Answer to Question #272279 in Statistics and Probability for Glen Lincon

Question

Bluereef real estate agent wants to form a relationship between the prices of houses, how many bedrooms, House size in sq ft and Lot Size in sq ft. The data pertaining to 100 houses were processed using MINITAB and the following is an extract of the output obtained: The regression equation is 𝑃𝑟𝑖𝑐𝑒 = 𝛽 + 𝜙𝐵𝑒𝑑𝑟𝑜𝑜𝑚 + 𝛾𝐻𝑜𝑢𝑠𝑒 𝑆𝑖𝑧𝑒 + 𝜆𝐿𝑜𝑡 𝑆𝑖𝑧𝑒 Predictor Coef SE Coef T P Constant 37718 14177 2.66 ** Bedrooms 2306 6994 0.33 0.742 House Size 74.3 52.98 * 0.164 Lot Size -4.36 17.02 -0.26 0.798 S= 25023 R-Sq=56.0% R-Sq(adj)=54.6% Source DF SS MS F P Regression 3 76501718347 25500572782 *** **** Residual Error 96 60109046053 626135896 Total 99 a) Write out the regression equation. [1] b) Fill in the missing values *, **, *** and ****. [6] d) Is 𝛾 significantly different from -0.5? [4] e) Perform the F test at the 1% level, making sure to state the null and alternative hypotheses. [4] f) Give an interpretation to the term “R-sq” and comment on its value .

Answer 1

a) The estimated regression equation is

price=37718+2306xbedroom+74.3xhousesize – 4.36xlotsize

b) * denotes the T value of house size.

T =coef/ SE coef

T = 74.3/52.98

T = 1.40

Now, ** denotes the P- value of constant.

Here the df = n-4 = 100-4 = 96. The T value of constant is 2.66.

So, the P-value corresponding to 96 df and T=2.66 from the t distribution table is 0.009

P-value = 0.009

Now, *** denotes the F statistic value.

F = MS reg /MS res

F = 25500572782 / 626135896

F = 40.73

Now, **** denotes the P-value of the F statistic.

Here the P-value corresponding to (3,96) df from the F distribution table is approximately 0.

P-value = 0.000

c)Here, ø is the coefficient of variable, bedroom.

The decision rule based on p value is that the regression coefficient is significant if p value is less than level of significance.

Here, P-value of bedroom = 0.742 and level of significance = 0.05

Clearly, 0.742 > 0.05

So, ø is not significant at 5% significance level.

d) Here the test hypothesis are

$H_0 : γ=-0.5 \\ H_1 : γ≠-0.5$

The test statistic is

$t=\frac{coeF -γ}{s.e(coeF)}\\ t=\frac{74.3-(-0.5)}{52.98}=\frac{74.3+0.5}{52.98}\\ t=1.41$

Now, the P-value corresponding to t =1.41 and 96df from the t distribution table is 0.162

The decision rule based on P-value is to Reject the null hypothesis if p value is less than alpha (0.05)

Clearly, 0.162 > 0.05

So, We fail to Reject the null hypothesis.

Thus we conclude that γis not significantly different from -0.5.

e) Here we are asked to carry out a F test. The test hypothesis are

H 0 : The Regression model is not significant

H 1 : The Regression model is significant.

We have already computed the value of F statistic and its P-value.

F = 40.73 and P-value = 0.000

Here level of significance is 0.01 (1%)

The decision rule based on P-value is to Reject the null hypothesis if P-value is less than the level of significance.

Here, 0.000 < 0.01

So, we Reject the null hypothesis.

Thus we conclude that the overall regression model is significant at 1% level of significance.

f) R-sq is called the coefficient of determination. Here the value of R-sq = 56.0%

This can be interpreted as " 56.0% of Variability in the variable , price can be explained by all the independent variables together through this regression model.

Here, only 56.0% of Variability is explained by the model. Rest 44% of Variability is left unexplained. So this cannot be considered as a good model.