Question

A researcher estimated the following regression model with a sample size of 36. ?? = ?0 + ?1??1 + ?2??2 + ? The researcher wanted to find out whether there is heteroscedasticity in the error variance and so applied the White’s heteroscedasticity test. The result is as follows: ?? = −5.8417 + 2.5629??1 + 0.6918??2 − 0.4081??1 2 − 0.0491??2 2 + 0.0015??1??2 R 2 = 0.2143 What conclusion can you assist the researcher to draw at 5 percent and 10 percent?

Answer 1

Answer,

For given regression equation of ui, we need to check whether the obtained equation is overall significant.

We can set the null hypothesis as

H0: b1=b2=b3=b4=b5=0

H1: at least one bi =! 0 (for i =1, 2, 3, 4, 5)

Symbol "=!" means "not equal to".

Now we will calculate F-statistic.

Formula for F-statistic = R-sqaured/(1-R-squared)*(n-p)/p

n= sample size, p =number of parameters

n=36 and p=5

F-statistic = 0.2143/(1-0.2143)*(36-5)/(5-1) = 0.2143*31/(0.7857*4)= 2.1138

Now, we will check for critical value of F-statistic for 5% and 10% significance level

Critical F-statistic for 10% level (4,31) = 2.136

Critical F-statistic for 5% significance level(4,31) = 2.679

For 5% significance, Critical F=2.68 is greater than Calculated F =2.11, therefore we cannot reject the null hypothesis

For 10% significance, Critical F=2.136 is greater than Calculated F =2.11, therefore we cannot reject the null hypothesis

hence, we can admit that there is no heteroskedasticity in this model.