Scenario one: Do University of Manitoba students enrolled in Distance Education have different GPAs than students enrolled in regular programs?
The populations being compared here are people living in Winnipeg and the people living in the rest of Canada.
Hypothesis test
H_{0}: (Null hypothesis): People living in Winnipeg make the same amount of money people living in the rest of Canada make.
H_{1}: (Alternative hypothesis): People living in Winnipeg make more money than people living in the rest of Canada.
The appropriate hypothesis for this test would be a one–tailed test since the alternative hypothesis has given a one direction outcome. That is, people living in Winnipeg can only be making MORE money THAN people living in the rest of Canada.
The populations being compared here are University of Manitoba students enrolled in regular programs and those enrolled in distance education.
Hypothesis test
H_{0}: (Null hypothesis): Students of University of Manitoba enrolled in Distance Education and regular programs have equal GPAs.
H_{1}: (Alternative hypothesis): Students of University of Manitoba enrolled in Distance Education and regular programs have different GPAs.
The appropriate hypothesis for this test would be a two–tailed test since the alternative hypothesis has given a twosided impression possibility. That is, the students enrolled in distance education could be having higher or lower GPAs than the regular students hence a twotailed test is used.
Scenario three: Do males laugh more than females?
The populations being compared here are males and females..
H_{0}: (Null hypothesis): Male and females laugh equally.
H_{1}: (Alternative hypothesis): Males laugh more than females.
The appropriate hypothesis for this test would be a one–tailed test since the hypothesis has given a direction to the outcome, that is, male laugh more than females.
Question two
Study A
 Critical value = 5%
 Critical value = 0.5% on both sides
 Critical value = 5% on both sides
Student Participant 
1 
2 
3 
4 
5 
6 
7 
8 
9 
10 
Skateboarder Coolness 
5 
5 
7 
7 
7 
3 
5 
8 
9 
10 
Table 1
Step 1
Population 1: High school students who skateboard to school
Population 2: High school students who do not skateboard to school
Hypothesis test
H_{0}: (Null hypothesis): Students who skateboard to school and those who do not skateboard have the same coolness.
H_{1}: (Alternative hypothesis): Students who skateboard to school are cooler than the rest.
Step 2
Calculating the standard error;
Step 3
The study sets the level of significance to be 5%. Since this is a directional hypothesis, only the positive side of the normal curve will be used. For 5% level of significance, the value of Z_{score} will be 1.64 as read from the normal tables.
Step 4
Determining the sample score of the study we have;
Step 5
Decision rule
The least Z score cut off to reject the null hypothesis was 1.64. The survey’s calculated Z score is 5. This value is greater than the cutoff value of 1.64. The decision therefore is that the alternative hypothesis is statistically significant at 0.05 level of significance.
Scenario two: Does body size affect walking pace?
Question 1
One sample ttest
A research was conducted to establish whether the mean body mass index (BMI) in kg/m3 was 25 among 30 patients who had signs of high blood pressure in a hospital.
The research employed a onesample ttest since there was only one sample being tested against a mean of 25.
Hypothesis test
H_{0}: (Null hypothesis): Mean body mass index among the patients is 25 kg/m3.
H_{1}: (Alternative hypothesis): Mean body mass index among the patients is not 25 kg/m3.
Data
BMI in Kg/m3 

27.4 
25.6 
15.9 
20.9 
28.5 
20.9 
21 
34.2 
20 
20.8 
23.7 
25.8 
18.1 
26.1 
30 
29.2 
24.8 
18.1 
25.4 
21.5 
27 
28.3 
29.3 
21 
22.4 
24.3 
15.1 
28.3 
25.4 
27.9 
Table 2
Since the sample was 30, the central limit theorem was applied and an assumption of normality arrived at.
OneSample Test 

Test Value = 25 

t 
df 
Sig. (2tailed) 
Mean Difference 
95% Confidence Interval of the Difference 

