Question

How do you conduct a hypothesis test when you have independent random samples of n1 = 233 and n2 = 312 from two populations, and you want to test whether the difference between the means (μ1 - μ2) is less than 0 against the alternative hypothesis (Ha: μ1 - μ2 < 0)?

Answer 1

The hypothesis test will involve conducting a one-tailed t-test for the difference between means. With independent samples of n1 = 233 and n2 = 312 from two populations, the aim is to determine if the difference between the means (μ1 - μ2) is less than 0. This would be done by computing the test statistic and comparing it to the critical value from the t-distribution to assess the evidence against the null hypothesis.

Step-by-step explanation:

To test whether the difference between two population means (μ1 - μ2) is less than 0, a one-tailed t-test for independent samples is appropriate. This involves computing the test statistic t and comparing it to the critical value from the t-distribution.

Given two independent sample sizes, n1 = 233 and n2 = 312, from populations with unknown variances but assumed to be equal, the formula for the test statistic is:

`\[t` =
`\frac{(\bar{x}_1 - \bar{x}_2) - 0}{\sqrt{(s_p^2)/(n_1) + (s_p^2)/(n_2)}}\]`

Here,
`\(\bar{x}_1\)`and
`\(\bar{x}_2\)`are the sample means,
`\(s_p^2\)` is the pooled sample variance, and
`\(n_1\)` and
`\(n_2\)`are the respective sample sizes.

Once t is calculated, it is compared against the critical value from the t-distribution at the specified significance level α (usually 0.05 for one-tailed tests) with degrees of freedom equal to
`\(n_1 + n_2 - 2\)` to determine statistical significance.

If the calculated t-value falls in the rejection region (i.e., it is smaller than the critical value), it provides evidence to reject the null hypothesis in favor of the alternative, suggesting that the mean of the first population is significantly less than the mean of the second population.