Question

Hypothesis test (2- sample)

The manager of an health insurance company in Brampton wonders whether the members needed more chiropractic help in 2017 than in 2016, on average. He found a random sample of those who were treated by chiropractic doctors in both years. Data provided in TERMTESTDATA file. Perform an Hypothesis test to check whether annual expenses increased, on average. For the test use Data Analysis t-Test: Paired Two Sample for Means and 2% significance level.

Annual expenses
2016	2017
340	374
560	476
816	850
238	136
374	340
646	748
612	510
136	162
850	816
1054	1190
298	238
272	366
578	510
340	408
476	510
748	680
204	170
774	816
510	570
408	468

Answer 1

	Annual expenses 2017	Annual expenses 2016	Difference = Annual expenses 2017 - Annual expenses 2016
	374	340	34
	476	560	-84
	850	816	34
	136	238	-102
	340	374	-34
	748	646	102
	510	612	-102
	162	136	26
	816	850	-34
	1190	1054	136
	238	298	-60
	366	272	94
	510	578	-68
	408	340	68
	510	476	34
	680	748	-68
	170	204	-34
	816	774	42
	570	510	60
	468	408	60
Average	516.9	511.7	5.2
St. Dev.	270.037	247.558	72.223
n	20	20	20

From the sample data, it is found that the corresponding sample means are: Ij = 516.9 X2 = 511.7 Also, the provided sample standard deviations are: $1 = 270.037 82 = 247.558 and the sample size is n = 20. For the score differences we have D=5.2 sp = 72.223 (1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: Hd = 0 Ha: pp > 0 This corresponds to a right-tailed test, for which a t-test for two paired samples be used.

(2) Rejection Region Based on the information provided, the significance level is a = 0.02, and the degrees of freedom are df = 19. Hence, it is found that the critical value for this right-tailed test is ta = 2.205, for a=0.02 and df = 19. The rejection region for this right-tailed test is R=t:t> 2.205. (3) Test Statistics The t-statistic is computed as shown in the following formula: t= D sp/n 5.2 72.223//20 Do = 0.322 (4) Decision about the null hypothesis Since it is observed that t = 0.322 <tc = 2.205, it is then concluded that the null hypothesis is not rejected. Using the P-value approach: The p-value is p=0.3755, and since p=0.3755 > 0.02, 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 Mi is greater than um, at the 0.02 significance level.

No, annual expenses not increased, on average