Question

Test the indicated claim about the means of two populations. Assume that the two samples are independent simple random samples selected from normally distributed populations. Do not assume that the population standard deviations are equal. Use the traditional method or P-value method as indicated.

A researcher was interested in comparing the salaries of female and male employees at a particular company. Independent simple random samples of 8 female employees and 15 male employees yielded the following weekly salaries (in dollars).

Female: 495, 760, 556, 904, 520, 1005, 743, 660

Male: 722, 562, 880, 520, 500, 1250, 750, 1640, 518, 904, 1150, 805, 480, 970, 605

Use a 0.05 significance level to test the claim that the mean salary of female employees is less than the mean salary of male employees. Use the traditional method of hypothesis testing.

Answer 1

Sample 1 be female and sample 2 be male

The sample size is n = 8 The provided sample data along with the data required to compute the sample mean and sample variance are shown in the table below:

	X	X2
	495	245025
	760	577600
	556	309136
	904	817216
	520	270400
	1005	1010025
	743	552049
	660	435600
Sum =	5643	4217051

The sample mean is computed as follows:

Also, the sample variance is

Therefore, the sample standard deviation s is

The sample size is n = 15. The provided sample data along with the data required to compute the sample mean and sample variance are shown in the table below:

	X	X2
	722	521284
	562	315844
	880	774400
	520	270400
	500	250000
	1250	1562500
	750	562500
	1640	2689600
	518	268324
	904	817216
	1150	1322500
	805	648025
	480	230400
	970	940900
	605	366025
Sum =	12256	11539918

The sample mean is computed as follows:

Also, the sample variance is

Therefore, the sample standard deviation s is

(1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

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) Rejection Region

Based on the information provided, the significance level is α=0.05, and the degrees of freedom are df = 20.889 . In fact, the degrees of freedom are computed as follows, assuming that the population variances are unequal:

Hence, it is found that the critical value for this left-tailed test is t_c = -1.721, for α=0.05 and df = 21.

(3) Test Statistics

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

t = -1.042

(4) Decision about the null hypothesis

Since it is observed that t = -1.042 ≥tc​=−1.721, it is then concluded that the null hypothesis is not rejected.

Using the P-value approach: The p-value is p = 0.1547 , and since p = 0.1547 ≥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 less than μ2​, at the 0.05 significance level.