# 5.5 - Randomization Test Examples in StatKey

5.5 - Randomization Test Examples in StatKey

The following pages contain examples of conducting randomization tests using StatKey

# 5.5.1 - Single Proportion Example: PA Residency

5.5.1 - Single Proportion Example: PA Residency

This example uses data collected from World Campus STAT 200 students at the beginning of the Fall 2016 semester. You can download this Minitab Express file here:

Research question: Are less than half of all World Campus STAT 200 students Pennsylvania residents?

This research question is asking if there is evidence that the population proportion is less than 0.50 which can be translated to the following hypotheses:

$H_0: p=0.50$

$H_a: p \lt 0.50$

Step 1: We are comparing the proportion in one group to 0.50. This is a one sample proportion test.

$H_0: p=0.50$

$H_a: p \lt 0.50$

Step 2: We used StatKey to construct a sampling distribution.

Step 3: $p<0.001$

Step 4: $p \leq 0.05$, reject the null hypothesis

Step 5: There is evidence that the proportion of all World Campus STAT 200 students who are Pennsylvania residents is less than 0.50.

# 5.5.2 - Single Mean Example: Height

5.5.2 - Single Mean Example: Height

Research question: On average, are husbands older than their wives?

Step 1: The data are paired by couple. This is a paired means test.

$H_0: \mu_d=0$

$H_a: \mu_d > 0$

Step 2: We constructed a randomization distribution in StatKey using the built in dataset.

Step 3: $p < 0.001$

Step 4: Reject the null hypothesis

Step 5: There is evidence that in the population, on average, husbands are older than their wives.

# 5.5.3 - Difference in Means Example: Exercise by Biological Sex

5.5.3 - Difference in Means Example: Exercise by Biological Sex

Do males and females differ in terms of how many hours per week they exercise? This example uses a dataset that is built in to StatKey.

Step 1: Hours exercised per week is a quantitative variable and we are comparing two independent groups. We should conduct a hypothesis test for the differences in means.

$H_0: \mu_m = \mu_f$

$H_a: \mu_m \ne \mu_f$

Step 2: We constructed the randomization distribution given that there is not a difference between the means of males and females.

Step 3: $p = 0.114+0.114=0.228$

Step 4: $p>0.05$, we should fail to reject the null hypothesis

Step 5: There is not evidence that the mean number of hours per week exercised by males and females is different in the population. Our results are not statistically significant.

# 5.5.4 - Correlation Example: Quiz & Exam Scores

5.5.4 - Correlation Example: Quiz & Exam Scores

Using the sample data in:

We want to know if there is evidence of a positive relationship between quiz scores and final exam scores in the population of all World Campus STAT 200 students. If there is a positive relationship, then the population correlation would be greater than zero. This can be translated to the following hypotheses:

$H_0: \rho = 0$

$H_a: \rho > 0$

Step 1: We are examining the relationship between two quantitative variables. We should compute and test Pearson's r which is a correlation coefficient.

$H_0: \rho = 0$

$H_a: \rho > 0$

Step 2: We constructed a randomization distribution given that the correlation in the population is 0.

Step 3: $p<0.001$

Step 4: $p\leq 0.05$, we should reject the null hypothesis

Step 5: There is evidence of a positive relationship between quiz scores and final exam scores in the population of all World Campus STAT 200 students.

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