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| 2 |
>Q
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| 1 |
a
Don't use plagiarized sources. Get Your Custom Essay on
economics
Just from $13/Page
n |
d Q2
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| Ho |
use Price
|
| 67
| 0 |
00
Question 1: Calculate the
| 9 |
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| 5 |
% confidence limits for the population mean of house prices.
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| 6
| 8 |
000
It is given that the population standard deviation is equal to $
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| 4 |
5000.
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| 68000 |
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| 69000 |
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| 72000 |
8820 |
lower limit |
11506
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| 1.5 |
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| 75000 |
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| 76000 |
| Upper |
Limit
1
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| 3 |
| 27 |
01.5
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| 76900 |
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| 77000 |
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| 78000 |
If Sigma was not given, use data anlaysis to get this (below) |
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| 79000 |
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| 80000 |
80000
| House Price |
| 8
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| 100 |
0
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| 82000 |
| Mean |
| 123881 |
.5
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| 83000 |
|
| Standard Error |
4457.1477412549 |
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| 84000 |
Median |
106500 |
84000
| Mode |
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| 102000 |
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| 86250 |
Standard Deviation |
44571.4774125491 |
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| 87000 |
Sample
|
| Variance |
1
| 98 |
6616598.73737
| 8
| 95 |
00
Kurtosis |
1.0548503827 |
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| 90400 |
Skewness |
1.1646598578 |
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| 90500 |
Range |
188000 |
Upper
115037.551896077 |
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| 91000 |
Minimum |
| 67000 |
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| 91500 |
Maximum |
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| 255000 |
Lower |
1
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| 32 |
725.448103923
91500
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| Sum |
12388150 |
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| 92500 |
| Count |
100
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| 93500 |
Confidence Level(95.0%) |
8843.9481039231 |
93500
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| 94000 |
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| 95500 |
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| 96000 |
96000
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| 97900 |
Question 2: It is claimed that the average house price is $125K or less. Set up the null and alternate hypotheses and |
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| 98000 |
conduct the test at the alpha of 0.05. It is given that the population standard deviation is equal to $45000. |
98000
98000
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| 99 |
000
Ho = |
Mu LE 125K |
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| 99000 |
99000
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| H1 |
=
Mu GT 125K |
102000
102000
102000
| Xbar |
123881
102000 n 100
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| 103000 |
103000
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| 103500 |
Zstat |
-0.
| 24 |
86666667
103500
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| 105000 |
Z critical |
1.646 |
for the alpha of 0.05 |
105000
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| 108000 |
Accept the null hypothesis. |
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| 112000 |
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| 112500 |
Conclusion: The claim that the mean house price is 125K 0r below is correct. |
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| 114900 |
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| 115500 |
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| 120500 |
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| 122000 |
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| 125500 |
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| 127000 |
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| 128000 |
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| 129900 |
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| 130350 |
