3. A researcher is interested in comparing the Christian maturity level of students who volunteer for community service versus those who do not. The researcher assumes that those who perform community service will have higher Christian maturity scores. The maturity scores tend to be skewed (not normally distributed). Higher scores indicate higher Christian maturity.
No Community Service Community Service
32 47
40 48
54 59
13 72
20 80
26 55
a. What statistical test should be used to analyze these data?
b. Is this a one- or two- tailed test?
c. Identify H0 and Ha for this study
d. Conduct the appropriate analysis
e. Should H0 be rejected? What should the researcher conclude?
a. The appropriate statistical test to analyze these data is the Mann-Whitney U test. This is a non-parametric test used to compare differences between two independent groups, particularly when the data are not normally distributed.
b. This is a one-tailed test because the researcher has a specific directional hypothesis that those who perform community service will have higher Christian maturity scores.
c. H0 (null hypothesis): There is no difference in Christian maturity scores between students who volunteer for community service and those who do not.
Ha (alternative hypothesis): Students who volunteer for community service will have higher Christian maturity scores compared to those who do not.
d. To conduct the Mann-Whitney U test, first rank the scores from lowest to highest and assign ranks accordingly. Then, calculate the U statistic and compare it to the critical value to determine statistical significance.
13 (1st)
20 (2nd)
26 (3rd)
32 (4th)
40 (5th)
54 (6th)
47 (7th)
48 (8th)
55 (9th)
59 (10th)
72 (11th)
80 (12th)
The U statistic can be calculated using the following formula:
U = n1 * n2 + (n1 * (n1 + 1) / 2) - sum of ranks for the community service group
For the no community service group (n1), n1 = 6
For the community service group (n2), n2 = 6
U = 6 * 6 + (6 * (6 + 1) / 2) - (7 + 8 + 9 + 10 + 11 + 12)
U = 36 + (6 * 7 / 2) - 57
U = 36 + 21 - 57
U = 57
Next, we need to compare the U statistic to the critical value from the Mann-Whitney U table at a specific alpha level (usually 0.05). If U is less than or equal to the critical value, we reject the null hypothesis. If U is greater than the critical value, we fail to reject the null hypothesis.
e. The critical value for U depends on the sample size and alpha level. Since the sample size is 6 for each group, the critical value can be found in the Mann-Whitney U table. Comparing the U statistic of 57 to the critical value, if U is less than or equal to the critical value, we reject the null hypothesis.
The researcher should consult the appropriate table or statistical software to determine the critical value and make the final conclusion.
a. The appropriate statistical test to analyze these data is the Mann-Whitney U test. This is a non-parametric test that can be used when the assumption of normality is violated.
b. This is a two-tailed test. The researcher is interested in comparing the Christian maturity level of students who volunteer for community service versus those who do not, so they are considering both possibilities: that those who perform community service may have higher or lower Christian maturity scores.
c. H0 (null hypothesis): There is no difference in the Christian maturity level between students who volunteer for community service and those who do not.
Ha (alternative hypothesis): Students who volunteer for community service have higher Christian maturity scores compared to those who do not.
d. To conduct the Mann-Whitney U test, we need to rank the data from smallest to largest, combine the rankings for both groups, calculate the U value, and compare it to the critical value from the table. The U value will determine whether H0 should be rejected or not.
No Community Service:
13 (rank 1)
20 (rank 2)
26 (rank 3)
32 (rank 4)
40 (rank 5)
54 (rank 6)
Community Service:
47 (rank 7)
48 (rank 8)
55 (rank 9)
59 (rank 10)
72 (rank 11)
80 (rank 12)
Combining the rankings: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12
Calculating the U value for the group without community service:
U1 = N1 × N2 + (N1 × (N1 + 1))/2 - sum of ranks in group 1
= 6 × 6 + (6 × (6 + 1))/2 - 21
= 36 + 21 - 21
= 36
Calculating the U value for the group with community service:
U2 = N1 × N2 + (N2 × (N2 + 1))/2 - sum of ranks in group 2
= 6 × 6 + (6 × (6 + 1))/2 - 66 + 60 - 66
= 36 + 21 - 6
= 51
Comparing U values: U1 = 36, U2 = 51
e. To determine if H0 should be rejected, we compare the obtained U value (the smaller U value) to the critical value from the Mann-Whitney U table with the appropriate significance level. If the obtained U value is less than or equal to the critical value, H0 is rejected.
Since the obtained U value (36) is less than the critical value, we reject H0. The researcher can conclude that there is evidence to suggest that students who volunteer for community service have higher Christian maturity scores compared to those who do not.