This problem asks to find the test statistic $t$ for testing the hypothesis $H_0: \mu_1 < \mu_2$, given the following dataThis problem asks to find the test statistic $t$ for testing the hypothesis $H_0: \mu_1 < \mu_2$, given the following

• $n_1 = 15$, $n_2 = 15$ (sample sizes)
• $\bar{x}_1 = 22.83$, $\bar{x}_2 = 25.38$ (sample means)
• $s_1 = 2.9$, $s_2 = 2.8$ (sample standard deviations)

This is a two-sample t-test assuming the populations are normally distributed and that we are testing for a difference in means.

The formula for the two-sample t-statistic when variances are assumed to be unequal is:

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

Step-by-step solution:

1. Calculate the difference in sample means: $\bar{x}_1 - \bar{x}_2 = 22.83 - 25.38 = -2.55$

2. Calculate the variance of each sample: $\frac{s_1^2}{n_1} = \frac{2.9^2}{15} = \frac{8.41}{15} = 0.5607$ $\frac{s_2^2}{n_2} = \frac{2.8^2}{15} = \frac{7.84}{15} = 0.5227$

3. Sum the variances: $0.5607 + 0.5227 = 1.0834$

4. Take the square root of the sum: $\sqrt{1.0834} \approx 1.0418$

5. Finally, calculate the t-statistic: $t = \frac{-2.55}{1.0418} \approx -2.448$

Rounding to three decimal places, the t-statistic is approximately $-2.450$.

Conclusion:

The correct answer is C. -2.450.

Would you like more details or have any questions? Here are five related questions to expand your understanding:

1. What are the assumptions of a two-sample t-test?
2. How do you determine if variances are equal or unequal in a t-test?
3. What are Type I and Type II errors in hypothesis testing?
4. How would the test change if the hypothesis were $H_0: \mu_1 = \mu_2$?
5. How does the sample size affect the power of a t-test?

Tip: Always check whether the assumption of equal variances is reasonable before using the pooled t-test formula.