(Eighteenth in a series)
Last week, we discussed the violation of the homoscedasticity assumption of regression analysis: the assumption that the error terms have a constant variance. When the error terms do not exhibit a constant variance, they are said to be heteroscedastic. A model that exhibits heteroscedasticity produces parameter estimates that are not biased, but rather inefficient. Heteroscedasticity most often appears in cross-sectional data and is frequently caused by a wide range of possible values for one or more independent variables.
Last week, we showed you how to detect heteroscedasticity by visually inspecting the plot of the error terms against the independent variable. Today, we are going to discuss three simple, but very powerful, analytical approaches to detecting heteroscedasticity: the Goldfeld-Quandt test, the Breusch-Pagan test, and the Park test. These approaches are quite simple, but can be a bid tedious to employ.
Reviewing Our Model
Recall our model from last week. We were trying to determine the relationship between a census tract’s median family income (INCOME) and the ratio of the number of families who own their homes to the number of families who rent (OWNRATIO). Our hypothesis was that census tracts with higher median family incomes had a higher proportion of families who owned their homes. I snatched an example from my college econometrics textbook, which pulled INCOME and OWNRATIOs from 59 census tracts in Pierce County, Washington, which were compiled during the 1980 Census. We had the following data:
Housing Data |
||
Tract |
Income |
Ownratio |
601 |
$24,909 |
7.220 |
602 |
$11,875 |
1.094 |
603 |
$19,308 |
3.587 |
604 |
$20,375 |
5.279 |
605 |
$20,132 |
3.508 |
606 |
$15,351 |
0.789 |
607 |
$14,821 |
1.837 |
608 |
$18,816 |
5.150 |
609 |
$19,179 |
2.201 |
609 |
$21,434 |
1.932 |
610 |
$15,075 |
0.919 |
611 |
$15,634 |
1.898 |
612 |
$12,307 |
1.584 |
613 |
$10,063 |
0.901 |
614 |
$5,090 |
0.128 |
615 |
$8,110 |
0.059 |
616 |
$4,399 |
0.022 |
616 |
$5,411 |
0.172 |
617 |
$9,541 |
0.916 |
618 |
$13,095 |
1.265 |
619 |
$11,638 |
1.019 |
620 |
$12,711 |
1.698 |
621 |
$12,839 |
2.188 |
623 |
$15,202 |
2.850 |
624 |
$15,932 |
3.049 |
625 |
$14,178 |
2.307 |
626 |
$12,244 |
0.873 |
627 |
$10,391 |
0.410 |
628 |
$13,934 |
1.151 |
629 |
$14,201 |
1.274 |
630 |
$15,784 |
1.751 |
631 |
$18,917 |
5.074 |
632 |
$17,431 |
4.272 |
633 |
$17,044 |
3.868 |
634 |
$14,870 |
2.009 |
635 |
$19,384 |
2.256 |
701 |
$18,250 |
2.471 |
705 |
$14,212 |
3.019 |
706 |
$15,817 |
2.154 |
710 |
$21,911 |
5.190 |
711 |
$19,282 |
4.579 |
712 |
$21,795 |
3.717 |
713 |
$22,904 |
3.720 |
713 |
$22,507 |
6.127 |
714 |
$19,592 |
4.468 |
714 |
$16,900 |
2.110 |
718 |
$12,818 |
0.782 |
718 |
$9,849 |
0.259 |
719 |
$16,931 |
1.233 |
719 |
$23,545 |
3.288 |
720 |
$9,198 |
0.235 |
721 |
$22,190 |
1.406 |
721 |
$19,646 |
2.206 |
724 |
$24,750 |
5.650 |
726 |
$18,140 |
5.078 |
728 |
$21,250 |
1.433 |
731 |
$22,231 |
7.452 |
731 |
$19,788 |
5.738 |
735 |
$13,269 |
1.364 |
Data taken from U.S. Bureau of Census 1980 Pierce County, WA; Reprinted in Brown, W.S., Introducing Econometrics, St. Paul (1991): 198-200.
And we got the following regression equation:
Ŷ= 0.000297*Income – 2.221
With an R^{2}=0.597, an F-ratio of 84.31, the t-ratios for INCOME (9.182) and the intercept (-4.094) both solidly significant, and the positive sign on the parameter estimate for INCOME, our model appeared to do very well. However, visual inspection of the regression residuals suggested the presence of heteroscedasticity. Unfortunately, visual inspection can only suggest; we need more objective ways of determining the presence of heteroscedasticity. Hence our three tests below.
