scale data. The data must be linearly related for Pearson’s correlation to be
valid however the data need only be monotonic in case of Spearman’s correlation to be valid.
2. Describe how you would assess the strength and direction of relationships with correlation
coefficients.
The strength of the relationship or degree of association between the two variables is represented
by the magnitude of the correlation coefficient itself. If the measure is close to or equal to zero little
or no relationship is indicated respectively. The magnitude being close or equal to one indicates high
correlation between the two variables. The sign of the measure, whether negative or positive
indicates the direction of the relationship. Positive sign signifies that with increase in one variable
corresponds with increase in value of the other and negative sign indicates that the increase in one
corresponds with the decrease in the other. +1 indicates strong positive relationship and -1 indicates
strong negative relationship. So deviation from zero indicates strength and the direction of deviation
is the nature of the association.
3. Develop a (real or fictitious) research question that could be addressed by a correlation procedure.To determine whether weight of adolescent boys in a class of 30 is related to their heights or not.
Then the data consists of heights and weights of 30 adolescent boys in a class.The problem can be
tackled using Pearson’s correlation coefficient. The fictitious dataset is attached.
4. State the Null and Alternative Hypotheses:The null hypothesis (H0) and the alternative hypothesis (H1) can be written as follows:
H0: r2 = 0 vs.H1: r2 not equal to 0
Where r= correlation coefficient
5. The Pearson Correlation to be valid, the following conditions must apply:a. The two variables are measured either in interval or ratio scale.
16

b. The variables are distributed normally.
c. The variables are linearly related.
d. There are no outliers in the data.
6. For the Spearman test of correlation is valid under all the conditions in which the Pearson test of
correlation is valid. In addition, the Spearman test of correlation is valid when:a. One or both variables are on ordinal measurement scale.
b. Either one or both variables are non-normally distributed.
c. The two variables are monotonically related.
d. Outliers do not affect the coefficient as much as Pearson’s correlation coefficient.
7. Choosing between the Pearson and the Spearman Correlation Coefficients:
The Pearson correlation coefficient can be used when the following assumptions are met:a. The data points are from independent random samples.
b. Only two variables are to be compared.
c. The measures are normal.
d. The variables are measured in interval or ratio scale.
e. The variables have linear relationship.
f. Outliers are not that influential in the samples.
g. For each value of a sampled variable the corresponding value of the other variable is
normally distributed.
h. For every value of x variable the corresponding value of the covariate y has equal variance
and vice versa, that is, the assumption of homogeneity is satisfied.
The Spearman correlation coefficient can be used when the following assumptions are met:a. The data points are from independent random samples.
b. Two measures of a variable are to be compared.
c. The variables ought to be measured in either ordinal, interval or ratio scale.
d. The two variables must be monotonically related.
e. Outliers are less likely to have impact on the test.
8. Make up a set of data to analyze in SPSS that would quality for the use of a correlation procedure.
Make sure that the data meets all of the necessary assumptions for either the Pearson or the Spearman
test.
9. What is your alpha level?
The alpha level used in 0.05 or 5% level of significance.
10. Enter the data into SPSS.
 Follow the Step-by-Step Procedure for Using SPSS on pages 273 – 280. Save the data entry file
and send your data entry file (input) from SPSS to the course instructor via email. Cut and paste
your SPSS output results below:The output to the correlation analysis done via SPSS in given as follows:
17

Correlations
inches pounds
inches
Pearson Correlation 1 .377*
Sig. (2-tailed) .040
N 30 30
pounds
Pearson Correlation .377* 1
Sig. (2-tailed) .040
N 30 30
*. Correlation is significant at the 0.05 level (2-tailed).
11. Determine the Statistical Significance and State a Conclusion.The Pearson correlation coefficient is -0.282 however the p-value for the test was found to be
0.131. Since level of significance is 0.05 and the p-value is 0.040, the null is rejected at 5% level of
significance, that is, there is indeed a significant positive correlation between height and weight.
Exercise 6. Chi-Square and McNemar Test
Chi-Squire Test
Review pages 289 – 305 (K& K)
1. Develop a (real or fictitious) research question that could be addressed by the Chi-Square Test.To test whether smoking habits of a typical college student is associated with gender.
2. State the (real or fictitious) Null and Alternative Hypotheses:H0: Gender is not associated with smoking habits
Against
H1: Gender is associated with smoking habits
3. Make up a set of data to analyze in SPSS that would qualify for the use of the Chi-Square Test.
Make sure that the data meets all of the necessary assumptions of the Chi-Square Test.
A data set of responses from 30 college students was designed with two variables, namely, the
gender of the respondent, and smoking habit variable, which is whether he/she is a smoker which is
category 1 or non-smoker which is category 0.
4. What is your alpha-level?
18

