Let's break down this hypothesis testing problem and work through the steps as presented in the image:

Problem Summary:

We have two groups:

• StudyFocus users (sample size $n_1 = 16$) with a sample mean $\bar{x}_1 = 491.3$ and sample variance $s_1^2 = 1180.5$.
• Prepara users (sample size $n_2 = 15$) with a sample mean $\bar{x}_2 = 532.3$ and sample variance $s_2^2 = 6817.7$.

The goal is to determine whether there is a statistically significant difference between the mean scores of StudyFocus users and Prepara users on a nationwide standardized test, using a two-tailed test at the $\alpha = 0.10$ significance level.

(a) State the null hypothesis ($H_0$) and alternative hypothesis ($H_1$):

• $H_0$: There is no difference in the population means of the test scores between the users of StudyFocus and Prepara. Mathematically: $H_0: \mu_1 = \mu_2$

• $H_1$: There is a difference in the population means of the test scores between the users of StudyFocus and Prepara. Mathematically: $H_1: \mu_1 \neq \mu_2$

(b) Determine the type of test statistic to use:

Since we are comparing the means of two independent samples, and we are given the sample variances, we use the two-sample t-test for the difference of means. The test statistic formula is: $t = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}}$ where:

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

(c) Find the value of the test statistic:

Let's plug in the values provided: $\bar{x}_1 = 491.3, \quad \bar{x}_2 = 532.3$ $s_1^2 = 1180.5, \quad s_2^2 = 6817.7$ $n_1 = 16, \quad n_2 = 15$

The test statistic formula is: $t = \frac{491.3 - 532.3}{\sqrt{\frac{1180.5}{16} + \frac{6817.7}{15}}}$ First, compute the variances over the sample sizes: $\frac{1180.5}{16} = 73.78125, \quad \frac{6817.7}{15} = 454.51333$ Add these two values: $73.78125 + 454.51333 = 528.29458$ Now, take the square root: $\sqrt{528.29458} = 22.986$ Finally, compute the t-statistic: $t = \frac{491.3 - 532.3}{22.986} = \frac{-41}{22.986} = -1.784$

(d) Find the two critical values:

Since this is a two-tailed test with $\alpha = 0.10$, we split the significance level into two tails: $0.05$ in each tail. The degrees of freedom for this test can be approximated using a formula for unequal variances. We will approximate it as: $df \approx \min(n_1 - 1, n_2 - 1) = \min(16 - 1, 15 - 1) = 14$ Using a t-distribution table or calculator for $df = 14$ and $\alpha = 0.05$ in each tail, the critical values are approximately: $t_{critical} = \pm 1.761$

(e) Conclusion:

Since the calculated test statistic $t = -1.784$ is slightly more extreme than $-1.761$, we reject the null hypothesis at the 0.10 level of significance. This suggests there is evidence of a statistically significant difference between the means of StudyFocus and Prepara users.

Would you like further details or clarifications? Here are some related questions to expand on the topic:

1. How do you calculate degrees of freedom for unequal variances in a two-sample t-test?
2. What is the significance of using a two-tailed test instead of a one-tailed test?
3. How do you interpret a p-value in the context of hypothesis testing?
4. What is the impact of sample size on the outcome of a t-test?
5. How do assumptions of normality affect the validity of a t-test?

Tip: Always check the sample sizes and variances before choosing between a pooled-variance t-test and a Welch's t-test for better accuracy.