The Goldfeld-Quandt Test
The Goldfeld-Quandt test is a computationally simple, and perhaps the most commonly used, method for detecting heteroscedasticity. Since a model with heteroscedastic error terms does not have a constant variance, the Goldfeld-Quandt test postulates that the variances associated with high values of the independent variable, X, are statistically significant from those associated with low values. Essentially, you would run separate regression analyses for the low values of X and the high values, and then compare their F-ratios.
The Goldfeld-Quandt test has four steps:
Step #1: Sort the data
Take the independent variable you suspect to be the source of the heteroscedasticity and sort your data set by the X value in low-to-high order:
Housing Data |
||
Tract |
Income |
Ownratio |
616 |
$4,399 |
0.022 |
614 |
$5,090 |
0.128 |
616 |
$5,411 |
0.172 |
615 |
$8,110 |
0.059 |
720 |
$9,198 |
0.235 |
617 |
$9,541 |
0.916 |
718 |
$9,849 |
0.259 |
613 |
$10,063 |
0.901 |
627 |
$10,391 |
0.410 |
619 |
$11,638 |
1.019 |
602 |
$11,875 |
1.094 |
626 |
$12,244 |
0.873 |
612 |
$12,307 |
1.584 |
620 |
$12,711 |
1.698 |
718 |
$12,818 |
0.782 |
621 |
$12,839 |
2.188 |
618 |
$13,095 |
1.265 |
735 |
$13,269 |
1.364 |
628 |
$13,934 |
1.151 |
625 |
$14,178 |
2.307 |
629 |
$14,201 |
1.274 |
705 |
$14,212 |
3.019 |
607 |
$14,821 |
1.837 |
634 |
$14,870 |
2.009 |
610 |
$15,075 |
0.919 |
623 |
$15,202 |
2.850 |
606 |
$15,351 |
0.789 |
611 |
$15,634 |
1.898 |
630 |
$15,784 |
1.751 |
706 |
$15,817 |
2.154 |
624 |
$15,932 |
3.049 |
714 |
$16,900 |
2.110 |
719 |
$16,931 |
1.233 |
633 |
$17,044 |
3.868 |
632 |
$17,431 |
4.272 |
726 |
$18,140 |
5.078 |
701 |
$18,250 |
2.471 |
608 |
$18,816 |
5.150 |
631 |
$18,917 |
5.074 |
609 |
$19,179 |
2.201 |
711 |
$19,282 |
4.579 |
603 |
$19,308 |
3.587 |
635 |
$19,384 |
2.256 |
714 |
$19,592 |
4.468 |
721 |
$19,646 |
2.206 |
731 |
$19,788 |
5.738 |
605 |
$20,132 |
3.508 |
604 |
$20,375 |
5.279 |
728 |
$21,250 |
1.433 |
609 |
$21,434 |
1.932 |
712 |
$21,795 |
3.717 |
710 |
$21,911 |
5.190 |
721 |
$22,190 |
1.406 |
731 |
$22,231 |
7.452 |
713 |
$22,507 |
6.127 |
713 |
$22,904 |
3.720 |
719 |
$23,545 |
3.288 |
724 |
$24,750 |
5.650 |
601 |
$24,909 |
7.220 |
Step #2: Omit the middle observations
Next, take out the observations in the middle. This usually amounts between one-fifth to one-third of your observations. There’s no hard and fast rule about how many variables to omit, and if your data set is small, you may not be able to omit any. In our example, we can omit 13 observations (highlighted in orange):
Housing Data |
||
Tract |
Income |
Ownratio |
616 |
$4,399 |
0.022 |
614 |
$5,090 |
0.128 |
616 |
$5,411 |
0.172 |
615 |
$8,110 |
0.059 |
720 |
$9,198 |
0.235 |
617 |
$9,541 |
0.916 |
718 |
$9,849 |
0.259 |
613 |
$10,063 |
0.901 |
627 |
$10,391 |
