Question

Answer the following hypothesis testing questions:
1. Determine if average prices of units “within 5KM of the CBD” exceed average prices
of units “located within 5 to 10KM from the CBD”.
2. Determine if average prices for House within 5 to 10KM of the CBD exceeds the
average price of Unit within 5KM of the CBD.
3. Determine if average prices for “Houses with three bed rooms within 5 to 10KM of
the CBD” exceeds the average price of “Units within 5KM of the CBD”.
4. Test the difference between population means of houses of the following groups:
prices for “One bedroom”, “Two bedrooms” and “Three bedrooms (or more)”
properties.
5. Test the difference between population means of units of the following groups: prices
for “One bathroom”, “Two bathrooms”, “Three bathrooms (or more)” properties.

Notes:
 You must follow the hypothesis testing steps to test the above hypotheses.
 You must clearly specify all population parameters and the chosen test procedures.
 Use 0.05 level of significance in your analyses and assume that we have normal
distributions and unequal variances of populations.
 Use and interpret Excel to conduct hypothesis testing.
 Each part requires an Excel output. The statistical output must be provided.

TYPE	1	House
	2	Unit
PROXIMITY	1	up to 5KM
	2	Between 5KM and 10KM
BEDROOM	1	one bedroom
	2	two bedrooms
	3	three bedrooms and above
BATHROOM	1	one bathroom
	2	two bathrooms
	3	three bathrooms and above

PRICE (in 10,000 AU Dollars)	TYPE	PROXIMITY	BEDROOM	BATHROOM
310	1	1	2	1
307	1	1	2	1
305	1	1	2	1
300	1	1	2	1
290	1	1	2	1
287	1	1	2	1
280	1	1	1	1
279	1	1	1	1
278	1	1	1	1
277	1	1	1	1
277	1	1	2	1
276	1	1	2	1
269	1	1	2	1
268	1	1	1	1
267	1	1	1	1
267	1	1	1	1
266	1	1	1	1
265	1	1	2	1
257	1	1	1	1
256	1	1	3	1
252	1	1	3	1
250	1	1	1	1
249	1	1	2	1
247	1	1	1	1
247	1	1	1	1
247	1	1	2	1
245	1	1	1	1
244	1	1	1	1
243	1	1	3	1
240	1	1	3	1
235	1	1	3	1
230	1	1	1	1
223	1	1	3	1
217	1	1	1	1
213	1	1	3	1
209	1	1	3	1
208	1	1	2	1
207	1	1	2	1
207	1	1	2	1
207	1	1	2	1
205	1	1	2	1
202	1	1	3	1
201	1	1	2	1
199	1	1	3	1
190	1	1	2	2
189	1	2	2	1
188	1	1	2	2
188	1	1	1	2
186	1	1	1	2
185	1	2	2	1
183	1	1	3	1
181	1	1	3	1
179	1	2	2	2
177	1	2	2	2
173	1	1	3	1
171	1	1	3	1
164	1	1	2	2
163	1	1	3	1
159	1	1	2	2
145	1	1	1	2
144	1	1	3	1
143	1	1	3	1
143	1	1	1	2
140	1	1	1	2
139	1	1	1	2
133	1	1	2	2
132	1	1	2	2
130	1	1	3	1
120	1	2	3	2
119	1	2	3	2
118	1	2	2	2
117	1	2	2	2
115	1	1	1	2
112	1	1	1	2
111	1	2	2	2
104	1	2	2	2
85	1	1	2	2
83	1	1	2	2
77	1	2	3	3
74	1	2	3	3
74	1	1	2	2
73	1	1	2	2
72	1	2	1	3
71	1	1	1	3
69	1	1	1	3
69	1	2	1	3
29	1	2	1	3
27	1	2	1	3
199	2	2	3	1
193	2	2	3	1
186	2	2	2	1
185	2	2	2	1
184	2	2	3	1
183	2	2	2	1
183	2	2	3	1
182	2	2	2	1
177	2	2	3	1
175	2	2	3	1
163	2	1	1	2
162	2	1	1	2
161	2	2	2	2
159	2	2	2	2
159	2	2	2	2
157	2	2	2	2
156	2	2	3	1
155	2	2	3	1
141	2	1	2	2
139	2	1	2	2
138	2	2	2	2
137	2	2	1	2
135	2	2	2	2
133	2	2	1	2
129	2	1	3	3
126	2	1	3	3
125	2	1	1	2
124	2	1	1	2
123	2	2	1	2
122	2	2	1	2
119	2	1	3	2
117	2	1	3	2
116	2	2	2	3
111	2	2	2	3
106	2	1	1	3
104	2	1	1	3
99	2	2	2	2
97	2	2	2	2
89	2	1	2	2
87	2	1	2	2
79	2	1	2	3
75	2	1	2	3
71	2	2	3	2
70	2	2	2	2
69	2	2	2	2
69	2	2	3	2
69	2	1	2	3
69	2	2	3	3
68	2	1	2	3
68	2	1	1	3
67	2	1	3	3
66	2	1	3	3
66	2	2	3	3
65	2	1	1	3
65	2	1	2	3
65	2	1	2	3
65	2	2	2	3
64	2	1	2	3
64	2	2	2	3
63	2	1	2	3
60	2	2	2	2
59	2	1	2	3
58	2	1	2	3
57	2	2	2	2
56	2	1	2	3
55	2	1	2	3
55	2	1	2	3
55	2	1	2	3
55	2	1	1	3
54	2	1	1	3
53	2	1	2	3
52	2	1	3	3
51	2	1	1	3
51	2	1	1	3
51	2	1	2	3
51	2	1	3	3
50	2	1	2	3
49	2	1	2	3
48	2	1	3	3
48	2	1	2	3
46	2	1	3	3
45	2	1	2	3
45	2	1	1	3
45	2	1	2	3
44	2	1	2	3
43	2	1	1	3
43	2	1	2	3
43	2	1	2	3
41	2	1	2	3
41	2	1	2	3
40	2	1	3	2
40	2	1	3	3
39	2	1	3	2
39	2	1	3	3
39	2	1	3	3
38	2	1	3	3
38	2	2	2	3
36	2	2	2	3
36	2	1	3	3
34	2	1	3	3
31	2	2	1	3
29	2	2	1	3

