This scenario involves conducting a two-sample t-test for independent samples, specifically to test if the treatment group (Population 1) has a greater mean decrease in cholesterol than the control group (Population 2). The test assumes unequal population variances.

Given Data:

• Treatment group (Population 1):

  • Sample size ($n_1$) = 32
  • Mean ($\bar{x}_1$) = 20.4
  • Standard deviation ($s_1$) = 4.5

• Control group (Population 2):

  • Sample size ($n_2$) = 42
  • Mean ($\bar{x}_2$) = 18.6
  • Standard deviation ($s_2$) = 2.3

• Significance level ($\alpha$) = 0.01

Step 1: State the hypotheses

• Null Hypothesis ($H_0$): $\mu_1 \leq \mu_2$ (no greater decrease in cholesterol for the treatment group)
• Alternative Hypothesis ($H_A$): $\mu_1 > \mu_2$ (treatment group has a greater decrease)

Step 2: Compute the test statistic

The formula for the test statistic $t$ in a two-sample t-test with unequal variances (Welch's t-test) is:

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

Where:

• $\bar{x}_1 = 20.4$, $\bar{x}_2 = 18.6$ are the sample means
• $s_1 = 4.5$, $s_2 = 2.3$ are the sample standard deviations
• $n_1 = 32$, $n_2 = 42$ are the sample sizes

Let's calculate the value of the test statistic.The value of the test statistic is $t = 2.066$, rounded to three decimal places.

Do you have any questions or would you like to continue with further steps in this hypothesis test?

Here are 5 related questions:

1. How do you find the degrees of freedom for Welch’s t-test?
2. How is the critical value for a one-tailed test determined?
3. How would the test statistic change if the sample sizes were different?
4. What are the implications if the test statistic had been negative?
5. What would happen if the population variances were assumed to be equal?

Tip: When using Welch’s t-test, the degrees of freedom are often non-integer and depend on sample variances. Always calculate or refer to an approximation.