0.410 |
619 |
$11,638 |
1.019 |
602 |
$11,875 |
1.094 |
626 |
$12,244 |
0.873 |
612 |
$12,307 |
1.584 |
620 |
$12,711 |
1.698 |
718 |
$12,818 |
0.782 |
621 |
$12,839 |
2.188 |
618 |
$13,095 |
1.265 |
735 |
$13,269 |
1.364 |
628 |
$13,934 |
1.151 |
625 |
$14,178 |
2.307 |
629 |
$14,201 |
1.274 |
705 |
$14,212 |
3.019 |
607 |
$14,821 |
1.837 |
634 |
$14,870 |
2.009 |
610 |
$15,075 |
0.919 |
623 |
$15,202 |
2.850 |
606 |
$15,351 |
0.789 |
611 |
$15,634 |
1.898 |
630 |
$15,784 |
1.751 |
706 |
$15,817 |
2.154 |
624 |
$15,932 |
3.049 |
714 |
$16,900 |
2.110 |
719 |
$16,931 |
1.233 |
633 |
$17,044 |
3.868 |
632 |
$17,431 |
4.272 |
726 |
$18,140 |
5.078 |
Tract |
Income |
Ownratio |
701 |
$18,250 |
2.471 |
608 |
$18,816 |
5.150 |
631 |
$18,917 |
5.074 |
609 |
$19,179 |
2.201 |
711 |
$19,282 |
4.579 |
603 |
$19,308 |
3.587 |
635 |
$19,384 |
2.256 |
714 |
$19,592 |
4.468 |
721 |
$19,646 |
2.206 |
731 |
$19,788 |
5.738 |
605 |
$20,132 |
3.508 |
604 |
$20,375 |
5.279 |
728 |
$21,250 |
1.433 |
609 |
$21,434 |
1.932 |
712 |
$21,795 |
3.717 |
710 |
$21,911 |
5.190 |
721 |
$22,190 |
1.406 |
731 |
$22,231 |
7.452 |
713 |
$22,507 |
6.127 |
713 |
$22,904 |
3.720 |
719 |
$23,545 |
3.288 |
724 |
$24,750 |
5.650 |
601 |
$24,909 |
7.220 |
Step #3: Run two separate regressions, one for the low values, one for the high
We ran separate regressions for the 23 observations with the lowest values for INCOME and the 23 observations with the highest values. In these regressions, we weren’t concerned with whether the t-ratios of the parameter estimates were significant. Rather, we wanted to look at their Error Sum of Squares (ESS). Each model has 21 degrees of freedom.
Step #4: Divide the ESS of the higher value regression by the ESS of the lower value regression, and compare quotient to the F-table.
The higher value regression produced an ESS of 61.489 and the lower value regression produced an ESS of 5.189. Dividing the former by the latter, we get a quotient of 11.851. Now, we need to go to the F-table and check the critical F-value for a 95% significance level and 21 degrees of freedom, which is a value of 2.10. Since our quotient of 11.851 is greater than that of the critical F-value, we can conclude there is strong evidence of heteroscedasticity in the model.
The Breusch-Pagan Test
The Breusch-Pagan test is also pretty simple, but it’s a very powerful test, in that it can be used to detect whether more than one independent variable is causing the heteroscedasticity. Since it can involve multiple variables, the Breusch-Pagan test relies on critical values of chi-squared (χ^{2}) to determine the presence of heteroscedasticity, and works best with large sample sets. There are five steps to the Breusch-Pagan test:
Step #1:
Run the regular regression model and collect the residuals
We already did that.
