To help you with your project, we'll first formulate the null and alternative hypotheses and then proceed to calculate the test statistic and p-value. Here are the steps:

1. Formulating the Hypotheses

Null Hypothesis (H₀): There is no difference in the pass rates between students taking online classes and those taking in-person classes. $H_0: p_{\text{online}} = p_{\text{in-person}}$

Alternative Hypothesis (H₁): There is a difference in the pass rates between students taking online classes and those taking in-person classes. $H_1: p_{\text{online}} \neq p_{\text{in-person}}$

2. Data Summary

• Online Class:

  • Passed: $x_1 = 45$
  • Didn't pass: $n_1 - x_1 = 55$
  • Total: $n_1 = 100$

• In-person Class:

  • Passed: $x_2 = 85$
  • Didn't pass: $n_2 - x_2 = 15$
  • Total: $n_2 = 100$

3. Calculating the Test Statistic

The test statistic for comparing two proportions can be calculated using the formula: $z = \frac{\hat{p}_1 - \hat{p}_2}{\sqrt{\hat{p}(1 - \hat{p}) \left( \frac{1}{n_1} + \frac{1}{n_2} \right)}}$

Where:

• $\hat{p}_1 = \frac{x_1}{n_1}$
• $\hat{p}_2 = \frac{x_2}{n_2}$
• $\hat{p} = \frac{x_1 + x_2}{n_1 + n_2}$

First, calculate the sample proportions: $\hat{p}_1 = \frac{45}{100} = 0.45$ $\hat{p}_2 = \frac{85}{100} = 0.85$

Next, calculate the pooled proportion: $\hat{p} = \frac{45 + 85}{100 + 100} = \frac{130}{200} = 0.65$

Then, compute the test statistic: $z = \frac{0.45 - 0.85}{\sqrt{0.65 \times (1 - 0.65) \left( \frac{1}{100} + \frac{1}{100} \right)}} = \frac{-0.40}{\sqrt{0.65 \times 0.35 \times 0.02}}$

$z = \frac{-0.40}{\sqrt{0.00455}} = \frac{-0.40}{0.0675} \approx -5.93$

4. Finding the p-value

The p-value for a two-tailed test can be found using the standard normal distribution. You can use standard normal distribution tables or statistical software to find the p-value corresponding to the calculated z-score.

$p\text{-value} = 2 \times P(Z < -5.93)$

Given that the z-score is very large in magnitude, the p-value will be very small, close to 0.

Summary

• Null Hypothesis (H₀): $p_{\text{online}} = p_{\text{in-person}}$
• Alternative Hypothesis (H₁): $p_{\text{online}} \neq p_{\text{in-person}}$
• Test Statistic (z): $-5.93$
• p-value: Very close to 0

Given that the p-value is extremely small, you would reject the null hypothesis and conclude that there is a significant difference in pass rates between students taking online classes and those taking in-person classes.

Do you want more details or have any questions about this? Here are some questions you might consider asking next:

1. How do I interpret the p-value in hypothesis testing?
2. What are the assumptions of the test for comparing two proportions?
3. Can you explain the steps to calculate the test statistic in more detail?
4. How do I use statistical software to calculate the p-value?
5. What should I include in my report to explain these results?

Tip: Always ensure your data meets the assumptions of the statistical test you are using to ensure valid results.