Question 1171438: "The mean number of English courses taken in a two-year time period by male and female college students is believed to be about the same. An experiment is conducted and data are collected from 29 males and 16 females. The males took an average of four English courses with a standard deviation of 0.7. The females took an average of five English courses with a standard deviation of 1.1. Are the means statistically the same? (Use 𝛼 = 0.05)
NOTE: If we are using a Student's t-distribution for the problem, including for paired data, we may assume that the underlying population is normally distributed. (In general, we must first prove that assumption, though.)"
I need help stating the distribution to use for the test (I have to enter my answer in the form z or tdf, where df is the degrees of freedom. I need to round my answer to two decimal places as well)
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! 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.
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