Presentation on theme: "C 3.7 Use the data in MEAP93.RAW to answer this question (i)Estimate the model And report results in the usual form, including the sample size and R-squared."— Presentation transcript:
C 3.7 Use the data in MEAP93.RAW to answer this question (i)Estimate the model And report results in the usual form, including the sample size and R-squared. Are the signs of the slope coefficients what you expected? The signs of the estimates suggest more expenditure increases the pass rate, holding lnchrpg fixed (A 1% increase in expenditure is predicted to increase the math pass rate by 6.229/100= percentage points, holding lnchrpg fixed). Also, a higher poverty rate (as represented by lnchrpg) decreases the pass rate, holding spending fixed. (A 1% increase in the number of students receiving free lunch reduces the math pass rate by.304 percentage points, holding expenditure fixed).
(ii) What do you make of the intercept you estimated in (i)? In particular, does it make sense to set the two explanatory variables to zero. Recall log(1)=0. -The intercept estimates the predict value of math10 when all regressors take a value of zero. -Setting lnchprg=0 makes sense, as there are many schools with low poverty rates. -Setting log(expend)=0 doesn’t make sense, because it’s the same as setting expend=1, and spending is measured in dollars per student….1 is really low. -As a consequence, the intercept of -20 is nonsensical given the data.
(iii) Now run the simple regression of math10 on log(expend), and compare the slope coefficient with the estimate obtained in part (1). Is the estimated spending effect now larger or smaller than in part (i) The estimated effect of expenditure is much higher than in the multiple linear regression.
(iv) Find the correlation between lexpend and lnchprg. Does its sign make sense to you? The correlation is negative. This indicates that schools with less spending have more students who receive free lunch. This is a result of the fact that school spending is mostly determined by local property tax revenue.
(v) Use part (iv) to explain your findings in part (iii) We know the estimate from the simple linear regression has the following relationship with the estimate from the multiple linear regression: -We don’t have to calculate to figure out it’s sign…the correlation we calculated in part (iv) tells us <0. -Looking at the result of the multiple linear regression, we see <0. -That means >0, so that we are adding a positive term to. That means >, so we get an overestimate of the effect of expenditure on math pass rates from the simple linear regression.
C4.6 (MODIFIED) Use the data in WAGE2.RAW for this exercise (i)Consider the standard wage equation State the null hypothesis that another year of age has no effect on wage, holding education, experience, tenure, and marital status constant. State the two-sided alternative. H 0 : H 1 :
T-statistic Approach t age =.007/.004=1.75 Let α=.05, n-k-1= =929 degrees of freedom Using TG.2, c=1.960 Rule: We reject the null if |t|>c We fail to reject the null if |t|
P-value Approach P-value for t age : P>|t|=0.133 P-value is the probability of observing a t-statistic as large as we do, if the null hypothesis is true (H 0 : ) Let α=.05 Rule: Small p-values are evidence against the null Large p-values provide little evidence against the null. If the null is true, we’d observe a t-statistic as large as we do, 13% of the time. Reject null if p-value< α Fail to reject null if p-value> α.133>.05 Fail to reject the null. Indicates that the estimate on age is not statistically different from 0 at the 5% level.
Confidence Interval Approach Confidence interval for estimate on age is: ( , ) This comes from: H 0 : H 1 : H 0 : H 1 : Let α=.05, n-k-1= =929 degrees of freedom From TG.2 c=1.96 (can use 2) Rule: If random samples obtained over and over, could calculate these bounds each time. In (1- α)% of these samples, the true value of the estimate falls in these bounds. Reject null if is not in the CI. Fail to reject null if is in the CI. 0 is in the CI. Fail to reject the null. Indicates that the estimate on age is not statistically different from 0 at the 5% level.