Step #2: Estimate the variance of the regression residuals
To do this, we square each residual, sum it up and then divide it by the number of observations. Our formula is:
Our residuals and their squares are as follows:
Observation |
Predicted Ownratio |
Residuals |
Residuals Squared |
1 |
5.165 |
2.055 |
4.222 |
2 |
1.300 |
(0.206) |
0.043 |
3 |
3.504 |
0.083 |
0.007 |
4 |
3.821 |
1.458 |
2.126 |
5 |
3.749 |
(0.241) |
0.058 |
6 |
2.331 |
(1.542) |
2.378 |
7 |
2.174 |
(0.337) |
0.113 |
8 |
3.358 |
1.792 |
3.209 |
9 |
3.466 |
(1.265) |
1.601 |
10 |
4.135 |
(2.203) |
4.852 |
11 |
2.249 |
(1.330) |
1.769 |
12 |
2.415 |
(0.517) |
0.267 |
13 |
1.428 |
0.156 |
0.024 |
14 |
0.763 |
0.138 |
0.019 |
15 |
(0.712) |
0.840 |
0.705 |
16 |
0.184 |
(0.125) |
0.016 |
17 |
(0.917) |
0.939 |
0.881 |
18 |
(0.617) |
0.789 |
0.622 |
19 |
0.608 |
0.308 |
0.095 |
20 |
1.662 |
(0.397) |
0.158 |
21 |
1.230 |
(0.211) |
0.045 |
22 |
1.548 |
0.150 |
0.022 |
23 |
1.586 |
0.602 |
0.362 |
24 |
2.287 |
0.563 |
0.317 |
25 |
2.503 |
0.546 |
0.298 |
26 |
1.983 |
0.324 |
0.105 |
27 |
1.410 |
(0.537) |
0.288 |
28 |
0.860 |
(0.450) |
0.203 |
29 |
1.911 |
(0.760) |
0.577 |
30 |
1.990 |
(0.716) |
0.513 |
31 |
2.459 |
(0.708) |
0.502 |
32 |
3.388 |
1.686 |
2.841 |
33 |
2.948 |
1.324 |
1.754 |
34 |
2.833 |
1.035 |
1.071 |
35 |
2.188 |
(0.179) |
0.032 |
36 |
3.527 |
(1.271) |
1.615 |
37 |
3.191 |
(0.720) |
0.518 |
38 |
1.993 |
1.026 |
1.052 |
39 |
2.469 |
(0.315) |
0.099 |
40 |
4.276 |
0.914 |
0.835 |
41 |
3.497 |
1.082 |
1.171 |
42 |
4.242 |
(0.525) |
0.275 |
43 |
4.571 |
(0.851) |
0.724 |
44 |
4.453 |
1.674 |
2.802 |
45 |
3.589 |
0.879 |
0.773 |
46 |
2.790 |
(0.680) |
0.463 |
47 |
1.580 |
(0.798) |
0.637 |
48 |
0.699 |
(0.440) |
0.194 |
49 |
2.800 |
(1.567) |
2.454 |
50 |
4.761 |
(1.473) |
2.169 |
51 |
0.506 |
(0.271) |
0.074 |
52 |
4.359 |
(2.953) |
8.720 |
53 |
3.605 |
(1.399) |
1.956 |
54 |
5.118 |
0.532 |
0.283 |
55 |
3.158 |
1.920 |
3.686 |
56 |
4.080 |
(2.647) |
7.008 |
57 |
4.371 |
3.081 |
9.492 |
58 |
3.647 |
2.091 |
4.373 |
59 |
1.714 |
(0.350) |
0.122 |
Summing the last column, we get 83.591. We divide this by 59, and get 1.417.
Step #3: Compute the square of the standardized residuals
Now that we know the variance of the regression residuals – 1.417 – we compute the standardized residuals by dividing each residual by 1.417 and then squaring the results, so that we get our square of standardized residuals, s_{i}^{2}:
Obs. |
Predicted Ownratio |
Residuals |
Standardized Residuals |
Square of Standardized Residuals |
1 |
5.165 |
2.055 |
1.450 |
2.103 |
2 |
1.300 |
(0.206) |
(0.146) |
0.021 |
3 |
3.504 |
0.083 |
0.058 |
0.003 |
4 |
3.821 |
1.458 |
1.029 |
1.059 |
5 |
3.749 |
(0.241) |
(0.170) |
0.029 |
6 |
2.331 |
(1.542) |
(1.088) |
1.185 |
7 |
2.174 |
(0.337) |
(0.238) |
0.057 |
8 |
3.358 |
1.792 |
1.264 |
1.599 |
9 |
3.466 |
(1.265) |
(0.893) |
0.797 |
10 |
4.135 |
(2.203) |
(1.555) |
2.417 |
11 |
2.249 |
(1.330) |
(0.939) |
0.881 |
12 |
2.415 |
(0.517) |
(0.365) |
0.133 |
13 |
1.428 |
0.156 |
0.110 |
0.012 |
14 |
0.763 |
0.138 |
0.097 |
0.009 |
15 |
(0.712) |
0.840 |
0.593 |
0.351 |
16 |
0.184 |
(0.125) |
(0.088) |
0.008 |
17 |