Lower 
Upper 

BMI 
.936 
29 
.357 
.77000 
2.4517 
.9117 
Table 3
The survey’s calculated pvalue score is .36. This value is greater than the level of significance which is 0.05. The decision therefore is to accept the null hypothesis and reject the alternative hypothesis. The conclusion is that the null hypothesis is statistically significant at 0.05 level of significance.
Question 2
Ttest for repeated measures (paired sample ttest)
A research is conducted to test whether heart beat rates are different before having a walk of 300 metres and after.
The appropriate test for the research is a paired sample ttest since we are comparing two variables. That is heart beat rate before and after 300 meters of walk. Since the sample was more than 30, the central limit theorem was applied and an assumption of normality arrived at.
Hypothesis test
H_{0}: (Null hypothesis): There is no significant difference in heart rate before and after walking for 300 metres.
H_{1}: (Alternative hypothesis): There is a significant difference in heart rate before and after walking for 300 metres.
Data (first 10)
Heart Rate at rest (bpm) 
Heart Rate Post 300m Walk (bpm) 
67 
83 
81 
97 
87 
105 
70 
90 
73 
98 
70 
104 
87 
98 
80 
106 
73 
107 
61 
85 
Table 4
Results table
Paired Samples Test 

Paired Differences 
t 
df 
Sig. (2tailed) 

Mean 
Std. Deviation 
Std. Error Mean 
95% Confidence Interval of the Difference 

Lower 
Upper 

Pair 1 
heart beat rate at rest – heart beat rate after walking 400 meters 
23.22500 
12.85119 
2.03195 
27.33501 
19.11499 
11.430 
39 
.000 
Table 5
The survey’s calculated pvalue score is .00. This value is less than the level of significance which is 0.05. The decision therefore is to reject the null hypothesis and accept the alternative hypothesis. The conclusion is that the alternative hypothesis is statistically significant at 0.05 level of significance.
Question 3
A sample of 46 intern accountants employed in a bank had been trained differently. The first 23 had gone through PC based training while the other 23 went through traditional lectures. The human resources department wanted to establish whether there is a significant difference in their mean aptitude scores.
The appropriate test was an independent samples ttest since the two samples are independent. They have been drawn from different populations.
Since the sample was drawn from a normally distributed population an assumption of normality was arrived at.
Scenario three: Do males laugh more than females?
Hypothesis test
H_{0}: (Null hypothesis): There is no difference in mean aptitude test scores between the two groups.
H_{1}: (Alternative hypothesis): There is a significant difference in mean aptitude test scores between the two groups.
Independent Samples Test 

Levene’s Test for Equality of Variances 
ttest for Equality of Means 

F 
Sig. 
t 
df 
Sig. (2tailed) 
Mean Difference 
Std. Error Difference 
95% Confidence Interval of the Difference 

Lower 
Upper 

PC_training 
Equal variances assumed 
11.158 
.002 
2.570 
44 
.014 
10.23478 
3.98200 
18.25997 
2.20960 
Equal variances not assumed 
2.570 
24.771 
.017 
10.23478 
3.98200 
18.43970 
2.02987 

Traditional_lectures 
Equal variances assumed 
12.152 
.001 
1.057 
44 
.296 
4.40826 
4.17203 
12.81643 
3.99991 
Equal variances not assumed 
1.057 
25.322 
.301 
4.40826 
4.17203 
12.99517 
4.17865 
Table 6
Compare the pvalue under Lavene’s test for equality of variances with the level of significance (0.05). As can be observed, the pvalue calculated (0.002) is less than the level of significance (0.05). The decision is therefore to reject the null hypothesis. The conclusion is that the alternative hypothesis is statistically significant at 0.05 level of significance.
Question 1
The appropriate test was an analysis of variance since the variables were more than two.
Dependent variable: Walking time
Independent variables: Body size
Overweight 
Heavyweight 
Obese 
normal 
298.4 
343.37 
289.21 
288.4 
305.97 
363.72 
290.53 
295.97 
345.31 
299.41 
308.4 
335.31 
359.47 
276.85 
294.22 
266.85 
260.63 
368.37 
350.44 
358.37 
288.9 
337.46 
319.5 
327.46 
347.55 
377.72 
244.19 
357.72 
330.91 
311.94 
285.09 
310.91 
416.5 
277.38 
248.31 
386.5 
321.76 
247.8 
299.28 
301.76 
Table 7
Statistics 

overweight 
heavyweight 
obese 
normal 

N 
Valid 
10 
10 
10 
10 
Missing 
36 
36 
36 
36 

Skewness 
.607 
.249 
.037 
.266 

Std. Error of Skewness 
.687 
.687 
.687 
.687 

Kurtosis 
1.005 
1.276 
.543 
.686 

Std. Error of Kurtosis 
1.334 
1.334 
1.334 
1.334 
Table 8
As can be observed above, the kurtosis values are near zero indicating normality.
H_{0}: (Null hypothesis): All the mean times in all the four groups are equal.
H_{1}: (Alternative hypothesis): At least one mean is different
Table of results
ANOVA 

Sum of Squares 
df 
Mean Square 
F 
Sig. 

overweight 
Between Groups 
16978.747 
9 
1886.527 
. 
. 
Within Groups 
.000 
0 
. 

