Question

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?

Answer 1

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?

Solution:

The student's t test for equality means can be used.

Let $\mu_{1}$ and $\mu_{2}$ denote the mean scores of the no community service and community service groups respectively.

H0: $\mu_{1} = \mu_{2}$ (the mean scores are the same in both groups)

H1: $\mu_{1} < \mu_{2}$ (the mean score of the community service is greater than that of the non community service group)

$$\begin{array}{l} n_{1} = n_{2} = 6, \sum x_{1} = 185, \sum x_{2} = 361, \sum x_{1}^{2} = 6785, \sum x_{2}^{2} = 22603 \\ \bar{x}_{1} = \frac{\sum x_{1}}{n_{1}} = \frac{185}{6} = 30.83, \bar{x}_{2} = \frac{\sum x_{2}}{n_{2}} = \frac{361}{6} = 60.17 \\ s_{1}^{2} = \frac{\sum x_{1}^{2} - (\sum x_{1})^{2} / n_{1}}{n_{1} - 1} = \frac{6785 - 185^{2} / 6}{5} = 216.17 \\ s_{2}^{2} = \frac{\sum x_{2}^{2} - (\sum x_{2})^{2} / n_{2}}{n_{2} - 1} = \frac{22603 - 361^{2} / 6}{5} = 176.57 \\ \end{array}$$ $$t = \frac{\bar{x}_{1} - \bar{x}_{2}}{\sqrt{\frac{(n_{1} - 1)s_{1}^{2} + (n_{2} - 1)s_{2}^{2}}{n_{1} + n_{2} - 2} \left(\frac{1}{n_{1}} + \frac{1}{n_{2}}\right)}} = \frac{30.83 - 60.17}{\sqrt{\frac{5 \times 216.17 + 5 \times 176.57}{6 + 6 - 2} \left(\frac{1}{6} + \frac{1}{6}\right)}} = -3.6257$$

Degrees of freedom = 6+6-2=10

The p-value is 0.0023 (left tail test)

Since the p-value is less than 0.01, the t is significant at the 1% level.

So we reject the null hypothesis.

Conclusion: those who perform community service will have higher Christian maturity scores.

When the assumption of normality is not satisfied, a non-parametric alternative is Wilcoxon Rank-Sum Test.

Wicoxon Rank-Sum test

The pooled data is ranked.

$$W_{1} = \text{sum of the ranks for the first group} = 4 + 5 + 8 + 1 + 2 + 3 = 23$$ $$W_{2} = \text{sum of the ranks for the second group} = 6 + 7 + 10 + 11 + 12 + 9 = 55$$ $$U _ {1} = W _ {1} - \frac {n _ {1} (n _ {1} + 1)}{2} = 23 - \frac {6 \times 7}{2} = 23 - 21 = 2$$

The critical region is $U_{1} \leq 4$

The computed value falls in the critical region. So we reject the null hypothesis.

**Answer:** Those who perform community service will have higher Christian maturity scores.