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UW-Madison ECON 310 - practice_final_with_solutions

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Practice Problems for Final Exam. Econ 310, Spring 2012Problem 1To determine the effectiveness of a new type of fertilizer, researchers took 8 two-acre plots of landand divided each plot into two equal-sized subplots. They then treated one of the subplots with thecurrent fertilizer and the other one with the new fertilizer. Wheat was planted, and the crop yieldswere measured. The results are summarized in the following table.Plot 1 2 3 4 5 6 7 8Current fertilizer 56 45 68 72 61 69 57 55New fertilizer 60 49 66 73 59 67 61 60(a) Can we conclude at the 5% significance level that the new fertilizer is more effective than thecurrent one?ANS: Let µ1= average crop yield with the new fertilizer and µ2= average crop yield with thecurrent fertilizer. Null hypothesis and alternative hypothesis is as follows; H0: µ1− µ2= 0,H1: µ1− µ2> 0. As in the sketch of the proof, samples we cannot assume that two samplescome from two independent populations. Since each sample from current fertilizer is matchedwith a sample from new fertilizer, as a result we should study uses a pairwise matched sampledesign. Let µD= µ1−µ2, mean of the population of differences. Then hypotheses to be testedare H0: µD= 0, H1: µD> 0. Test statistic for µD, t =¯xD−µDsD/√nD=1.5−03.117/√8= 1.361. Therejection region is t > t0.05,7= 1.895, thus we can not reject the null hypothesis.(b) Estimate with 95% confidence the difference in mean crop yields between the two fertilizers.ANS: Confidence interval for µDis ¯xD± tα/2sD√nD= [1.5 ± 2.364 ∗3.117√8] = [−1.106, 4.106]Problem 2How do drivers react to sudden large increases in the price of gasoline? To help answer this, astatistician recorded the speeds of cars as they passed a large service station. He recorded thespeeds (mph) in the same location after the service station sign showed that the price of gasolinehad risen by 15 cents.Speeds Recorded Before Price Increase43 36 32 29 28 36 27 36 35 30 32 36 33Speeds Recorded After Price Increase32 33 31 32 29 28 39 26 32 301(a) Can we conclude, at a 5% significance level, that the average speed differs before and after theprice increase?ANS: First, test Ho: σ21/σ22= 1. The observed test statistic is s21/s22= 18.997/12.178 = 1.552while the critical values are F.025,12,9= .29 and F.925,12,9= 3.87. We don’t reject the null.Second, based on the previous result, we assume the population variances are equal. We testHo: µ1− µ2= 0 against H1: µ1− µ26= 0. Our test statistic is:t =¯x1− ¯x2rs2p1n1+1n2=33.308 − 31.2q16.018113+110= 1.252The critical value is t.025,21= 2.414 so we don’t reject the null.(b) Can we conclude, at a 5% significance level, that the average speed falls after the price increase?ANS: This is a one-tail test, the alternative hypothesis is now H1: µ1− µ2> 0. The observedt stat is the same as before, but the value from the table is different, t.05,21= 2.080, and onceagain, we do not reject the null, i.e. we do not have enough evidence to reject the hypothesisthat the average speeds are the same before and after the price increase.Problem 3Is aspirin effective in reducing the incidence of heart attacks? To study this question, 22,000 in-dividuals took part of a study. Half of them took an aspirin tablet three times per week, and theother half took a placebo on the same schedule. Over time, the following was observed: Amongthose who took aspirin, 104 suffered heart attacks, while 189 individuals taking the placebo suffereda heart attack.1. Determine whether these results indicate, at a 1% significance level, that aspirin is effectivein reducing the incidence of heart attacks?ANS: The key is to recognize that this is a test about the difference between two populationproportions: p1= proportion of aspirin-taking individuals who suffer a heart attack, and p2= proportion of placebo-taking individuals who suffer a heart attack. Our null hypothesisis then: H0: p1− p2= 0 and our alternative is H1: p1− p2< 0. Remember that in thecase of a population test between two proportions, we are also implicitly assuming that thepopulation variances are the same. This stems from the fact that the population variancesare σ21= p1· (1 − p1) and σ22= p2· (1 − p2); hence the null hypothesis is also a test for equalpopulation variances. We accordingly proceed by using the pooled variance estimator. First,we compute our estimates for the sample proportions ˆp1=10411,000= 0.009, ˆp2=18911,000= 0.017.We then defineˆP =104+18922,000= 0.0133. The pooled variance estimator is thus s2p=ˆP (1 −ˆP ) =0.0133 ∗ 0.9967 → 0.0131. Our test statistic under the null is then:ˆp1− ˆp2rs2p1n1+1n2=0.009 − 0.017r0.0131111,000+111,000→ −5.002As a rule of thumb, recall that whenever we do inference on proportions, the Standard Normal(and not the Student T) is the approximate distribution we use. Since this is a one-sided test,the critical value is −2.33. Hence we reject the null.2. Construct a 90% confidence interval for the difference in the proportion of aspirin-takingindividuals who suffer heart attacks and the proportion of placebo-taking individuals whosuffer heart attacks.ANS: Relying on the Normal approximation, a Confidence Interval for p1−p2with (1 −α)%coverage probability is given by(ˆp1− ˆp2) ± zα/2·sˆp1(1 − ˆp1)n1+ˆp2(1 − ˆp2)n2Hence, we have:"(0.009 − 0.017) ± 1.645 ·s0.009(1 − 0.009)11, 000+0.017(1 − 0.017)11, 000#→[−0.01027, −0.0052]Problem 4In order to find out if the performance of publicly traded companies affects the way their boardmembers are paid, a study was conducted where publicly traded companies were divided into fourcategories based on the rate of return in their stocks. The annual payment to their board memberswas recorded. The summary statistics for each group are below.Sample Size Sample Mean Sample VarianceGroup 1 28 74.10 249.96Group 2 23 78.24 159.94Group 3 25 78.54 247.96Group 4 24 80.20 240.58(a) Can we infer, at a 5% significance level, that the amount of payment differs between the fourgroups of companies?ANS: This is a straightforward ANOVA test problem. Simply perform an ANOVA F-test forthe null hypothesis H0: µ1= µ2= µ3= µ4.¯¯X =Pkj=1Pnji=1Xijn=P4j=1(nj·¯Xj)n1+ n2+ n3+ n4= 77.6262SST =4Xj=1nj· (¯Xj−¯¯X)2= 536.682 =⇒ M ST =SSTk − 1= 178.893SSE =4Xj=1(nj− 1) · s2j= 21751.98 =⇒ M SE =SSEn − k= 226.58Test statistic: F =MSTMSE= 0.79. Because under


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UW-Madison ECON 310 - practice_final_with_solutions

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