1 Econ 101 Professor Wallace Fall 2013 Lab Section Handout Answers (Week 9) Instructions: The questions below serve as a practice exam for the coming midterm. Answer them as best you can in the allotted time during the first 1:15 of the section. The final 40 minutes of lab is reserved for question and answer. 1. (20 points) Definitions. Provide definitions for the following A. (5 points) Null hypothesis – a tentative assumption made about a population parameter. B. (5 points) Type 1 error – the probability of rejecting the null when it is true C. (5 points) Statistical power – the power of (correctly) rejecting the null when it is false. D. (5 points) p-value – likelihood of drawing a value of the sample statistic at least as extreme as the one that was observed assuming the null is correct. 2. (20 points) True, False, Uncertain A. (5 points) When elements of a sample 12, ,...,nX X X are not independent  2Var Xn The answer I intended: True – In this case we cannot ensure that 21niiVar X n or, put differently, the variance of the sum in the numerator of X is not generally equal to the sum of the variances. * But, just because 12, ,...,nX X X are not independent doesn’t mean that ( , ) 0ijCov X X  for ij. If 12, ,...,nX X X are not independent but ( , ) 0ijCov X X  for all ij then 2Var Xn. B. (5 points) When the sample is too small for the central limit theorem to apply X is not generally normally distributed. True – for normality we rely on the random large sample size or, alternatively, a normally distributed population (which cannot generally be assumed).2 C. (5 points) The statistical power is bounded below by  True – when the distribution under the null and the true distribution are infinitesimally different the statistical power is infinitesimally greater. D. (5 points) In the hypothesis testing framework we tentatively assume null is false and look for evidence from the sample to refute this assumption. False – in the hypothesis testing framework we tentatively assume the null is true and then look for evidence from the sample to refute this tentative assumption. 3. (20 points) Short Answer A. (10 points) Describe the circumstances under which a 95% confidence interval for  will not contain the value  (Hint: You may want to utilize a graph). A 95% confidence forinterval will not contain the value of  when we are unfortunate enough to draw a value of the sample statistic X that is in the bottom 2.5% or top 2.5% of the distribution. B. (10 points) Describe three factors that influence the probability of type 2 error. The probability of a type 1 error (the significance level) – if we are willing to accept a higher probability of a type 1 error the probability of a type 2 error can be decreased The sample size – increasing the sample size will decrease the probability of type 2 error (as long as it is not already zero), all else equal. The larger the difference between the hypothesized and true (alternate) values of the population parameter the smaller the probability of a type 2 error, all else equal. 4. (60 points) Has Support for Obamacare Increased? By all accounts the rollout to President Obama’s signature legislative achievement has been botched. The federally run healthcare exchanges, electronic marketplaces where consumers can shop for subsidized health insurance policies, has been plagued by computer glitches that have made the purchase of policy difficult if not impossible for the vast majority of consumers. Moreover, despite Obama’s assurances that people that like the insurance policy that the currently have will be able to keep it, some consumers who have in the past bought policies in the individual marketplace have begun receiving cancellation notices from their insurers. Despite these problems with the rollout support for Obamacare in some polls has increased. In an NBC News, Wall Street Journal Poll of 800 Americans taken in early September (before the rollout problems) 31 percent of those polled indicated they thought the health care law was a “good idea”. In a more recent poll of 800 Americans, conducted by the same pollsters in late October (after the rollout problems) found that 37 percent of those polled indicated they thought he health care law was a good idea. In this question you will be asked to use information from the two NBC News, Wall Street Journal Polls to draw inferences about the change in the support for Obamacare between early-September and late-October.3 A. (5 points) What is the population parameter of interest? The population parameter is the change in the fraction of Americans who think the health care law was a good idea. Mathematically this is given by Oct Septp p p   B. (5 points) Provide a point estimate of the population parameter from part A. A point estimate of this parameter is given by the difference in the sample proportion of poll respondents who indicated that the healthcare law is a good idea between the October and September polls. Thus, 0.06 is a point estimate of p C. (10 points) Construct 95% confidence interval for the parameter from part A. Use your constructed confidence interval in a sentence.  0.37(1 0.37) 0.31(1 0.31)0.06 1.96 0.014,0.106800 800   We are 95% confident that the change in the fraction of Americans that think the healthcare law was a good idea between September and October is between 0.014 and 0.106 D. (5 points) What is the most appropriate hypothesis to test to determine whether there is strong statistical evidence that support for the health care law has grown between early-September and late-October? Be sure to state the null and alternative hypothesis. 0:0:0AHpHp E. (5 points) What is the distribution of the appropriate sample statistic for conducting the hypothesis test in part D. under the null? P is distributed normally with mean zero and variance 2 (1 )800pp where Oct Septp p p4 F. (10 points) Conduct the hypothesis test in part E. At what standard significance levels can you reject the null? Note that p-tilde=0.34  0.06| 00.06 2.53 0.012(0.34)(1 0.34)800p value P p pP Z z          Thus, we can reject the null hypothesis at significance levels of 0.10,