Answer 1

1. Using two sample t-test assuming unequal variance at 0.05 level of significance

H0: units “within 5KM of the CBD” are less than average prices
of units “located within 5 to 10KM from the CBD”.

Ha: units “within 5KM of the CBD” exceed average prices
of units “located within 5 to 10KM from the CBD”.

2. Using two sample t-test assuming unequal variance at 0.05 level of significance

Ha: average prices for House within 5 to 10KM of the CBD is less than the
average price of Unit within 5KM of the CBD.

Ha: average prices for House within 5 to 10KM of the CBD exceeds the
average price of Unit within 5KM of the CBD.

Conclusion: Reject Null is favour of the alternate as p-value < 0.05

3. Using two sample t-test assuming unequal variance at 0.05 level of significance

H0: average prices for “Houses with three bed rooms within 5 to 10KM of
the CBD” is less than the average price of “Units within 5KM of the CBD”.

Ha: average prices for “Houses with three bed rooms within 5 to 10KM of
the CBD” exceeds the average price of “Units within 5KM of the CBD”.

Conclusion: Reject Null is favour of the alternate as p-value < 0.05

4. Using Single Factor ANOVA at 0.05 level of significance

H0: population means of houses of the following groups:
prices for “One bedroom”, “Two bedrooms” and “Three bedrooms (or more)” are equal

Ha: atleast one of the population means of houses of the following groups:
prices for “One bedroom”, “Two bedrooms” and “Three bedrooms (or more)” is different

Conclusion: Fail to reject Null as p-value > 0.05

5. Using Single Factor ANOVA at 0.05 level of significance

H0: population means of units of the following groups:
prices for “One bathroom”, “Two bathrooms” and “Three bathrooms (or more)” are equal

Ha: atleast one of the population means of units of the following groups:
prices for “One bathroom”, “Two bathrooms” and “Three bathrooms (or more)” is different

Conclusion: Reject Null in favour of the alternate as p-value < 0.05