This is the result for Part one. We use the full sample, which has 177 observations. From this estimation, we obtain the student ized residuals and we call them as t r. Supply. The number of student ized residuals, which are above 1.96 in absolute value, is nine. If the student ized residuals were independently drawn from a standard normal distribution, we would expect about by percent of our sample. Sir, 177 times 5%. You will get a number between eight and nine. It is 8.56 something we would expect between eight and nine cases of student ties residuals to be above two. It is so because in a standard normal distribution, about 95.5% of the observations are within two standard errors off standard deviation are within two standard deviation. And in a standard normal distribution, the standard deviation is one. So 95.5% of the observation are within two equivalently. It means 5% up to 5% of the observations are either above two or less than minus two. That's why we have the 5% number here and as just say, um, right here to be above two in absolute value, you can check. You can fact check this statement and you will find that there are eight observation with student ties. Residuals above two in absolute value for three. The student ties residuals are used to detect are liars. We will drop. There are liars, which are defined as observation with student ties. Residuals above 1.9 16 absolute value so we wouldn't drop there NYT, cases we find out in Part two, we will re estimate the model in part one again using 169 observations. This is the result. Compare with the regression. In Part one, we find that the main coefficient become significant at the 1% level. Let me come back to part one, so the first one lakh of sales is significant at the 1% level I windy note with three stars. Lakoff M. Kate Evil is significant at the 5% level, so two Stars CEO 10. Thank you significant at the 1% level, and CEO 10 square is significant at the 5% level. Back Thio, Part three, lock of sales is significant at 1% level. Still lock of em Katie Value before it was significant at the 5% level. Now it is significant. At the 1% level, nothing changed in terms of significance level for C E. 0. 10 and for CEO 10 square. Okay, nothing changed. It is still significant at the 5% level. Yeah, so we have beta head of lock of sales Mhm and CEO 10. They have the same level of significance. The exactly value is different, but not too substantial to give them a new level of significance. The estimates on em, Katie Value increase in magnitude and significance level. Okay, You may also notice that the magnitude of the estimates on sales and CEO 10 decrease, but not too much. And the coefficient of CEO 10 square does not change in terms of magnitude. Now we will use least absolute deviation to estimate the regression in part One again, this is part four. We re news all the data Here is the result. The l. A. D method is, um, estimated with with a different, um methodology. So it doesn't have the are square. It is estimated by maximum likelihood So to measure the fit of the model, you will need to look at their results generated by their statistical software. And you will look at the lock likelihood value. I would not report that here because we don't care about the fit of the model in this problem, we care about the estimates of their explanatory variables. Compare this regression with previous regression where we use l s. We see that beta one. The coefficient of lack of cells is closer to that of the restricted sample and the restricted. Simple is the regression. In part three, where we drop the outlier observations. We don't have the same observation for beta three beta three hat. The coefficient of CO 10 is actually closer to the estimate in the full sample. Even these results part five, we will be able to evaluate this statement dropping our lawyers based on extreme values of student ties. Residuals makes the resulting L s estimates closer to the L A G estimates on the phone sample. This statement is not always true. It is not true to every estimate