(0.917) |
0.939 |
0.662 |
0.439 |
18 |
(0.617) |
0.789 |
0.557 |
0.310 |
19 |
0.608 |
0.308 |
0.217 |
0.047 |
20 |
1.662 |
(0.397) |
(0.280) |
0.079 |
21 |
1.230 |
(0.211) |
(0.149) |
0.022 |
22 |
1.548 |
0.150 |
0.106 |
0.011 |
23 |
1.586 |
0.602 |
0.425 |
0.180 |
24 |
2.287 |
0.563 |
0.397 |
0.158 |
25 |
2.503 |
0.546 |
0.385 |
0.148 |
26 |
1.983 |
0.324 |
0.229 |
0.052 |
27 |
1.410 |
(0.537) |
(0.379) |
0.143 |
28 |
0.860 |
(0.450) |
(0.318) |
0.101 |
29 |
1.911 |
(0.760) |
(0.536) |
0.288 |
30 |
1.990 |
(0.716) |
(0.505) |
0.255 |
31 |
2.459 |
(0.708) |
(0.500) |
0.250 |
32 |
3.388 |
1.686 |
1.190 |
1.415 |
33 |
2.948 |
1.324 |
0.935 |
0.874 |
34 |
2.833 |
1.035 |
0.730 |
0.534 |
35 |
2.188 |
(0.179) |
(0.127) |
0.016 |
36 |
3.527 |
(1.271) |
(0.897) |
0.805 |
37 |
3.191 |
(0.720) |
(0.508) |
0.258 |
38 |
1.993 |
1.026 |
0.724 |
0.524 |
39 |
2.469 |
(0.315) |
(0.222) |
0.049 |
40 |
4.276 |
0.914 |
0.645 |
0.416 |
41 |
3.497 |
1.082 |
0.764 |
0.584 |
42 |
4.242 |
(0.525) |
(0.370) |
0.137 |
43 |
4.571 |
(0.851) |
(0.600) |
0.361 |
44 |
4.453 |
1.674 |
1.182 |
1.396 |
45 |
3.589 |
0.879 |
0.621 |
0.385 |
46 |
2.790 |
(0.680) |
(0.480) |
0.231 |
47 |
1.580 |
(0.798) |
(0.563) |
0.317 |
48 |
0.699 |
(0.440) |
(0.311) |
0.097 |
49 |
2.800 |
(1.567) |
(1.106) |
1.223 |
50 |
4.761 |
(1.473) |
(1.040) |
1.081 |
51 |
0.506 |
(0.271) |
(0.192) |
0.037 |
52 |
4.359 |
(2.953) |
(2.084) |
4.344 |
53 |
3.605 |
(1.399) |
(0.987) |
0.974 |
54 |
5.118 |
0.532 |
0.375 |
0.141 |
55 |
3.158 |
1.920 |
1.355 |
1.836 |
56 |
4.080 |
(2.647) |
(1.868) |
3.491 |
57 |
4.371 |
3.081 |
2.175 |
4.728 |
58 |
3.647 |
2.091 |
1.476 |
2.179 |
59 |
1.714 |
(0.350) |
(0.247) |
0.061 |
Step #4: Run another regression with all your independent variables using the sum of standardized residuals as the dependent variable
In this case, we had only one independent variable, INCOME. We will now run a regression substituting the last column of the table above for OWNRATIO, and making it the dependent variable. Again, we’re not interested in the parameter estimates. We are, however, interested in the regression sum of squares (RSS), which is 15.493.
Step #5: Divide the RSS by 2 and compare with the χ^{2} table’s critical value for the appropriate degrees of freedom
Dividing the RSS by 2, we get 7.747. We look up the critical χ^{2} value for one degree of freedom and in the table, for a 5% significance level, we get 3.84. Since our χ^{2} value exceeds our critical, we can conclude there is strong evidence of heteroscedasticity present.
The Park Test
Last, but certainly not least comes the Park test. I saved this one for last because it is the simplest of the three methods and unlike the other two, provides information that can help eliminate the heteroscedasticity. The Park Test assumes there is a relationship between the error variance and one of the regression model’s independent variables. The steps involved are as follows:
Step #1: Run your original regression model and collect the residuals
Done.
Step #2: Square the regression residuals and compute the logs of the squared residuals and the values of the suspected independent variable.