The level of significance used was 0.05 or 5%.
5. Enter the data into SPSS. Follow the Step-by-Step Procedure for Using SPSS to compute the Chi-
Square Test. Save the data entry file and send your data entry file (input) from SPSS to the course
instructor via email. Cut and paste your SPSS output results below.
Case Processing Summary
Cases
Valid Missing Total
N Percent N Percent N Percent
Smoker * Gender 30 100.0% 0 0.0% 30 100.0%
Smoker * Gender Crosstabulation
Gender Total
Female Male
Smoker
Non smoker
Count 8 10 18
% within Smoker 44.4% 55.6% 100.0%
% within Gender 47.1% 76.9% 60.0%
% of Total 26.7% 33.3% 60.0%
Smoker
Count 9 3 12
% within Smoker 75.0% 25.0% 100.0%
% within Gender 52.9% 23.1% 40.0%
% of Total 30.0% 10.0% 40.0%
Total
Count 17 13 30
% within Smoker 56.7% 43.3% 100.0%
% within Gender 100.0% 100.0% 100.0%
% of Total 56.7% 43.3% 100.0%
Chi-Square Tests
Value df Asymp. Sig. (2-
sided)
Exact Sig. (2-
sided)
Exact Sig. (1-
sided)
Pearson Chi-Square 2.738a 1 .098
Continuity Correctionb 1.635 1 .201
Likelihood Ratio 2.827 1 .093
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Fisher's Exact Test .141 .100
Linear-by-Linear Association 2.646 1 .104
N of Valid Cases 30
a. 0 cells (0.0%) have expected count less than 5. The minimum expected count is 5.20.
b. Computed only for a 2x2 table
Symmetric Measures
Value Approx. Sig.
Nominal by Nominal Contingency Coefficient .289 .098
N of Valid Cases 30
a. Not assuming the null hypothesis.
b. Using the asymptotic standard error assuming the null hypothesis.
6. Determine the Statistical Significance and State a Conclusion.
The Chi-squared test returned a p-value of 0.098 which suggests that gender does not have any
association with smoking habits.
McNemar Test
Review pages 305 – 309 (K & K)
1. Develop a (real or fictitious) research question that could be addressed by the McNemar Test.To check whether a certain anti-smoking campaign program is effective to convince smokers
to give up smoking or not.
2. State the (real or fictitious) Null and Alternative Hypotheses:H0: The probability of an individual in post program group to be smoker is same as that in pre
program group
Against
H1: The probability of an individual in post program group to be smoker is less than that in pre
program group
3. Make up a set of data to analyze in SPSS that would qualify for the use of the McNemar Test.
Make sure that the data meets all of the necessary assumptions of the McNemar Test.The dataset consists of recorded smoker status of the same individual before the program and
after the program from 30 respondents who participated in the program.
4. What is your alpha-level?
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The level of significance used was 0.05 or 5%.
5. Enter the data into SPSS. Follow the Step-by-Step Procedure for Using SPSS to compute the
McNemar Test. Save the data entry file and send your data entry file (input) from SPSS to the course
instructor via email. Cut and paste your SPSS output results below.
PreSmoke & PostSmoke
PreSmoke PostSmoke
Non Smoker Smoker
Non Smoker 7 5
Smoker 16 2
Test Statisticsa
PreSmoke &
PostSmoke
N 30
Exact Sig. (2-tailed) .027b
a. McNemar Test
b. Binomial distribution used.
5. Determine the Statistical Significance and State a Conclusion.The p-value was found to be 0.027 for the two tailed test. The p-value for the one tailed test is
thus 0.0135 which is less than level of significance 0.05 and thus null is rejected. Therefore the
program is inferred to have indeed had a positive effect in reducing smoking.
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