Economics 310 Menzie D. Chinn Fall 2004 Social Sciences 7418 University of Wisconsin-Madison Problem Set 4 Answer Key This problem set is due in lecture on Wednesday, December 1st. No late problem sets will be accepted. Be sure to show your work (that is, do not use a spreadsheet or statistical program to generate your answers), and to write your name, ID number, as well as the name of your Teaching Assistant, on your problem set. Numbered problems refer to the textbook problems. • 8.16 • 8.20 • 8.30 • 8.50 • 8.56 • 8.62 • 8.72 • 8.80 • 8.82 • 9.2 • 9.22 • 9.28 • 9.36 • 9.44 • 9.60 • 9.7023456789• Problem X.1 During the late 1990s, there was a belief that Europe lagged behind the United States in productivity growth, especially over the 1995q1-2000q4 period (shaded area in the graph below). -.02-.01.00.01.02.03.04.05.0682 84 86 88 90 92 94 96 98 00D(LPRODUS90,0,4) D(LPRODEU90,0,4)US (4 qtr growth rate)Eurozone(4 qtr growth rate) Figure 1: US and Eurozone 4 quarter growth rates. Here are some statistics below on quarter-on-quarter annualized growth rates and differences in growth rates between the two economies. 012345-0.02 0.00 0.02 0.04 0.06 0.08Series: D(LPRODUS90)*4Sample 1995:1 2000:4Observations 24Mean 0.026989Median 0.023134Maximum 0.088328Minimum -0.016824Std. Dev. 0.025717Skewness 0.575041Kurtosis 3.120793Jarque-Bera 1.337277Probability 0.512406 Figure 2: Histogram for United States qoq annualized GDP growth1001234567-0.04 -0.02 0.00 0.02 0.04 0.06 0.08Series: D(LPRODEU90)*4Sample 1995:1 2000:4Observations 24Mean 0.012612Median 0.008859Maximum 0.089501Minimum -0.036604Std. Dev. 0.025384Skewness 0.843972Kurtosis 4.866237Jarque-Bera 6.331998Probability 0.042172 Figure 3: Histogram for Eurozone qoq annualized GDP growth 012345-0.06 -0.04 -0.02 0.00 0.02 0.04 0.06Series: DIFSample 1995:1 2000:4Observations 24Mean 0.014377Median 0.013389Maximum 0.061469Minimum -0.055100Std. Dev. 0.032564Skewness -0.357949Kurtosis 2.343087Jarque-Bera 0.944045Probability 0.623739 Figure 4: Histogram for DIF = (d(lprodus90)-d(lprodeu90))*4 X.1.a. Conduct a test for difference in means between the US and Eurozone, assuming independent sampling, and setting the probability of Type I error equal to 2.5%. The hypothesis test that corresponds to this text is: Hdy dyUS Eu0:= Hdy dyAUS Eu:≠ The general formula for this test, assuming independent random sampling, is given by: tXX Dsnnsp=−−+⎛⎝⎜⎞⎠⎟()1021211 where snsnsnnp211222212112=−+−+−()() or in this case where 1 refers to the US and 2 refers to Eu, and D0 = 0, tXXsnnUS EupUS Eu=−−+⎛⎝⎜⎞⎠⎟=−+⎛⎝⎜⎞⎠⎟==()(. . )(. )...0110270 01260 0013061241240 01440 01043138211The critical value for ttα/.2 0125= for 46 (=24+24-2) degrees of freedom cannot be read off the t-table directly, for two reasons. First, there are no 46 df entries, just 40 and 60 df entries. Use 40 df; then there is no entry for t.0125 (remember this is a two-tailed test). One way to proceed is to realize that even using a lower probability of Type I, error, say 0.025, means the critical t value is 2.021, and 1.38 < 2.021, one fails to reject the null hypothesis in favor of the alternative. The other approach is to be more careful, and to linearly interpolate between the two values; .0025 (=0.0125-0.01) is 1/6 of the distance between .00250 and 0.010, so one could say that 2.021 + (5/6)× (2.423-2.021) = 2.356 is the critical t value. X.1.b. Conduct a test for difference in means between the US and Eurozone assuming paired sampling, and once again setting the probability of Type I error equal to 2.5%. The hypothesis test is now: Hdy00:$= where dy dy dyUS Eu$=− HdyA:$≠0 txsnDD=−=−≈00 0144 00 0326 23212/../. Now it becomes obvious that being careful is important. Using the t.025 critical value for 23 df (=24-1), one would reject the null. However, linearly interpolating, one finds that the t.0125 critical value is approximately 2.069+(5/6)×(2.500-2.069) = 2.428. Notice, if one performed the one tail test corresponding to the hypothesis test: Hdy00:$= HdyA:$>0 then one could use the entry in the t-tables, where the critical value is 2.069 < 2.428, and one would reject the null hypothesis in favor of the alternative. X.1.c. Which testing procedure is more appropriate? Why? The paired differences test is more appropriate because the data are sampled in a systematically related manner. • Problem X.2 Consider the following data pertaining to annualized quarter-on-quarter growth rates of real US GDP (in 1996 chained dollars) over the last 25 years.12-.10-.05.00.05.1080 82 84 86 88 90 92 94 96 98 00 02 04D(LGDP96C)*4 Figure 5: US qoq annualized GDP growth rate, 1979q1-2004q3 X.2.a. Suppose one wanted to test the hypothesis that the variability of US GDP growth was greater than 2.5%. Write out the hypothesis test, given the following information. 024681012-0.05 0.00 0.05Series: D(LGDP96C)*4Sample 1979:3 2004:4Observations 101Mean 0.029616Median 0.032474Maximum 0.089218Minimum -0.081572Std. Dev. 0.029318Skewness -0.971839Kurtosis 5.257074Jarque-Bera 37.33745Probability 0.000000 Figure 6: Histogram for United States qoq annualized GDP growth, 1979q3-2004q3. H0220025:(.)σ= H0220 025:(.)σ> X.2.b. Conduct the hypothesis test, using the data provided. What is your conclusion, setting α = 0.05? We calculate the χ2statistic: χσ2221137 36=−=().ns The critical value for 100 df (=101-1) is 124.32; hence, we reject the null hypothesis that the standard deviation of the annualized growth rate is 2.5% in favor of the alternative that it is greater than 2.5%.13X.2.c. What assumptions do you need to impose in order to arrive at your conclusion? It is necessary to assume that the growth rate is normally distributed. • X.3. There is a widespread view that the US macroeconomy has been more “stable” since 1984. X.3.a. Write out a hypothesis test that the economy has been more stable in the post-1984 era. Let sample 1 correspond to the pre-1984 era, and sample 2 correspond to the post-1984 era. Then Hs0122:σσ= H01222:σσ> X.3.b. Conduct the appropriate test, using the statistics provided below, and setting Type I error equal to 1%. 048121620-0.10 -0.05 -0.00 0.05 0.10 0.15Series: D(LGDP96C)*4Sample 1953:1 1984:4Observations 127Mean 0.032815Median