To determine if there is a significant difference in test scores between the two groups (in-person instruction and online instruction), we can perform an independent samples t-test. This test is appropriate when comparing the means of two independent groups to assess if the means are statistically significantly different.

Step-by-Step Solution:

1. State the Hypotheses

• Null hypothesis (H₀): There is no significant difference in the mean test scores between the two groups. $H₀: \mu_1 = \mu_2$
• Alternative hypothesis (H₁): There is a significant difference in the mean test scores between the two groups. $H₁: \mu_1 \neq \mu_2$ where $\mu_1$ is the mean of the in-person group and $\mu_2$ is the mean of the online group.

2. Determine the Test Statistic

The formula for the independent samples t-test is:

$t = \frac{\bar{X_1} - \bar{X_2}}{\sqrt{\frac{S_1^2}{n_1} + \frac{S_2^2}{n_2}}}$ where:

• $\bar{X_1} = 80$ (mean of in-person group)
• $\bar{X_2} = 75$ (mean of online group)
• $S_1 = 5$ (standard deviation of in-person group)
• $S_2 = 7$ (standard deviation of online group)
• $n_1 = n_2 = 25$ (sample sizes for both groups)

3. Compute the Test Statistic

Let's compute the t-value.

$t = \frac{80 - 75}{\sqrt{\frac{5^2}{25} + \frac{7^2}{25}}}$ $t = \frac{5}{\sqrt{\frac{25}{25} + \frac{49}{25}}}$ $t = \frac{5}{\sqrt{1 + 1.96}} = \frac{5}{\sqrt{2.96}} = \frac{5}{1.72}$ $t \approx 2.91$

4. Determine the Degrees of Freedom (df)

The degrees of freedom for the independent t-test can be calculated as:

$df = n_1 + n_2 - 2 = 25 + 25 - 2 = 48$

5. Compare the t-value to the Critical Value

At a 5% significance level ($\alpha = 0.05$) for a two-tailed test, we look up the critical value for $df = 48$ in a t-distribution table. For $df = 48$ and $\alpha = 0.05$, the critical value of $t_{\text{critical}}$ is approximately 2.01.

6. Decision

• If $|t_{\text{calculated}}| > t_{\text{critical}}$, we reject the null hypothesis.
• Here, $t = 2.91 > 2.01$.

Since $t_{\text{calculated}} > t_{\text{critical}}$, we reject the null hypothesis. This indicates that there is a significant difference in test scores between the in-person and online groups.

Conclusion:

At the 5% significance level, the difference in test scores between students who received in-person instruction and those who received online instruction is statistically significant.

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Would you like more details on the interpretation or any further clarifications? Here are some related questions you might find interesting:

1. How is the t-distribution affected by sample size?
2. How can we perform this test if the sample sizes were unequal?
3. What is the meaning of the p-value in hypothesis testing?
4. What assumptions are made in an independent samples t-test?
5. How would a different significance level (e.g., 1%) affect the result?

Tip: In hypothesis testing, always ensure the assumptions (normality, equal variances) for the chosen test are met to ensure the results are valid.