Given Data:

• Sample 1 (population 1):

  • Sample size: $n_1 = 49$
  • Sample mean: $\bar{x}_1 = 16$
  • Sample standard deviation: $s_1 = 5$

• Sample 2 (population 2):

  • Sample size: $n_2 = 64$
  • Sample mean: $\bar{x}_2 = 19$
  • Sample standard deviation: $s_2 = 6$

• Significance level: $\alpha = 0.01$

(a) Distribution of the test statistic:

We are comparing the means of two independent populations. Since the population standard deviations are unknown but sample sizes are relatively large ($n_1 = 49$ and $n_2 = 64$), the appropriate test is the Student's t-distribution.

Answer: The correct option is:

• The Student's t. We assume that both population distributions are approximately normal with unknown standard deviations.

(b) Hypotheses:

The claim is that the population means are different, which means we are performing a two-tailed test. This leads to the following hypotheses:

• Null hypothesis ($H_0$): $\mu_1 = \mu_2$ (The population means are equal)
• Alternative hypothesis ($H_1$): $\mu_1 \neq \mu_2$ (The population means are different)

Answer: The correct option is:

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

(c) Compute $\bar{x}_1 - \bar{x}_2$:

The difference between the sample means is: $\bar{x}_1 - \bar{x}_2 = 16 - 19 = -3$

Next, we calculate the standard error (SE) for the difference between two independent sample means:

$SE = \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}} = \sqrt{\frac{5^2}{49} + \frac{6^2}{64}} = \sqrt{\frac{25}{49} + \frac{36}{64}} = \sqrt{0.5102 + 0.5625} = \sqrt{1.0727} \approx 1.036$

Now, we compute the test statistic $t$:

$t = \frac{\bar{x}_1 - \bar{x}_2}{SE} = \frac{-3}{1.036} \approx -2.896$

Summary of answers:

• $\bar{x}_1 - \bar{x}_2 = -3$
• The test statistic is approximately $t = -2.896$.

Would you like more details on how to interpret this result or any questions on the significance testing?

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Here are 5 related questions to deepen your understanding:

1. What would be the decision rule for this two-tailed test at the 0.01 significance level?
2. How do you calculate degrees of freedom for unequal variances in a t-test?
3. What is the p-value associated with a test statistic of -2.896 in a two-tailed test?
4. How would the test change if the sample sizes were smaller?
5. Why is the Student’s t-distribution preferred for unknown population standard deviations?

Tip: Always ensure that assumptions like normality or equal variances are valid before applying a specific statistical test to avoid misleading conclusions.