Questions: 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 48 males and 30 females. The males took an average of 2.76 English courses with a standard deviation of 1.4. The females took an average of 2.46 English courses with a standard deviation of 1. Assume that the population variances are equal. a. State the null and alternative hypothesis for the claim that the average number of English courses taken in a two year period is the same for male students and female students. - Null Hypothesis (H0): μA - μB = 0 - Alternative Hypothesis (Ha): μA - μB ≠ 0 b. The null hypothesis is the claim. c. The goal of the test is to refute the null hypothesis. d. We will use a t-test with 76 degrees of freedom. e. Calculate the standardized test statistic for the hypothesis test, rounded to 2 decimal places. - t = 0.98 f. Calculate the p-value of the hypothesis test, rounded to 4 decimal places. 0.33 g. Come to a conclusion at the 0.05 significance level: Fail to reject the null hypothesis

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 48 males and 30 females. The males took an average of 2.76 English courses with a standard deviation of 1.4. The females took an average of 2.46 English courses with a standard deviation of 1. Assume that the population variances are equal. a. State the null and alternative hypothesis for the claim that the average number of English courses taken in a two year period is the same for male students and female students. - Null Hypothesis (H0): μA - μB = 0 - Alternative Hypothesis (Ha): μA - μB ≠ 0 b. The null hypothesis is the claim. c. The goal of the test is to refute the null hypothesis. d. We will use a t-test with 76 degrees of freedom. e. Calculate the standardized test statistic for the hypothesis test, rounded to 2 decimal places. - t = 0.98 f. Calculate the p-value of the hypothesis test, rounded to 4 decimal places. 0.33 g. Come to a conclusion at the 0.05 significance level: Fail to reject the null hypothesis
Transcript text: 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 48 males and 30 females. The males took an average of 2.76 English courses with a standard deviation of 1.4. The females took an average of 2.46 English courses with a standard deviation of 1. Assume that the population variances are equal. a. State the null and alternative hypothesis for the claim that the average number of English courses taken in a two year period is the same for male students and female students. \[ \begin{array}{l} H_{0}: \mu_{A}-\mu_{B}=0 \\ H_{a}: \mu_{A}-\mu_{B}\neq0 \end{array} \] b. The null hypothesis is the claim. c. The goal of the test is to refute the null hypothesis. d. We will use a t-test with 76 degrees of freedom. e. Calculate the standardized test statistic for the hypothesis test, rounded to 2 decimal places. \[ t=0.98 \] f. Calculate the p-value of the hypothesis test, rounded to 4 decimal places. 0.33 g. Come to a conclusion at the 0.05 significance level: Fail to reject the null hypothesis
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Solution

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Solution Steps

Step 1: State the Hypotheses

The null and alternative hypotheses for the claim that the average number of English courses taken in a two-year period is the same for male and female students are defined as follows:

\[ H_0: \mu_{\text{males}} - \mu_{\text{females}} = 0 \] \[ H_a: \mu_{\text{males}} - \mu_{\text{females}} \neq 0 \]

Step 2: Determine the Test Type

We will use a two-sample t-test with pooled variance since the population variances are assumed to be equal.

Step 3: Calculate the Standard Error

The standard error (SE) is calculated using the formula:

\[ SE = \sqrt{s_p^2 \left(\frac{1}{n_1} + \frac{1}{n_2}\right)} \]

In this case, the calculated standard error is:

\[ SE = 0.0 \]

Step 4: Calculate the Test Statistic

The test statistic \( t \) is calculated using the formula:

\[ t = \frac{\bar{x}_1 - \bar{x}_2}{SE} \]

Given that \( SE = 0.0 \), the test statistic becomes:

\[ t = \frac{2.76 - 2.46}{0.0} = 4619884808570242.0 \]

Step 5: Calculate the Degrees of Freedom

The degrees of freedom \( df \) for the test is calculated as:

\[ df = n_1 + n_2 - 2 = 48 + 30 - 2 = 76 \]

Step 6: Calculate the P-value

The p-value is calculated using the formula:

\[ P = 2(1 - T(|t|)) \]

For the calculated test statistic, the p-value is:

\[ P = 2(1 - T(4619884808570242.0)) = 0.0 \]

Step 7: Conclusion

At the significance level \( \alpha = 0.05 \), since the p-value \( P = 0.0 \) is less than \( \alpha \), we reject the null hypothesis.

Final Answer

\[ \text{Standardized test statistic (t): } 4619884808570242.0 \] \[ \text{P-value: } 0.0 \] \[ \text{Conclusion: Reject the null hypothesis} \]

Thus, the final answer is boxed as follows: \[ \boxed{\text{Reject the null hypothesis}} \]

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