Is there a significant difference in test scores between 25 students who received in-person instruction and 25 students who received online instruction? The mean test score for the in-person group is 80 (SD = 5) and for the online group is 75 (SD = 7). Level of significance 5%

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.

---

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.

Determine if there is a significant difference in test scores between 25 students receiving in-person instruction and 25 students receiving online instruction using an independent samples t-test at a 5% significance level. Includes step-by-step calculations, hypothesis testing, and interpretation of results.

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