Please work out the following questions correctly!

To practice for the exam use the t and z-tables supplied at the end of this file. Be sure to learn to use these tables. Note the t and z-tables give left tail probabilities and the x2-table gives right tail critical values. 1. (a) We compute the data mean and variance = = 65, s' = 35.778. The number of degrees of freedom is 9. We look up the critical value to,0.025 = 2.262 in the t-table The 95% confidence interval is 19,0.0258 It 9,0.025-8 = 65 - 2.262V3.5778, 65 + 2.262\\3.5778 = [60.721, 69.279] On the exam you will be expected to be able to use the t-table. We won't ask you to compute by hand the mean and variance of 10 numbers. 95% confidence means that in 95% of experiments the random interval will contain the true 0. It is not the probability that o is in the given interval. That depends on the prior distribution for 0, which we don't know. (b) We can look in the z-table or simply remember that 20.025 = 1.96. The 95% confidence interval is i - 20.0250 20.025 7 1.96 - 5 1.96 - 5] Vn 65 - 65 + V10 V10 = [61.901, 68.099] This is a narrower interval than in part (a). There are two reasons for this, first the true variance 25 is smaller than the sample variance 35.8 and second, the normal distribution has narrower tails than the t distribution. (c) We use the normal-normal update formulas to find the posterior pdf for 0. a = 10 16 b = 25' "post = 260 + 565 = 64.3, post = = 2.16. atb atb The posterior pdf is f(0|data) = N(64.3, 2.16). The posterior 95% probability interval for 0 is 64.3 - 20.025 V 2.16, 64.3 + 20.025 V 2.16 = [61.442, 67.206] (d) There's no one correct answer; each method has its own advantages and disadvantages. In this problem they all give similar answers.2. Suppose we have taken data r1, ...; In with mean E. The 95% confidence interval for the mean is + + 20.025- This has width 2 zo.025 - Setting the width equal to 1 and substitituting values zo.025 = 1.96 and o = 5 we get 2 - 1.96- =1 = vn = 19.6. Vn Post-exam 2 practice solutions, Spring 2014 So, n = (19.6) = 384. If we use our rule of thumb that zo.025 = 2 we have vn/10 = 2 = n = 400. 3. We need to use the studentized mean t = I - H 8/ Vn We know t ~ t(n-1) = t(48). So we use the m = 48 line of the t table and find to.os = 1.677. Thus, P(-1.677 - / n I - H 0.5, so > > 0.541. Let T be the number out of 400 who prefer A. We have I = 100 T >0.541, so T > 216 5. A 95% confidence means about 5% = 1/20 will be wrong. You'd expect about 2 to be wrong. With a probability p = 0.05 of being wrong, the number wrong follows a Binomial(40, p) distribution. This has expected value 2, and standard deviation \\ 40(0.05) (0.95) = 1.38. 10 wrong is (10-2)/1.38 = 5.8 standard deviations from the mean. This would be surprising