The problem in the image is asking to perform a hypothesis test comparing two sample means using a t-test, specifically for the case where we want to test if $\mu_1 < \mu_2$.

Here’s the data given:

• Sample sizes:

  • $n_1 = 15$
  • $n_2 = 15$

• Sample means:

  • $\bar{x}_1 = 22.83$
  • $\bar{x}_2 = 25.38$

• Sample standard deviations:

  • $s_1 = 2.9$
  • $s_2 = 2.8$

Steps to Calculate the Test Statistic $t$:

1. Formulate the Hypotheses:

   • Null hypothesis $H_0$: $\mu_1 = \mu_2$
   • Alternative hypothesis $H_a$: $\mu_1 < \mu_2$

2. Calculate the Pooled Standard Deviation $s_p$: $s_p = \sqrt{\frac{(n_1 - 1)s_1^2 + (n_2 - 1)s_2^2}{n_1 + n_2 - 2}}$ Substituting the given values: $s_p = \sqrt{\frac{(15 - 1)(2.9)^2 + (15 - 1)(2.8)^2}{15 + 15 - 2}}$ $s_p = \sqrt{\frac{14(8.41) + 14(7.84)}{28}} = \sqrt{\frac{117.74 + 109.76}{28}} = \sqrt{\frac{227.5}{28}} \approx 2.85$

3. Calculate the Test Statistic $t$: $t = \frac{\bar{x}_1 - \bar{x}_2}{s_p \sqrt{\frac{1}{n_1} + \frac{1}{n_2}}}$ Substituting the values: $t = \frac{22.83 - 25.38}{2.85 \sqrt{\frac{1}{15} + \frac{1}{15}}} = \frac{-2.55}{2.85 \sqrt{0.1333 + 0.1333}} = \frac{-2.55}{2.85 \times 0.5164} = \frac{-2.55}{1.472}$ $t \approx -1.733$

Thus, the test statistic $t$ is approximately -1.733.

Let me know if you'd like further details on this solution or if you have any questions!

Related Questions:

1. What is the critical value for this t-test at a specific significance level?
2. How do we calculate the degrees of freedom for a two-sample t-test?
3. What is the p-value associated with this test statistic?
4. How does pooling variances differ from assuming unequal variances?
5. How would the result change if the sample sizes were unequal?

Tip: Always check the assumption of normality or use non-parametric tests if normality is in doubt!