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Hypothesis Testing:
Hypothesis testing:
House prices depend on a number of factors; the price of a house depends on the size, number of rooms, location and other components such as a pool and garage, in this paper we establish whether the location of houses affect the house prices, the following table summarizes the mean price of houses and the standard deviation in the five townships:
1
2
3
4
5
total
2953.7
4549
5719.8
6290.9
3702.4
mean
196.9133
227.45
228.792
216.9275862
231.4
standard deviation
35.78405
44.19337
48.65464
49.97713689
48.79833
The mean price is summarized in the table below:
From the chart it is evident that township one the mean price is the lowest while in township 5 the mean price is the highest, we now try and establish whether there is a difference in the price of houses in the townships, our area of interest will be to determine whether the price of houses in township five is higher than price of houses in township three, we state our hypothesis and test the null hypothesis at 95% level.
Hypothesis:
The price of houses in township five is greater than the price of houses in township three:
Testing the hypothesis:
The first step is to state the null and the alternative hypothesis, we assume that the mean price of houses in township five is B5 while the mean price of houses in township three is B3, therefore we state the null and alternative hypothesis as follows:
Null hypothesis:
B5 = B3 = 0
Alternative hypothesis:
B5 > B3
The next step will be to determine the mean and the standard deviation in both township 3 and township five; we will test the hypothesis at 95% level of test,
We use the following formula:
Zc = (X1 –X2) – (μ1- μ2)
−−−−−−−−−−−−−−−−
[(σ12/ n1 + σ22 /n2)]½
Where X1 and X2 is the mean of the samples and μ1and μ2 are the population means of the two samples σ1 and σ2 is the standard deviation of the samples, in our case we rewrite the above formula as follows:
Zc = (B5 –B3)
−−−−−−−−−−−−−−−−
[(σ52/ n5 + σ32 /n3)]½
We ignore (μ1- μ2) because the population means are the same and therefore the value is zero, we substitute the values as follows:
Zc = (231.4 –228.792)
−−−−−−−−−−−−−−−−
[((48.79833)2/ 16 + (48.65464)2 /25)]½
Zcal = 0.167124
Therefore the value of our Z calculated is 0.167124
We test the hypothesis at 95% level one tail test, we obtain the critical value of a 95% level of test from the Z table, the value is 0.0199 therefore
Z crit = 0.0199
Decision:
We make our decision using the following diagram:
In our case
Zcal = 0.167124
Z crit =0.0199
Zcal > Z crit
Because the Z calculated is greater than the Z critical we reject the null hypothesis, when we reject the null hypothesis this means we accept the alternative hypothesis, the alternative hypothesis stated that B5 > B3. This means that the cost of houses in township five is greater than the cost of township in township three. This results shows that in some areas the cost of housing is higher than in other areas which may be caused by other factors.
References:
Michael Crawley (2005) Statistics: An Introduction, John Wiley and Sons, New York
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