Question

The annual maintenance bills of similar buildings in down town Nairobi were related to their

age with the following results.

First row age A. Age (yrs) and second row M. Maintenance bills (000 shs)

A 15 5 25 40 5 20 10 30 445 15 25 35 10 30

M 39 24 115 105 50 86 67 90 140 112 70 186 43 126

a) fit a regression line to this data (5 marks)

b) find the 95 % confidence interval estimate of the maintenance cost of the building for

age in years, assuming the data is homoscedastic (5 marks)

c) Test the significance of the regression coefficients b and a at the .05 significance level

Answer 1

a)

equation of regression line:

$y=ax+b$

where x is age,

y is maintenance cost

$a=\frac{\sum xy-\sum x\sum y}{n\sum x^2-(\sum x)^2}=0.154$

$b=\frac{\sum y\sum x^2-\sum x\sum xy}{n\sum x^2-(\sum x)^2}=81.69$

$y=0.154x+81.69$

b)

$t=a/SE$

$SE=\sqrt{\frac{\sum(y_i-\tilde{y}_i)^2}{(n-2)\sum(x_i-\overline{x})^2}}$

where y_i is the value of the dependent variable for observation i,

$\tilde{y}_i$ is estimated value of the dependent variable for observation i,

x_i is the observed value of the independent variable for observation i,

$\overline{x}$ is the mean of the independent variable,

n is the number of observations.

$n=14$

$\overline{x}=50.7$

$\sum(x_i-\overline{x})^2=14\cdot109.9^2$

estimated values $\tilde{y}_i$ : 84, 82.5, 85.5, 88, 82.5, 85, 83, 86, 150, 84, 85, 87, 83, 86

$\sum(y_i-\tilde{y}_i)^2=21175.5$

$SE=0.816$

$df=n-2=14-2=12$

critical values for 95 % confidence interval:

$t_{crit}=\pm 2.179$

$-2.179<a/0.816<2.179$

$-1.778<a<1.778$

c)

If there is a significant linear relationship between the independent variable X and the dependent variable Y, the slope will not equal zero.

H_o: a = 0

H_a: a ≠ 0

$t=a/SE=0.154/0.816=0.189$

Since $t<t_{crit}$ we accept the null hypothesis. There is no a significant linear relationship between the independent variable X and the dependent variable Y.

using Goodness of fit test:

H₀: Model (regression line) Fits

H₁: Model Doesn't Fit

test statistic:

$\chi^2=\sum \frac{(y_i-\tilde{y}_i)^2}{\tilde{y_i}}=\frac{(39-84)^2}{84}+\frac{(24-82.5)^2}{82.5}+\frac{(115-85.5)^2}{85.5}+\frac{(105-88)^2}{88}+$

$+\frac{(50-82.5)^2}{82.5}+\frac{(86-85)^2}{85}+\frac{(67-83)^2}{83}+\frac{(90-86)^2}{86}+\frac{(140-150)^2}{150}+\frac{(112-84)^2}{84}+\frac{(70-85)^2}{85}+$

$+\frac{(186-87)^2}{87}+\frac{(43-83)^2}{83}+\frac{(126-86)^2}{86}=343.502$

$df=n-1=13$

critical value:

$\chi^2_{crit}=22.362$

since $\chi^2>\chi^2_{crit}$ , H₀ is rejected. The statistical model (regression line) does not fit the observations. So, the regression coefficients b and a does not fit the observations.