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.

Show your tested hypotheses for each questions, decision rules and then provide both statistical and managerial conclusions for each part. Then, provide an overall conclusion. Be clear in the conclusion that you draw from your analysis, and provide useful suggestions to the company’s CEO. Conclusions must be based on the findings of your analysis only.

Notes:

- Use critical value approach

- Use 0.05 level of significance in your analyses, and assume we have normal distributions and unequal variances of populations.

- Use Excel to conduct your analysis and hypothesis testing.

- Each question requires an Excel output. Simply copy your outputs from Excel sheets and paste them into the Word file of your Business Report.

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

Solution :

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

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