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| 132350 |
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| 133000 |
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| 134500 |
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| 135500 |
135500
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| 136500 |
136500
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| 137400 |
137400
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| 139500 |
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| 144000 |
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| 145000 |
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| 149000 |
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| 155000 |
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| 154000 |
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| 155500 |
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| 156500 |
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| 163000 |
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| 165000 |
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| 167000 |
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| 168700 |
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| 169900 |
169900
169900
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| 176000 |
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| 179000 |
179000
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| 179500 |
179500
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| 187500 |
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| 203000 |
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| 220000 |
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| 222000 |
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| 250000 |
250000
255000
255000
Q3 Solution
| t-Test: Two-Sample Assuming Unequal Variances |
|
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|
| Location |
2
|
| Location 3 |
Ho
Mu2 =Mu3 |
Mean
|
| 88377.7777777778 |
| 10131
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| 2.5 |
H1
Mu2 NE Mu3 |
Variance
|
| 267547179.487179 |
| 201860967.741935 |
|
| Observations |
27 32
Two tail test |
| Hypothesized Mean Difference |
0
|
|
| df |
52 |
Tstat |
| -3.2119420179 |
|
| t Stat |
-3.2119420179
| P(T<=t) one-tail |
0.0011318082 |
Tcritical |
| 2.0066468051 |
-2.0066468051 |
| t Critical one-tail |
1.6746891537 |
| P(T<=t) two-tail |
0.0022636164 |
Tstat is outside the acceptance range, hence we reject the null hypothesis. |
| t Critical two-tail |
2.0066468051
| Conclusion: The two means are different. |
| Q4 |
Solution
Q4
| SUMMARY |
HO |
Mu1 =Mu2=Mu3=Mu4=Mu5 |
| Groups |
Count Sum
Average |
Variance
|
| Location 1 |
8
747500 |
93437.5 |
26941964.2857143 |
H1
Not all means are equal. |
Location 2 27
| 2386200 |
88377.7777777778 267547179.487179
Location 3 32
| 3242000 |
101312.5 201860967.741935
|
| Location 4 |
24
3746600 |
156108.333333333 |
458024275.362321 |
|
| Location 5 |
9
1963500 |
218166.666666667 |
1462250000 |
|
| ANOVA |
Since
|
| F |
stat is above
Fcritical |
, we reject the Ho,
| Source of Variation |
| SS |
df
| MS |
F
| P-value |
F crit |
| Between Groups |
161766824850 |
4
40441706212.5 |
107.8140782368 |
1.93306561967325E-34 |
2.4674936234 |
Hence not all the means are equal. |
| Within Groups |
35635068750 |
95
375105986.842105 |
|
| Total |
197401893600 |
99
Q3 and Q4
Location 1 Location 2 Location 3 Location 4 Location 5
86250 67000 75000 120500 165000
| Question 3: Using the t test, test the claim that the average housing prices of location 2 and location 3 |
86250 68000 78000 128000 167000
| are not different. Write the hypotheses, etc. |
| 90000 |
68000
| 81000 |
134500 179500
| 95000 |
69000 87000 135500 220000
96000 72000
|
| 89500 |
136500 222000
| 97000 |
76000 90500 136500 250000
98000 76900 91000 137400 250000
99000 77000 91500 137500 255000
79000 91500 144000 255000
80000 92500 145000
80000 95500 149000
82000 97900 155000
83000 98000 154000
84000 98000 156500
84000 99000 163000
90400 99000 169900
93500 102000 169900
93500 102000 169900
96000 102000 176000
96000 103000 179000
98000 103000 179000
99000 103500 179500
102000 103500 187500
108000 105000 203000
114900 105000
122000 112000
127000 112500
115500
125500
129900
130350
| Question 4: Using the ANOVA, test the claim that the average housing prices of all these locations are different. |
132350
| Wrtie the hypothese,etc.. |
Q5
solution
| SUMMARY OUTPUT |
Ho
| Slope B1 =0 |
|
| Regression |
Statistics
H1
Slope B1 NE 0 |
| Multiple R |
0.3961102897 |
| R Square |
0.1569033616 |
| Adjusted R Square |
0.1483003347 |
equation |
Standard Error
| 41133.9360170468 |
Observations 100
| Yhat = |
27754.9 +29396.5 X |
ANOVA
| It is significant since we reject the Ho (Fstat > Fcritical) |
df SS MS F
| Significance F |
Regression 1
|
| 30858975434.0591 |
30858975434.0591
18.2381576883 |
| 0.0000451375 |
Fcritical
~4 |
| Residual |
98
165816067840.941 |
