Hypothesis Testing Examples For Statistical Analysis

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

H₀: (Null hypothesis): People living in Winnipeg make the same amount of money people living in the rest of Canada make.

H₁: (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

H₀: (Null hypothesis): Students of University of Manitoba enrolled in Distance Education and regular programs have equal GPAs.

H₁: (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 two-sided impression possibility. That is, the students enrolled in distance education could be having higher or lower GPAs than the regular students hence a two-tailed test is used.

Scenario three: Do males laugh more than females?

The populations being compared here are males and females..

H₀: (Null hypothesis): Male and females laugh equally.

H₁: (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

1. Critical value = 5%

1. 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

H₀: (Null hypothesis): Students who skateboard to school and those who do not skateboard have the same coolness.

H₁: (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 cut-off 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 t-test

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 one-sample t-test since there was only one sample being tested against a mean of 25.

Hypothesis test

H₀: (Null hypothesis): Mean body mass index among the patients is 25 kg/m3.

H₁: (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.

One-Sample Test
	Test Value = 25
t	df	Sig. (2-tailed)	Mean Difference	95% Confidence Interval of the Difference
Lower	Upper
BMI	-.936	29	.357	-.77000	-2.4517	.9117

Table 3

The survey’s calculated p-value 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

T-test for repeated measures (paired sample t-test)

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 t-test 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

H₀: (Null hypothesis): There is no significant difference in heart rate before and after walking for 300 metres.

H₁: (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. (2-tailed)
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 p-value 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 t-test 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

H₀: (Null hypothesis): There is no difference in mean aptitude test scores between the two groups.

H₁: (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	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
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 p-value under Lavene’s test for equality of variances with the level of significance (0.05). As can be observed, the p-value 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.

H₀: (Null hypothesis): All the mean times in all the four groups are equal.

H₁: (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 p-value 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

H₀: (Null hypothesis): The average amount of time taken to walk between the normal and obese groups is the same.

H₁: (Alternative hypothesis): The average amount of time taken to walk between the normal and obese groups is not the same.

Hypothesis test 2

H₀: (Null hypothesis): The average amount of time taken to walk between the males and females is the same.

H₁: (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 Between-Subjects Effects
Dependent Variable:   time in minutes
Source	Type III Sum of Squares	df	Mean Square	F	Sig.
Corrected Model	6869.907a	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 t-test

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

NON-PARAMETRIC TESTS

Question 1

Chi-square 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?

Chi-square 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

H₀: (Null hypothesis): There was no effect on consumer preference of the three washing powders after the campaigns

H₁: (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
Chi-Square	4.200a
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 p-value (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

Chi-square 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 chi-square 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	R-PROG	Total
Bsc. Mathematics	8	6	14
Bsc. Statistics	9	7	16
Total	17	13	30

Table 14

Hypothesis test

H₀: (Null hypothesis): The two variables (degree course and analysis software) are independent

H₁: (Alternative hypothesis): The two variables (degree course and analysis software) are dependent

Results table

Chi-Square Tests
	Value	df	Asymp. Sig. (2-sided)	Exact Sig. (2-sided)	Exact Sig. (1-sided)
Pearson Chi-Square	.002a	1	.961
Continuity Correctionb	.000	1	1.000
Likelihood Ratio	.002	1	.961
Fisher’s Exact Test				1.000	.626
Linear-by-Linear 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 p-value (.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

H₀: (Null hypothesis): Advertising did not have an effect on the likelihood of purchase.

H₁: (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 Statisticsa
	LIKELIHOOD
Mann-Whitney U	107.500
Wilcoxon W	227.500
Z	-.213
Asymp. Sig. (2-tailed)	.831
Exact Sig. [2*(1-tailed Sig.)]	.838b
a. Grouping Variable: PRODUCT
b. Not corrected for ties.

Table 17

As can be observed from the results table, the p-value (.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.