Question 1171438
Let's conduct a hypothesis test to determine if the mean number of English courses taken by male and female college students is statistically the same.

**1. Define the Problem**

* **Null Hypothesis (H0):** μ_m = μ_f (The means are equal)
* **Alternative Hypothesis (H1):** μ_m ≠ μ_f (The means are not equal)
* **Significance Level (α):** 0.05
* **Male Data:**
    * n_m = 29
    * x̄_m = 4
    * s_m = 0.7
* **Female Data:**
    * n_f = 16
    * x̄_f = 5
    * s_f = 1.1

**2. Choose the Appropriate Test**

Since we are comparing the means of two independent groups with known sample standard deviations but unknown population standard deviations, we'll use a two-sample t-test. Also, since the sample sizes are different we will use the unequal variances t test.

**3. Calculate the Test Statistic (t-score)**

The formula for the t-score in a two-sample t-test with unequal variances is:

* t = (x̄_m - x̄_f) / sqrt((s_m^2 / n_m) + (s_f^2 / n_f))

Plugging in the values:

* t = (4 - 5) / sqrt((0.7^2 / 29) + (1.1^2 / 16))
* t = -1 / sqrt((0.49 / 29) + (1.21 / 16))
* t = -1 / sqrt(0.01689655 + 0.075625)
* t = -1 / sqrt(0.09252155)
* t = -1 / 0.30417355
* t ≈ -3.2875

**4. Calculate the Degrees of Freedom (df)**

For unequal variances, the degrees of freedom are calculated using the Welch-Satterthwaite equation:

* df = ((s_m^2 / n_m + s_f^2 / n_f)^2) / (((s_m^2 / n_m)^2 / (n_m - 1)) + ((s_f^2 / n_f)^2 / (n_f - 1)))

Plugging in the values:

* df = ((0.7^2 / 29 + 1.1^2 / 16)^2) / (((0.7^2 / 29)^2 / (29 - 1)) + ((1.1^2 / 16)^2 / (16 - 1)))
* df = (0.09252155)^2 / (((0.01689655)^2 / 28) + ((0.075625)^2 / 15))
* df = 0.0085599 / ((0.00028549 / 28) + (0.00571914 / 15))
* df = 0.0085599 / (0.000010196 + 0.00038128)
* df = 0.0085599 / 0.000391476
* df ≈ 21.866

We'll round the degrees of freedom down to 21 for the t-table.

**5. Determine the P-value**

Since this is a two-tailed test, we need to find the probability P(|t| > 3.2875) with df = 21.

Using a t-table or a calculator, we find:

* P(|t| > 3.2875) ≈ 0.0033

**6. Make a Decision**

* Compare the P-value (0.0033) to the significance level (0.05).
* Since 0.0033 < 0.05, we reject the null hypothesis.

**7. Conclusion**

There is sufficient evidence to conclude that the mean number of English courses taken by male and female college students is statistically different at the 5% significance level. The female students on average take more English courses than the male students.