1692000692.2545 |
for alpha of 0.05; df1 =1 and df2 = 98. |
Total 99
| 196675043275 |
| Coefficients |
Standard Error t Stat P-value
Lower 95% |
Upper 95% |
Lower 95.0% |
Upper 95.0% |
Rsquare = 0.1569. |
| Intercept |
27754.9005880707 |
22881.6048896187 |
1.212978754 |
0.2280536798 |
| -17652.8996222996 |
| 73162.700798441 |
-17652.8996222996 73162.700798441
15.69% of the total variation is explained by the |
|
| Bedrooms |
29396.5135816297 |
6883.4370185067 |
4.27061561 |
0.0000451375
| 15736.5568432443 |
| 43056.4703200152 |
15736.5568432443 43056.4703200152
regression equation. |
| Yhat for X=4 |
145340.95491459 |
Q5
House Price Location Bedrooms
| Bathrooms |
67000 2 2 1
| Question 5: It is believed the price of a house is related |
68000 2 3 1
| the number of bedrooms. |
68000 2 3 1
| (a) Perform a regression analysis to determine such a relationship. |
69000 2 2 1
| (b) Write down the equation. Is the relationship significant? Why? |
72000 2 4 2
| © What is the interpretation of R-Square? |
75000 3 2 1
76000 2 2 1
| (d) What will the predicted price of the house that has 4 bedrooms? |
76900 2 3 1
77000 2 2 3
78000 3 2 1
79000 2 3 2
80000 2 3 1.5
80000 2 3 1
81000 3 2 1
82000 2 3 1.5
83000 2 3 1
84000 2 3 1
84000 2 3 1.5
86250 1 4 2
87000 3 3 2
89500 3 3 2
90400 2 4 2
90500 3 3 1.5
91000 3 3 2
91500 3 4 2
91500 3 4 2
92500 3 3 1.5
93500 2 3 2
93500 2 4 2
94000 1 3 1.5
95500 3 3 2
96000 2 3 2
96000 2 3 2
97900 3 3 2
98000 3 3 2
98000 2 3 2
98000 3 3 2
99000 2 4 2
99000 3 4 2
99000 3 3 2
102000 3 4 2
102000 2 3 1.5
102000 3 4 2
102000 3 3 1.5
103000 3 3 2
103000 3 3 1.5
103500 3 3 2
103500 3 3 2
105000 3 3 2
105000 3 3 1.5
108000 2 3 2
112000 3 4 2
112500 3 3 2
114900 2 5 2
115500 3 4 2
120500 4 3 2
122000 2 3 3
125500 3 4 2.5
127000 2 3 2.5
128000 4 3 2
129900 3 4 2.5
130350 3 3 2
132350 3 3 2
133000 3 3 2
134500 4 3 2
135500 3 3 3
135500 4 3 3
136500 4 3 2
136500 4 3 2
137400 3 4 2.5
137400 4 4 2.5
137500 4 3 2
139500 3 4 2.5
144000 4 4 2.5
145000 4 3 2
149000 4 3 2
155000 4 4 2
154000 4 3 2
155500 3 3 2.5
156500 4 3 2
163000 4 4 2
165000 5 4 2
167000 5 4 2
168700 3 3 2.5
169900 4 4 2.5
169900 4 3 2.5
169900 4 3 2.5
176000 4 4 2.5
179000 4 4 2.5
179000 4 4 2.5
179500 4 3 2.5
179500 5 3 2.5
187500 4 4 2.5
203000 4 4 3
220000 5 4
|
| 3.5 |
222000 5 3 3.5
250000 5 4 2.5
250000 5 4 2.5
255000 5 4 2.5
255000 5 3 2.5
Q3
a
ble>
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A
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| |
utility compa
| n |
y wants to catgotize its customers in three categories – low bills, medium bills and high bills.
| The bills below $
| 5 |
0 are low, above 150 high and in between medium.
| The data was collected and a histogram was plotte
| d. |
It determined that the data is distributed as a Normal
| distribution whose mean and stadard devaition are
|
|
| 10 |
0
and
|
| 25 |
respectively.
| Mu |
100
| Sigma |
25
| a) What is the probability of a customerto be in the low category? |
P (X<50) |
| 0.02275 |
01319
| b) What proportions of the customers are in the medium category? |
P (50 < X<150) |
0.954
| 49 |
98681
| c) What percentage of the customers are in the high category? |
P (X>150 |
0.02275
| d) How much a customer must spend to be in the bottom 5%? |
P (X
| 58.8786593262 |
|
Q4
| In reference to Problem 3, a random sample of 25 customers was taken. |
Mu 100 n 25
| SigmaofXBAR |
5
| sigma |
25
a
| What is the probability that the sample mean is below $95? |
P (XBAR <95) |
0.1586552539 |
|
| b. |
What proportion of the sample mean are above $110?
P (XBAR>110) |
0.0227501319 |
| c. What percentage of the sample mean be between $95 and $110? |
0.8185946141 |
d.
| If the data was not distributed as a Normal Distribution, would you be able to answer the above questions? |
No,theample sizemust be at least30) |
Q5
| Data was collected to study the background of the people who participate in the stock market. |
| 49 customers were selected at random and asked about their annual savings and whether they lived in a suburb. |
| After the sample was collected, the data was analyzed to calculate sample mean annual savings , sample standard deviation and the sample proportion |
| of people living in a suburb |
n 49
| Sample mean = $10K |
Sample Standard Deviation = $3K |
Sample Proportion =
| 0.4 |
| Sample Mean Standard deviation |
0.4285714286 |
t for0.025and48df |
1.6772241961 |
| a. |
Calculate the 90% confidence limits for the Annual Savings. |
10
0.7188103698 |
9.2811896302 |
Upper Limit |
| 10.7188103698 |
| Lower Limit |
| The interval 9.28 and 10.72 contains the true population mean 90% ofthetime |
b.
| Calculate the 95% confidence limits for the population proportion living in suburbs |
Z |
1.96 |
| p |
0.4
upper limit |
0.5371714256 |
| p(1-p) |
0.24 |
Lower Limit
0.2628285744 |
| p (1-p)/n |
0.0048979592 |
| The true percentage of people living in suburbs is in the range 26 and 54 95% ofthetime. |
| SQRT() |
0.0699854212 |