Total 
16978.747 
9 

heavyweight 
Between Groups 
17812.225 
9 
1979.136 
. 
. 
Within Groups 
.000 
0 
. 

Total 
17812.225 
9 

obese 
Between Groups 
8742.267 
9 
971.363 
. 
. 
Within Groups 
.000 
0 
. 

Total 
8742.267 
9 
Table 8
As can be observed, the pvalue calculated (0.00) is less than the level of significance (0.05). The decision is therefore to reject the null hypothesis. The conclusion is that the alternative hypothesis is statistically significant at 0.05 level of significance.
Question 2
Factorial anova
A study is conducted to establish the effect of gender (male and female) and body size (obese and normal) on time used to walk 200 meters. It is hypothesized that gender and body size affect the speed at which people walk. So the research question is, does gender or body size affect speed of walking?
Dependent variable: Walking time
Independent variables: Gender and body size
Factorial anova was appropriate since the study focused on establishing whether there is a difference between mean time taken to walk 200 meters between the obese and normal individuals.
Hypothesis test 1
H_{0}: (Null hypothesis): The average amount of time taken to walk between the normal and obese groups is the same.
H_{1}: (Alternative hypothesis): The average amount of time taken to walk between the normal and obese groups is not the same.
Hypothesis test 2
H_{0}: (Null hypothesis): The average amount of time taken to walk between the males and females is the same.
H_{1}: (Alternative hypothesis): The average amount of time taken to walk between the males and females is not the same.
Descriptive Statistics 

Dependent Variable: time_in_minutes 

gender 
body_size 
Mean 
Std. Deviation 
N 
male 
obese 
327.5400 
43.43417 
10 
normal 
320.4020 
44.48748 
10 

Total 
323.9710 
42.94778 
20 

female 
obese 
292.2100 
32.97211 
9 
normal 
320.7755 
35.84174 
11 

Total 
307.9210 
36.69412 
20 

Total 
obese 
310.8047 
41.89179 
19 
normal 
320.5976 
39.15307 
21 

Total 
315.9460 
40.25702 
40 
Table 9
Tests of BetweenSubjects Effects 

Dependent Variable: time in minutes 

Source 
Type III Sum of Squares 
df 
Mean Square 
F 
Sig. 
Corrected Model 
6869.907^{a} 
3 
2289.969 
1.463 
.241 
Intercept 
3954871.018 
1 
3954871.018 
2527.318 
.000 
gender 
3039.549 
1 
3039.549 
1.942 
.172 
body_size 
1142.071 
1 
1142.071 
.730 
.399 
gender * body_size 
3170.827 
1 
3170.827 
2.026 
.163 
Error 
56334.558 
36 
1564.849 

Total 
4056079.462 
40 

Corrected Total 
63204.465 
39 

a. R Squared = .109 (Adjusted R Squared = .034) 
Table 10
The factorial ANOVA shows that there was no main effect of body size on time taken (.24, p > .05). There was also no main effect of gender on time taken (.17, p > .05).
Question one: One sample ttest
There was also a significant interaction between the two factors (gender and body size) (.16, p > .05).
Plot of means of interaction effects
Interpretation
 The main effect of time: Normal people do not have difference in walk time regardless of gender.
PART FOUR
NONPARAMETRIC TESTS
Question 1
Chisquare test for goodness of fit
A research was conducted on 30 individuals after 3 companies advertised their brands of new washing powders. The 30 individuals were asked which washing powder they would prefer. It is assumed that before the campaign preference for the washing powders was the same across the three. The research question therefore is, did the advertising campaign have an effect on the consumers’ preference on the washing powders?
Chisquare test for goodness of fit was appropriate in this research since it was comparing effect of a factor on proportions before and after.
The washing powders were A, B and C. They were chosen as in the table below;
Powder A 
Powder B 
Powder C 
9 
16 
5 
Table 11
Hypothesis test 1
H_{0}: (Null hypothesis): There was no effect on consumer preference of the three washing powders after the campaigns
H_{1}: (Alternative hypothesis): There was a significant effect on consumer preference of the three washing powders after the campaigns.
powder_type 

