PART TWO Solutions to Empirical Exercises

Chapter 3
Review of Statistics
Solutions to Empirical Exercises
1. (a) Average Hourly Earnings, Nominal $’s Mean AHE1992 AHE2004 AHE2004 − AHE1992 (b) Average Hourly Earnings, Real $2004 Mean AHE1992 AHE2004 AHE2004 − AHE1992 15.66 16.77 Difference 1.11 SE(Mean) 0.086 0.098 SE(Difference) 0.130 95% Confidence Interval 15.49−15.82 16.58−16.96 95% Confidence Interval 0.85−1.37 11.63 16.77 Difference 5.14 SE(Mean) 0.064 0.098 SE(Difference) 0.117 95% Confidence Interval 11.50−11.75 16.58−16.96 95% Confidence Interval 4.91−5.37

(c) The results from part (b) adjust for changes in purchasing power. These results should be used. (d) Average Hourly Earnings in 2004 Mean High School College
(d) · AHE = −0.23 + 0.69 × Age (1.54) (0.05) The t-statistic is 0.69/0.05 = 13.06, which has a p-value of 0.000, so the null hypothesis can be rejected at the 1% level (and thus, also at the 10% and 5% levels). ˆ ˆ (e) The difference in the estimated β1 coefficients is β1,College − β1, HighScool = 0.69 − 0.26 = 0.43. The ˆ ˆ standard error of for the estimated difference is SE ( β ) = (0.032 + 0.052)1/2 = −β
1,College 1, HighScool

0.06, so that a 95% confidence interval for the difference is 0.43 ± 1.96 × 0.06 = 0.32 to 0.54 (dollars per hour). 2.

· Course _ Eval = 4.00 + 0.13 × Beauty (0.03) (0.03) The t-statistic is 0.13/0.03 = 4.12, which has a p-value of 0.000, so the null hypothesis can be rejected at the 1% level (and thus, also at the 10% and 5% levels).

3.

¶ (a) Ed = 13.96 − 0.073 × Dist (0.04) (0.013)

The t-statistic is −0.073/0.013 = −5.46, which has a p-value of 0.000, so the null hypothesis can be rejected at the 1% level (and thus, also at the 10% and 5% levels). (b) The 95% confidence interval is −0.073 ± 1.96 × 0.013 or −0.100 to −0.047. ¶ (c) Ed = 13.94 − 0.064 × Dist (0.05) (0.018)

Solutions to Empirical Exercises in Chapter 5

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¶ (d) Ed = 13.98 − 0.084 × Dist (0.06) (0.013)

ˆ ˆ (e) The