# 11.4 - Non-Parametric

#### Case-Study: Fourth Article

Jin is surprised in reading his final article. The article compares two segments of town residence, senior citizens and teenagers measuring the number of times they violate mandated recycling requirements. Jin has learned enough statistics to understand that comparing two independent groups with a quantitative response variable requires a two sample t test for independent groups, however, this article reports using a Mann Whitney U test. Can we help Jin understand why the author’s use a Mann Whitney U?

• Two independent sample t test
• Normality assumption for t test
• Normality and non-normality
• Median and mean

Reading further into Jin’s article, we see that there were only 10 responses for each group. When the researcher looked at a histogram of the responses, they found that the responses were not normally distributed.

Pointing out the small number of responses and a non-normal distribution, Jin realizes that the assumption of normality for the t test is not met. We can explain to Jin that the Mann Whitney is a type of nonparametric test that does not require the assumption of normality. We also let Jin know that the Mann Whitney is one of many nonparametric alternatives to the many techniques in our introductory statistics calls (such as Fishers exact for chi square, Krusko Wallis for ANOVA, and one sample Wilcoxen for one sample t tests).

The good news is that the interpretation of most of the nonparametric tests is very similar to that of the parametric tests. Significant differences exist when test statistics are large and p values are small. Jin just has to be careful in noting that the nonparametric tests typically refer to the median, not the mean as the measure of central tendency being compared.

##### Mann-Whitney: Teenager, Senior
###### Method

$$\eta_1 \colon$$ Median of Teenager

$$\eta_2 \colon$$ Median of Senior

Difference: $$\eta_1 - \eta_2$$

Sampl N Median
Constant 10 5.5
Senior 10 11.5
##### Estimation for Difference
Difference CI for Difference Achieved Confidence
-5 (-7, -2) 95.48%
##### Test

Null Hypothesis: $$H_0\colon \eta_1 -\eta_2 = 0$$

Alternate Hypothesis: $$H_1\colon \eta_1 -\eta_2 \ne 0$$

Method W-Value P-Value
Not Adjusted for ties 63.00 0.002