5. Suppose we are studying the relationship between skipping class and college G

Question

5. Suppose we are studying the relationship between skipping class and college GPA controlling for ACT score and high school GPA. Defining "skipped" as the number of classes skipped per week, we assume the following population relationship: and we obtain the following estimated equation (corresponding standard errors are below the coefficient estimates): col GPA = 1 .39 0.412 hsGPA + 0.01 5 ACT _ 0.083 skipped. (0.330) (0.094) (0.011) (0.026) where n = 141 and R-0.234 In words, state what our estimate of , s telling us. Calculate the test statistic for the coefficient estimate on "skipped." Is "skipped" statistically significant at the 5 percent level? How about the 1 percent level? Can we reject Ho : 3-1 at the 5 percent level in favor of H, : 1? a. b. d. Can we reject H0 : 3-1 at the 5 percent level in favor of H, : -1 ? e. Find the 95 percent confidence interval for P;. f. Suppose that if we exclude hsGPA and ACT we obtain the following estimated equation: colGPA-3 ,16-0 226 skipped. (0.300) (0.022) where n = 141 and R2-0.112. Can we reject H, : / 0,2-0 at the 5 percent level in favor of H! : 0A Does it appear that omitting hsGPA and ACT has biased our coefficient estimator for skipped? (Consider both coefficient estimates and standard errors.) 0? g. h. Redo part (b), but assume that you only have 22 observations instead of 141. (Keep everything else the same, even though lowering the sample size would probably inflate the standard errors.) Does the lower sample size change your conclusions?

Explanation / Answer

a) b3^ = -0.083

which means if we increase skipped class per week by 1 unit , colGPA will decrease by 0.083 units

b) TS = (b3^ - 0)/ se (b3^)

= -0.083 / 0.026

= -3.192

critical value at 5 % = 2 (approx)

since |TS| > critical value ,we reject the null and conclude that skipped is significant

c) Ho : b3 = -1

TS = (-0.083 - (-1))/0.026

= 35.269

clearly we reject the null as |TS| > critical value (1.96)

d) TS remains same in this on-sided hypothesis

critcal value changes = 1.645  

since |TS| > critical value

we reject the null

e) 95 % confidence interval

df= n-k-1 = 141-3-1 = 137

t-critical = 1.97743

=(-0.083 - 1.97743*0.026,-0.083 + 1.97743*0.026)

= (-0.13441, -0.031586)

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