Question

The average income of 19 families who reside in a large metropolitan East Coast city is $62,245. The standard deviation is $9695. The average income of 12 families who reside in a rural area of the Midwest is $60,213, with a standard deviation of $2079. At α = 0.05, can it be concluded that the families who live in the cities have a higher income than those who live in the rural areas?

calculate test statistics and round to 2 decimal places

Answer 1

Let the people of large metropolitan East Coast city be sample 1

rural area resident be sample 2

The provided sample means are shown below:

Also, the provided sample standard deviations are:

and the sample sizes are n1​=19 and n2​=12.

(1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ho: μ1​ = μ2​

Ha: μ1​ > μ2​

This corresponds to a right-tailed test, for which a t-test for two population means, with two independent samples, with unknown population standard deviations will be used.

(2) Rejection Region

Based on the information provided, the significance level is α=0.05, and the degrees of freedom are df = 29 . In fact, the degrees of freedom are computed as follows, assuming that the population variances are equal:

Hence, it is found that the critical value for this right-tailed test is t_c = 1.699 , for α=0.05 and df = 29.

(3) Test Statistics

Since it is assumed that the population variances are equal, the t-statistic is computed as follows:

(4) Decision about the null hypothesis

Since it is observed that t = 0.712 < t_c = 1.699, it is then concluded that the null hypothesis is not rejected.

Using the P-value approach: The p-value is p = 0.2412, and since p = 0.2412 > 0.05 , it is concluded that the null hypothesis is not rejected.

(5) Conclusion

It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that the population mean μ1​ is greater than μ2​, at the 0.05 significance level.