The given image provides a hypothesis testing problem with a test statistic of $1.50$, rounded to two decimal places. To find the test statistic without rounding, we need to understand the context of this question.

This problem involves comparing two sample means, where:

• Population 1: Children treated with the herb
• Population 2: Children given a placebo

We are tasked with testing if the herb has an effect on the mean number of days with fever. The hypotheses are:

• $H_0: \mu_1 = \mu_2$ (null hypothesis, meaning no difference in the mean number of days with fever between the two groups)
• $H_1: \mu_1 \neq \mu_2$ (alternative hypothesis, meaning there is a difference in the mean number of days with fever)

The test statistic formula for comparing two independent means (assuming equal variances) is:

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

Where:

• $\bar{x}_1$ and $\bar{x}_2$ are the sample means for populations 1 and 2, respectively
• $s_1^2$ and $s_2^2$ are the sample variances for populations 1 and 2, respectively
• $n_1$ and $n_2$ are the sample sizes for populations 1 and 2, respectively

From the information provided:

• $\bar{x}_1 = 0.91$, $s_1 = 1.59$, and $n_1 = 315$ for the herb group
• $\bar{x}_2 = 0.76$, $s_2 = 1.05$, and $n_2 = 389$ for the placebo group

Let’s compute the unrounded test statistic using this data.The unrounded test statistic is approximately $1.44$.

Do you have any questions or would you like further details?

Here are 5 related questions for further exploration:

1. How does changing the sample sizes affect the test statistic?
2. What is the critical value for this test at a 0.01 significance level?
3. How would you interpret the result if the test statistic was much higher or lower?
4. What assumptions must be met for this t-test to be valid?
5. How do we compute the p-value based on the test statistic?

Tip: Always ensure the assumptions of normality and equal variances are met when performing t-tests with small sample sizes.