Exercise econometrics 5.9.
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5.9[pic 1]
Table 5.5 gives data on average public teacher pay (annual salary in dollars) and spending on public schools per pupil (dollars) in 1985 for 50 states and the District of Columbia.
To find out if there is any relationship between teacher’s pay and per pupil expenditure in public schools, the following model was suggested:
Payi = β1 + β2 Spendi + ui, where Pay stands for teacher’s salary and Spend stands for per pupil expenditure.
- Plot the data and eyeball a regression line.
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- Suppose on the basis of a you decide to estimate the above regression model. Obtain the estimates of the parameters, their standard errors, r2, RSS, and ESS.
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- Interpret the regression. Does it make economic sense?
d. Establish a 95% confidence interval for β2. Would you reject the hypothesis that the true slope coefficient is 3.0?
e. Obtain the mean and individual forecast value of Pay if per pupil spending is $5000. Also establish 95% confidence intervals for the true mean and individual values of Pay for the given spending figure.
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f. How would you test the assumption of the normality of the error term? Show the test(s) you use
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Distribución de frecuencias para uhat1, observaciones 1-52
número de cajas = 7, media = 8,55995e-013, desv.típ.=2324,78
intervalo punto medio frecuencia rel acum.
< -3066,5 -3848,0 4 7,84% 7,84% **
-3066,5 - -1503,6 -2285,1 10 19,61% 27,45% *******
-1503,6 - 59,240 -722,20 14 27,45% 54,90% *********
59,240 - 1622,1 840,68 9 17,65% 72,55% ******
1622,1 - 3185,0 2403,6 10 19,61% 92,16% *******
3185,0 - 4747,9 3966,5 1 1,96% 94,12%
>= 4747,9 5529,3 3 5,88% 100,00% **
Observaciones ausentes = 1 ( 1,92%)
Contraste de la hipótesis nula de distribución normal:
Chi-cuadrado(2) = 2,905 con valor p 0,23395
5.10. Refer to exercise 3.20 and set up the ANOVA tables and test the hypothesis that there is no relationship between productivity and real wage compensation. Do this for both the business and nonfarm business sectors.
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The ANOVA TABLE FOR A BUSINESS SECTOR: [pic 16][pic 17]
THE ANOVA TABLE FOR A NON FARM BUSINESS SECTOR:
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5.13. Refer to exercise 3.22.
a. Estimate the two regressions given there, obtaining standard errors and the other usual output.
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- Test the hypothesis that the disturbances in the two regression models are normally distributed.
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Distribución de frecuencias para uhat1, observaciones 1-15
número de cajas = 5, media = -1,32635e-014, desv.típ.=104,694
intervalo punto medio frecuencia rel acum.
< -96,781 -149,83 2 13,33% 13,33% ****
-96,781 - 9,3124 -43,734 7 46,67% 60,00% ****************
9,3124 - 115,41 62,359 5 33,33% 93,33% ************
115,41 - 221,50 168,45 0 0,00% 93,33%
>= 221,50 274,55 1 6,67% 100,00% **
Contraste de la hipótesis nula de distribución normal:
Chi-cuadrado(2) = 6,149 con valor p 0,04622
[pic 29]
Distribución de frecuencias para uhat2, observaciones 1-15
número de cajas = 5, media = 3,90799e-014, desv.típ.=19,8418
intervalo punto medio frecuencia rel acum.
< -26,850 -34,501 1 6,67% 6,67% **
-26,850 - -11,548 -19,199 4 26,67% 33,33% *********
-11,548 - 3,7537 -3,8972 2 13,33% 46,67% ****
3,7537 - 19,055 11,405 6 40,00% 86,67% **************
>= 19,055 26,706 2 13,33% 100,00% ****
Contraste de la hipótesis nula de distribución normal:
Chi-cuadrado(2) = 1,567 con valor p 0,45675
c. In the gold price regression, test the hypothesis that β2 = 1, that is, there is a one-to-one relationship between gold prices and CPI (i.e., gold is a perfect hedge). What is the p value of the estimated test statistic?
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d. Repeat step c for the NYSE Index regression. Is investment in the stock market a perfect hedge against inflation? What is the null hypothesis you are testing? What is its p value?
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t(13, .025) = 2,160
Variable | Coeficiente | Intervalo de confianza 95% | |
const | -102,061[pic 32] | (-153,406, -50,7155) |
|
CPI | 2,12944 | (1,63194, 2,62694) |
|
- Between gold and stock, which investment would you choose? What is the basis of your decision?
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5.17. Refer to the S.A.T. data given in exercise 2.16. Suppose you want to predict the male math (Y) scores on the basis of the female math scores (X) by running the following regression: Yt = β1 + β2Xt + ut
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