Part 1: Identifying Null and Alternative Hypotheses

The hypotheses for this test are as follows:

• Null Hypothesis ($H_0$): There is no difference in the mean number of days with fever between the two groups. Mathematically, $\mu_1 = \mu_2$.
• Alternative Hypothesis ($H_1$): The herb affects the number of days with fever, so the means are not equal. Mathematically, $\mu_1 \neq \mu_2$.

This corresponds to option D:

• $H_0: \mu_1 = \mu_2$
• $H_1: \mu_1 \neq \mu_2$

Part 2: Calculating the Test Statistic

We use the formula for the two-sample $t$-test statistic assuming unequal variances:

$t = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}}$

Where:

• $\bar{x}_1 = 0.88$, $\bar{x}_2 = 0.63$ are the sample means
• $s_1 = 1.57$, $s_2 = 1.16$ are the standard deviations
• $n_1 = 325$, $n_2 = 363$ are the sample sizes

Step 1: Calculate the numerator ($\bar{x}_1 - \bar{x}_2$):

$\bar{x}_1 - \bar{x}_2 = 0.88 - 0.63 = 0.25$

Step 2: Calculate the denominator:

$\text{Denominator} = \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}$

$s_1^2 = 1.57^2 = 2.4649, \quad s_2^2 = 1.16^2 = 1.3456$

$\frac{s_1^2}{n_1} = \frac{2.4649}{325} = 0.007584, \quad \frac{s_2^2}{n_2} = \frac{1.3456}{363} = 0.003708$

$\text{Denominator} = \sqrt{0.007584 + 0.003708} = \sqrt{0.011292} = 0.1063$

Step 3: Calculate the test statistic:

$t = \frac{\bar{x}_1 - \bar{x}_2}{\text{Denominator}} = \frac{0.25}{0.1063} = 2.35$

The test statistic is $t = 2.35$ (rounded to two decimal places).

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Conclusion

With $t = 2.35$, the next step would be to compare this test statistic to the critical value or calculate the $p$-value to decide whether to reject the null hypothesis at the 0.01 significance level.

Would you like to proceed with this calculation or have additional details clarified?