Question

Subject 4

George also believes that the mean of the average room rate in August for 4 stars hotels is smaller than the mean of the average room rate in August for 5 stars hotels.

4.1 State the null and the alternative hypotheses.

4.2 Test the null hypothesis at α = 5%. Use a Student-t statistic and unequal variances in the two populations.

4.3 On the basis of your results, do you agree with George? Explain briefly.

STARS	Total_Rooms	Region_ID	ARR_MAY	ARR_AUG	L_COST
5	412	1	95	160	2.165.000
5	313	1	94	173	2.214.985
5	265	1	81	174	1.393.550
5	204	1	131	225	2.460.634
5	172	1	90	195	1.151.600
5	133	1	71	136	801.469
5	127	1	85	114	1.072.000
4	322	1	70	159	1.608.013
4	241	1	64	109	793.009
4	172	1	68	148	1.383.854
4	121	1	64	132	494.566
4	70	1	59	128	437.684
4	65	1	25	63	83.000
3	93	1	76	130	626.000
3	75	1	40	60	37.735
3	69	1	60	70	256.658
3	66	1	51	65	230.000
3	54	1	65	90	200.000
2	68	1	45	55	199.000
1	57	1	35	90	11.720
4	38	1	22	51	59.200
4	27	1	70	100	130.000
3	47	1	60	120	255.020
3	32	1	40	60	3.500
3	27	1	48	55	20.906
2	48	1	52	60	284.569
2	39	1	53	104	107.447
2	35	1	80	110	64.702
2	23	1	40	50	6.500
1	25	1	59	128	156.316
4	10	1	90	105	15.950
3	18	1	94	104	722.069
2	17	1	29	53	6.121
2	29	1	26	44	30.000
1	21	1	42	54	5.700
1	23	1	30	35	50.237
2	15	1	47	50	19.670
1	8	1	31	49	7.888
1	20	1	35	45
1	11	1	40	55
1	15	1	40	55	3.500
1	18	1	35	40	112.181
3	23	1	40	55
4	10	1	57	97	30.000
2	26	1	35	40	3.575
5	306	2	113	235	2.074.000
5	240	2	61	132	1.312.601
5	330	2	112	240	434.237
5	139	2	100	130	495.000
4	353	2	87	152	1.511.457
4	324	2	112	211	1.800.000
4	276	2	95	160	2.050.000
4	221	2	47	102	623.117
4	200	2	77	178	796.026
4	117	2	48	91	360.000
3	170	2	60	104	538.848
3	122	2	25	33	568.536
5	57	2	68	140	300.000
4	62	2	55	75	249.205
3	98	2	38	75	150.000
3	75	2	45	70	220.000
3	62	2	45	90	50.302
5	50	2	100	180	517.729
4	27	2	180	250	51.000
3	44	2	38	84	75.704
3	33	2	99	218	271.724
3	25	2	45	95	118.049
2	42	2	28	40
2	30	2	30	55	40.000
1	44	2	16	35
3	10	2	40	70	10.000
2	18	2	60	100	10.000
1	18	2	16	20
2	73	2	22	41	70.000
2	21	2	55	100	12.000
1	22	2	40	100	20.000
1	25	2	80	120	36.277
1	25	2	80	120	36.277
1	31	2	18	35	10.450
3	16	2	80	100	14.300
2	15	2	30	45	4.296
1	12	2	40	65
1	11	2	30	50
1	16	2	25	70	379.498
1	22	2	30	35	1.520
4	12	2	215	265	45.000
4	34	2	133	218	96.619
2	37	2	35	95	270.000
2	25	2	100	150	60.000
2	10	2	70	100	12.500
5	270	3	60	90	1.934.820
5	261	3	119	211	3.000.000
5	219	3	93	162	1.675.995
5	280	3	81	138	903.000
5	378	3	44	128	2.429.367
5	181	3	100	187	1.143.850
5	166	3	98	183	900.000
5	119	3	100	150	600.000
5	174	3	102	211	2.500.000
5	124	3	103	160	1.103.939
4	112	3	40	56	363.825
4	227	3	69	123	1.538.000
4	161	3	112	213	1.370.968
4	216	3	80	124	1.339.903
3	102	3	53	91	173.481
4	96	3	73	134	210.000
4	97	3	94	120	441.737
4	56	3	70	100	96.000
3	72	3	40	75	177.833
3	62	3	50	90	252.390
3	78	3	70	120	377.182
3	74	3	80	95	111.000
3	33	3	85	120	238.000
3	30	3	50	80	45.000
3	39	3	30	68	50.000
3	32	3	30	100	40.000
2	25	3	32	55	61.766
2	41	3	50	90	166.903
2	24	3	70	120	116.056
2	49	3	30	73	41.000
2	43	3	94	120	195.821
4	9	3	100	180
2	20	3	70	120	96.713
2	32	3	19	45	6.500
2	14	3	35	70	5.500
2	14	3	50	80	4.000
1	13	3	25	45	15.000
1	13	3	30	50	9.500
2	53	3	55	80	48.200
3	11	3	95	120	3.000
1	16	3	25	31	27.084
1	21	3	16	40	30.000
1	21	3	16	40	20.000
1	46	3	19	23	43.549
1	21	3	30	40	10.000

Answer 1

4.1 State the null and the alternative hypotheses.

Let µ₁ be 4 STARS Hotels.

Let µ₂ be 5 STARS Hotels.

The hypothesis being tested is:

H₀: µ₁ = µ₂

H₁: µ₁ < µ₂

4.2 Test the null hypothesis at α = 5%. Use a Student-t statistic and unequal variances in the two populations.

The output is:

4 STARS Hotels	5 STARS Hotels
131.29	213.91	mean
106.73	96.79	std. dev.
28	23	n
	48	df
	-82.627	difference (4 STARS Hotels - 5 STARS Hotels)
	28.533	standard error of difference
	0	hypothesized difference
	-2.896	t
	.0028	p-value (one-tailed, lower)

The p-value from the output is 0.0028.

Since the p-value (0.0028) is less than the significance level (0.05), we can reject the null hypothesis.

4.3 On the basis of your results, do you agree with George? Explain briefly.

Since the p-value (0.0028) is less than the significance level (0.05), we can reject the null hypothesis.

Therefore, we can conclude that the mean of the average room rate in August for 4 stars hotels is smaller than the mean of the average room rate in August for 5 stars hotels.