Observed N 
Expected N 
Residual 

Powder A 
9 
10.0 
1.0 
Powder B 
15 
10.0 
5.0 
Powder C 
6 
10.0 
4.0 
Total 
30 
Table 12
Test Statistics 

powder_type 

ChiSquare 
4.200^{a} 
df 
2 
Asymp. Sig. 
.122 
a. 0 cells (0.0%) have expected frequencies less than 5. The minimum expected cell frequency is 10.0. 
Table 13
As can be observed from the results table, the pvalue (1.22) is greater than significance level (0.05). Therefore there is sufficient evidence to reject the null hypothesis. The conclusion is that the campaigns had a significant effect on consumers’ preference of the three washing powders.
Question 2
Chisquare test for independence
For a research to predict the demand for analysis package courses in a college, it was presumed or hypothesized that students’ undergraduate degree courses affected their choice of analysis packages. That is some degree courses would tend to have a preference for certain analysis packages. Therefore, are the two variables independent?
The chisquare test for independence is appropriate in this question since it involved two qualitative variables (the analysis software and degree courses).
The variables are degree course and analysis software.
The below contingency table gives a summary of the data;
Analysis software 

Degree course 
SPSS 
RPROG 
Total 
Bsc. Mathematics 
8 
6 
14 
Bsc. Statistics 
9 
7 
16 
Total 
17 
13 
30 
Table 14
Hypothesis test
H_{0}: (Null hypothesis): The two variables (degree course and analysis software) are independent
H_{1}: (Alternative hypothesis): The two variables (degree course and analysis software) are dependent
Results table
ChiSquare Tests 

Value 
df 
Asymp. Sig. (2sided) 
Exact Sig. (2sided) 
Exact Sig. (1sided) 

.002^{a} 
1 
.961 

Continuity Correction^{b} 
.000 
1 
1.000 

Likelihood Ratio 
.002 
1 
.961 

Fisher’s Exact Test 
1.000 
.626 

LinearbyLinear Association 
.002 
1 
.961 

N of Valid Cases 
30 

a. 0 cells (0.0%) have expected count less than 5. The minimum expected count is 6.07. 

b. Computed only for a 2×2 table 
Table 15
As can be observed from the results table, the pvalue (.63) is greater than significance level (0.05). Therefore there is sufficient evidence to reject the alternative hypothesis. The conclusion is that the two variables are independent.
Question 3
For a research to assess the effectiveness of advertising, two products were compared (product X and product Y). The products were advertised after which 30 participants allowed to rate their chances of purchasing the products. The rating was between 1 and 5 with 5 indicating “will definitely buy”. 15 participants rated X and the other rated Y.
This test was appropriate since the variables involved ordinal scale and they were also not normally distributed.
Dependent variable: Likelihood of purchasing
Independent variables: advertising
Hypothesis test
H_{0}: (Null hypothesis): Advertising did not have an effect on the likelihood of purchase.
H_{1}: (Alternative hypothesis): Advertising had a significant effect have an effect on the likelihood of purchase.
Table of results
Ranks 

PRODUCT 
N 
Mean Rank 
Sum of Ranks 

LIKELIHOOD 
X 
15 
15.17 
227.50 
Y 
15 
15.83 
237.50 

Total 
30 
Table 16
Test Statistics^{a} 

LIKELIHOOD 

MannWhitney U 
107.500 
Wilcoxon W 
227.500 
Z 
.213 
Asymp. Sig. (2tailed) 
.831 
Exact Sig. [2*(1tailed Sig.)] 
.838^{b} 
a. Grouping Variable: PRODUCT 

b. Not corrected for ties. 
Table 17
As can be observed from the results table, the pvalue (.83) is greater than significance level (0.05). Therefore there is sufficient evidence to reject the alternative hypothesis. The conclusion is that the two variables are independent.