Step-by-Step Guidance for Statistics Lab Worksheet

Q1. Suppose Bay Ridge is looking to improve student engagement with online homework. They randomly assign 100 students to use a new platform and 100 students to use the old platform. After one semester, 65 students using the new platform and 55 students using the old platform completed all assignments. Is there evidence that the new platform increases completion rates?

Background

Topic: Hypothesis Testing for Proportions

This question tests your ability to compare two proportions and determine if there is a statistically significant difference between them using hypothesis testing.

Key Terms and Formulas

• Proportion ($\hat{p}$): The fraction of students who completed all assignments in each group.

• Null Hypothesis ($H_0$): $p_1 = p_2$ (no difference in completion rates).

• Alternative Hypothesis ($H_a$): $p_1 > p_2$ (new platform increases completion rates).

• Test Statistic for Two Proportions:

$z = \frac{\hat{p}_1 - \hat{p}_2}{\sqrt{\hat{p}(1-\hat{p})\left(\frac{1}{n_1} + \frac{1}{n_2}\right)}}$

Where $\hat{p}$ is the pooled proportion:

$\hat{p} = \frac{x_1 + x_2}{n_1 + n_2}$

Step-by-Step Guidance

1. Calculate the sample proportions for each group: $\hat{p}_1 = \frac{65}{100}$ for the new platform, $\hat{p}_2 = \frac{55}{100}$ for the old platform.

2. Compute the pooled proportion: $\hat{p} = \frac{65 + 55}{100 + 100}$.

3. Set up the null and alternative hypotheses: $H_0: p_1 = p_2$, $H_a: p_1 > p_2$.

4. Calculate the standard error using the pooled proportion: $SE = \sqrt{\hat{p}(1-\hat{p})\left(\frac{1}{100} + \frac{1}{100}\right)}$.

5. Plug the values into the test statistic formula to find $z$.

Try solving on your own before revealing the answer!

Final Answer: z ≈ 1.41, p-value ≈ 0.079

Since the p-value is greater than 0.05, there is not enough evidence to conclude that the new platform increases completion rates at the 5% significance level.