Step 1 :Given that the sample means are \(\bar{x}_1 = 5318\) and \(\bar{x}_2 = 5284\), the sample standard deviations are \(s_1 = 142\) and \(s_2 = 190\), and the sample sizes are both 395.
Step 2 :We are asked to perform a hypothesis test to determine if there is a significant difference between the means of two populations. The null hypothesis (H0) is that the means are equal, and the alternative hypothesis (Ha) is that the means are not equal.
Step 3 :We can use a two-sample t-test to test this hypothesis. The formula for the test statistic for a two-sample t-test is \[t = \frac{{\bar{x}_1 - \bar{x}_2}}{{\sqrt{{s_1^2/n_1 + s_2^2/n_2}}}}\] where \(\bar{x}_1\) and \(\bar{x}_2\) are the sample means, \(s_1\) and \(s_2\) are the sample standard deviations, and \(n_1\) and \(n_2\) are the sample sizes.
Step 4 :We can calculate the test statistic using the given data and then compare it to the critical value for a t-distribution with \(n_1 + n_2 - 2\) degrees of freedom to determine whether to reject the null hypothesis.
Step 5 :Substituting the given values into the formula, we get \(t_{stat} = 2.8488013249525865\).
Step 6 :The degrees of freedom is \(df = n_1 + n_2 - 2 = 788\).
Step 7 :The critical value for a t-distribution with 788 degrees of freedom is \(crit_{val} = 1.962979031749005\).
Step 8 :Since the test statistic is greater than the critical value, we reject the null hypothesis. There is sufficient evidence that the population means are different.
Step 9 :Final Answer: \(\boxed{\text{Reject H0. There is sufficient evidence that the population means are different.}}\)