Based on the image provided, let's analyze and work through the hypothesis testing step by step. Here's the breakdown of the problem:

Given Data:

1. Male students:

   • Sample size ($n_1$) = 21
   • Sample mean ($\bar{X}_1$) = 4.3
   • Sample standard deviation ($s_1$) = 0.8

2. Female students:

   • Sample size ($n_2$) = 20
   • Sample mean ($\bar{X}_2$) = 3.9
   • Sample standard deviation ($s_2$) = 0.3

3. Significance level ($\alpha$) = 0.03

4. Hypothesis test type = Two-tailed test.

Step 1: State the Null and Alternative Hypotheses

• Null Hypothesis ($H_0$): $\mu_1 - \mu_2 = 0$ (mean difference is zero).
• Alternative Hypothesis ($H_a$): $\mu_1 - \mu_2 \neq 0$ (mean difference is not zero).

Step 2: Identify the Sampling Distribution

The sampling distribution for the difference in sample means is approximately a $t$-distribution since the sample sizes are small.

Standard Error (SE):

The standard error of the difference in sample means is calculated as:
$SE = \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}$

Substitute the given values:
$SE = \sqrt{\frac{0.8^2}{21} + \frac{0.3^2}{20}}$ $SE = \sqrt{\frac{0.64}{21} + \frac{0.09}{20}}$ $SE = \sqrt{0.03048 + 0.0045} = \sqrt{0.03498} \approx 0.187$

Thus, the standard error is approximately 0.187.

Step 3: Test Statistic

The test statistic $t$ is given by:
$t = \frac{(\bar{X}_1 - \bar{X}_2)}{SE}$

Substitute the values:
$t = \frac{(4.3 - 3.9)}{0.187} = \frac{0.4}{0.187} \approx 2.14$

Step 4: Find the P-value

Since this is a two-tailed test, calculate the probability of obtaining a $t$-value as extreme as 2.14 (or -2.14) under the null hypothesis.

Degrees of freedom ($df$) can be approximated using the formula: $df = \frac{\left(\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}\right)^2}{\frac{\left(\frac{s_1^2}{n_1}\right)^2}{n_1 - 1} + \frac{\left(\frac{s_2^2}{n_2}\right)^2}{n_2 - 1}}$

Substitute the values:
$df = \frac{\left(\frac{0.64}{21} + \frac{0.09}{20}\right)^2}{\frac{\left(\frac{0.64}{21}\right)^2}{20} + \frac{\left(\frac{0.09}{20}\right)^2}{19}}$ Approximating the calculation, $df \approx 30$.

Using a $t$-distribution table or calculator with $t = 2.14$ and $df = 30$, the two-tailed p-value is approximately 0.041.

Step 5: Conclusion

• Compare the p-value to $\alpha = 0.03$:
  Since $p = 0.041 > 0.03$, we fail to reject the null hypothesis.
• Conclusion: There is insufficient evidence to conclude that the mean number of English courses taken by males and females is significantly different.

Would you like further clarification or additional details?
Here are five related questions for exploration:

1. How is the $t$-distribution used in small-sample hypothesis testing?
2. How does changing the significance level ($\alpha$) affect the conclusion?
3. What would happen if the sample sizes were larger?
4. Can this test be conducted using a z-distribution instead of a t-distribution? Why or why not?
5. What are the assumptions behind using the two-sample $t$-test?

Tip: Always check the