We’ll square the regression residuals, and take their natural log. We will also take the natural log of INCOME:
Tract |
Residual Squared |
LnResidual Squared |
LnIncome |
601 |
4.222 |
1.440 |
10.123 |
602 |
0.043 |
(3.157) |
9.382 |
603 |
0.007 |
(4.987) |
9.868 |
604 |
2.126 |
0.754 |
9.922 |
605 |
0.058 |
(2.848) |
9.910 |
606 |
2.378 |
0.866 |
9.639 |
607 |
0.113 |
(2.176) |
9.604 |
608 |
3.209 |
1.166 |
9.842 |
609 |
1.601 |
0.470 |
9.862 |
609 |
4.852 |
1.579 |
9.973 |
610 |
1.769 |
0.571 |
9.621 |
611 |
0.267 |
(1.320) |
9.657 |
612 |
0.024 |
(3.720) |
9.418 |
613 |
0.019 |
(3.960) |
9.217 |
614 |
0.705 |
(0.349) |
8.535 |
615 |
0.016 |
(4.162) |
9.001 |
616 |
0.881 |
(0.127) |
8.389 |
616 |
0.622 |
(0.475) |
8.596 |
617 |
0.095 |
(2.356) |
9.163 |
618 |
0.158 |
(1.847) |
9.480 |
619 |
0.045 |
(3.112) |
9.362 |
620 |
0.022 |
(3.796) |
9.450 |
621 |
0.362 |
(1.015) |
9.460 |
623 |
0.317 |
(1.148) |
9.629 |
624 |
0.298 |
(1.211) |
9.676 |
625 |
0.105 |
(2.255) |
9.559 |
626 |
0.288 |
(1.245) |
9.413 |
627 |
0.203 |
(1.596) |
9.249 |
628 |
0.577 |
(0.549) |
9.542 |
629 |
0.513 |
(0.668) |
9.561 |
630 |
0.502 |
(0.689) |
9.667 |
631 |
2.841 |
1.044 |
9.848 |
632 |
1.754 |
0.562 |
9.766 |
633 |
1.071 |
0.069 |
9.744 |
634 |
0.032 |
(3.437) |
9.607 |
635 |
1.615 |
0.479 |
9.872 |
701 |
0.518 |
(0.658) |
9.812 |
705 |
1.052 |
0.051 |
9.562 |
706 |
0.099 |
(2.309) |
9.669 |
710 |
0.835 |
(0.180) |
9.995 |
711 |
1.171 |
0.158 |
9.867 |
712 |
0.275 |
(1.289) |
9.989 |
713 |
0.724 |
(0.323) |
10.039 |
713 |
2.802 |
1.030 |
10.022 |
714 |
0.773 |
(0.257) |
9.883 |
714 |
0.463 |
(0.770) |
9.735 |
718 |
0.637 |
(0.452) |
9.459 |
718 |
0.194 |
(1.640) |
9.195 |
719 |
2.454 |
0.898 |
9.737 |
719 |
2.169 |
0.774 |
10.067 |
720 |
0.074 |
(2.608) |
9.127 |
721 |
8.720 |
2.166 |
10.007 |
721 |
1.956 |
0.671 |
9.886 |
724 |
0.283 |
(1.263) |
10.117 |
726 |
3.686 |
1.305 |
9.806 |
728 |
7.008 |
1.947 |
9.964 |
731 |
9.492 |
2.250 |
10.009 |
731 |
4.373 |
1.476 |
9.893 |
735 |
0.122 |
(2.102) |
9.493 |
Step #3: Run the regression equation using the log of the squared residuals as the dependent variable and the log of the suspected independent variable as the dependent variable
That results in the following regression equation:
Ln(e^{2}) = 1.957(LnIncome) – 19.592
Step #4: If the t-ratio for the transformed independent variable is significant, you can conclude heteroscedasticity is present.
The parameter estimate for the LnIncome is significant, with a t-ratio of 3.499, so we conclude heteroscedasticity.
Next Forecast Friday Topic: Correcting Heteroscedasticity
Thanks for your patience! Now you know the three most common methods for detecting heteroscedasticity: the Goldfeld-Quandt test, the Breusch-Pagan test, and the Park test. As you will see in next week’s Forecast Friday post, the Park test will be beneficial in helping us eliminate the heteroscedasticity. We will discuss the most common approach to correcting heteroscedasticity: weighted least squares (WLS) regression, and show you how to apply it. Next week’s Forecast Friday post will conclude our discussion of regression violations, and allow us to resume discussions of more practical applications in forecasting.
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