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Use the data in PNTSPRD for this exercise.(i) The variable sprdcvis a binary variable equal to one if the Las Vegas point spread for a collegebasketball game was covered. The expected value of sprdcur, say $\mu$ , is the probability that thespread is covered in a randomly selected game. Test $\mathrm{H}_{0} : \mu=.5$ against $\mathrm{H}_{1} : \mu \neq .5$ at the 10$\%$ significance level and discuss your findings. (Hint: This is easily done using a $t$ test by regressing spracvron an intercept only.)(ii) How many games in the sample of 553 were played on a neutral court?(iii) Estimate the linear probability model$\quad$ sprdcvr $=\beta_{0}+\beta_{1}$ favhome $+\beta_{2}$ neutral $+\beta_{3} f a v 25+\beta_{4} u n d 25+u$and report the results in the usual form. (Report the usual OLS standard errors and theheteroskedasticity-robust standard errors.) Which variable is most significant, both practicallyand statistically?(iv) Explain why, under the null hypothesis $\mathrm{H}_{0} : \beta_{1}=\beta_{2}=\beta_{3}=\beta,$ there is no heteroskedasticity in the model.(v) Use the usual $F$ statistic to test the hypothesis in part (iv). What do you conclude?(vi) Given the previous analysis, would you say that it is possible to systematically predict whetherthe Las Vegas spread will be covered using information available prior to the game?

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The explanatory power of the model is given by R-squared which is $0.0034,$ indicating that theexplanatory power of the modelsprdcvr$=\beta_{0}+\beta_{1}$ favhome $+\beta_{2}$ neutral $+\beta_{3} f a v 25+\beta_{4} u n d 25+u$ is very lowSecondly, none of the explanatory variables are either individually or jointly statistically significantat 5$\%$ level of significanceFrom these two pints mentioned above, it would not be possible to systematically predict whetherthe Las Vegas spread will be covered

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Chapter 8

Heteroskedasticity

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Adán B.

June 9, 2021

thanks a lot. It´s correct

Dylan L.

November 7, 2021

Hello Are we able to see the video for the coding of this in STATA?It would be much appreciated!

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in part one. We need to test the null hypothesis. New equal 0.5. We will test this hypothesis by regressing the dependent variable. S p r D C V are on an intercept only we will get a new hat equal 0.515 And the standard error of this estimation is 0.21 The T statistic for the non high policies is you had minus meal over the standard error off Mu had plugging in the numbers. You have 0.515 minus 0.5 over 0.21 The value of the T statistic is 0.71 So the result is not significant at the 10% level. And we're unable to reject the non hypothesis party to ask you how maney games were played on a neutral court. We we need to look at the variable. Neutral. Neutral is a dummy variable, variable, neutral text, a value of one if the site is neutral and it is zero. Otherwise, who in count, how many cases we're neutral received the value of one, and if you did it correctly, you in fine. There are 35 cases or 35 games that were play on a neutral court. Part three. This is the regression results for the estimated linear probability model. The dependent variable is S p R D C v R. This variable take a value of one. If the spread is covered. Therefore, independent variables, the first one equals one. If the favorite team is at home, the second one neutral equals one. If the site is neutral, the third one equals one. If the favorite team is in top 25 and the last one equals one, if the underdog is in top 25 and we have one estimation for the intercept, regardless of their statistical software that you use, you will always receive their estimation for the coefficients and standard error. The program will also show you the number of observations or n here, which is 553 and the R square that measured their fitness of the model. This model has a very low are square is not even 1%. Looking at these results, you know that you can see that neutral is the most important variable. It has the largest effect. If you compare by the estimation of beta had the meaning of this coefficient is if the game is played on a neutral court, the probability that the spread is covered is estimated to be about point on to higher mhm. Besides having the largest estimated coefficient mutual also significant, it's Teesta district is the only one that is greater than one in absolute value. Now, I don't show you the T statistic here, but you can always rely on the statistical software that you use, or you can calculate the T statistic by yourself. Remember that the teeth thought is the ratio of beta head over the Senate era of beta head, and if you calculate currently, then the teeth thought of the neutral variable is one point 24 This is the new hypothesis given in the problem in past four. Under this now hypothesis, the response probability does not depend on any explanatory variables, which means it's mean no end the variance do not depend. Okay, okay, this is the nal hypotheses given by their problem under the's null hypothesis, The response probability does not depend on any explanatory variables, which means neither the means nor the variance depends on the explanatory variables, and that implies there is no hetero Scholastic City. Mhm. We contest this non hypothesis in Part five with an F test. This is a test used for joint significance. The F statistic we get when equal 0.47 and there are two degrees of freedom. One is four and the other is 548. The P value is 0.76 This means we are unable to reject the null hypotheses part six. It is not possible to predict whether the spread would be covered for two reasons. I Okay, First, their explanatory power of the model is very low. Recall from part tree, the R Square we got is only 0.34 and the second reason is the explanatory variables are jointly insignificant. This is just there. Repetition of their result we got from Part five. The coefficient on neutral may indicate something is going on with games played on a neutral court, but we would not want to bet money on it unless it could be confirmed with our separate larger sample

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