Question

HYPOTHESIS TEST FOR MEANS 8. Test the claim that, historically, the average man's salary is greater than the average woman's salary by at least $15,000. In other words, test if the average wage gap is greater than 15 thousand dollars. Use a 95% significance level. What is the null and alternative hypotheses? a. H, b. What is the test statistic? What is the p-value? c. d. Do you reject or fail to reject the null hypothesis? Explain in layman's terms what the test is telling us e.

year	average male income	average female income	year since 1965	wage gap
1965	42.6	25.5	0	17.1
1970	49.5	29.4	5	20.1
1975	51.7	30.4	10	21.3
1980	51.6	31.1	15	20.5
1985	51.5	33.2	20	18.3
1990	49.3	36.3	25	13
1995	49.3	35.2	30	14.1
2000	51.9	38.3	35	13.6
2005	50.9	39.2	40	11.7
2010	52.8	40.6	45	12.2
2015	57.9	41.3	50	16.6
sample mean	50.8181818	34.5909091	25	16.2272727
sample deviation	3.62	5.11	16.58	3.5

Please help answer questions above with data in table.

Answer 1

Here we have x-16.227, s=3.5, n =11 Hypotheses are H0:11-15 Ha: μ >15 Level of significance α =0.05 Test is one tailed (right tailed) df -n-1-10 Degree of freedom: Test Statistics x-μ = (16.227-15)/(3.5/sqrt(11)) -1.163 Critical value: 1.812 If |t| >1.812, Reject HO Rejection Region: Decision: Since test statistics does not lie in rejection region so we fail to reject the null hypothesis P-value-0.1359 P-value: Decision: Since p-value is not less than level of significance so we fail to reject the nul hypothesis Excel function for critical value TINV(0.1,10) Excel function for p-value =TDIST(1. 163,10,1)

Conclusion: We can not conclude that average wage gap is greater than 15 thousand dollars.