Question

A city commissioner in Brownsville believe his police officers are underpaid compared with a neighboring town. A random sample of 17 police officers in Brownsville has a mean annual income of $35,800 and a standard deviation of $7,800. In Greensville, a random sample of 15 police officers has a mean annual income of $35,100 and a standard deviation of $7,375. Test the claim at α = 0.01 that the mean annual income in Brownsville is less than Greensville. Assume the population variances are equal.

a.What is the standard error?

b.What is the test statistic?

c.Reject or fail to reject?

Answer 1

The answers are:

a.

b.

The t-statistic is computed as follows:

c.

Fail to reject H0.

The test:

T-test for two Means – Unknown Population Standard Deviations - Equal Variance

The following information about the samples has been provided: a. Sample Means : Xˉ1​=35800 and Xˉ2​=35100 b. Sample Standard deviation: s1=7800 and s2=7375 c. Sample size: n1=17 and n2=15 (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 Left-tailed test, for which a t-test for two population means, with two independent samples, with unknown population standard deviations will be used. (2) The degrees of freedom Assuming that the population variances are equal, the degrees of freedom are given by n1+n2-2=17+15-2=30. (3a) Critical Value Based on the information provided, the significance level is α=0.01, and the degree of freedom is 30. Therefore the critical value for this Left-tailed test is tc​=-2.4573. This can be found by either using excel or the t distribution table. (3b) Rejection Region The rejection region for this Left-tailed test is t<-2.4573 (4)Test Statistics The t-statistic is computed as follows: (5) The p-value The p-value is the probability of obtaining sample results as extreme or more extreme than the sample results obtained, under the assumption that the null hypothesis is true. In this case, the p-value is 0.6016 (6) The Decision about the null hypothesis (a) Using traditional method Since it is observed that t=0.2598 > tc​=-2.4573, it is then concluded that the null hypothesis is Not rejected. (b) Using p-value method Using the P-value approach: The p-value is p=0.6016, and since p=0.6016>0.01, it is concluded that the null hypothesis is Not rejected. (7) 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 less than μ2, at the 0.01 significance level.