STA 6166 UNIT 2 Section 2 Answers
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# Unit 2 Section 2 Answers

You can choose to work some or all of the problems listed below. We recommend that you at least work the problems listed in your major area of interest.

 General Questions. If two independent random variables, y1 and y2, are normally distributed with means and variances (m1, s21) and (m2,s22) respectively, the difference between the random variables has what distribution (be specific - give mean and variance). The difference of two independent normal random variables is itself a normal random variable having mean equal to the difference of the two means and variance equal to the sum of the two variances. Hence The sampling distribution of the difference between two sample means has an approximate normal distribution in large samples with mean equal to what? If each mean is normal, by the rule above, the mean for the distribution of the difference of two sample means should be the difference of the two population means: The standard error of the sampling distribution of the difference between two sample means has what value? Assuming the population variance for each random variable is known, each sample mean has corresponding standard error equal to . Hence, by 1 above, we sum the individual variances and then take the square root to get: . This works if the two population variances are assumed known and are truly different. If they are known to be the same, we can factor the population variance term out of the sum and even remove it from under the square root radical sign to get: The following estimate for the standard error of the difference between two means is used when what assumption about the two population variances can be made?This estimate is used when we can honestly make the assumption that the two populations have common variance (or common standard deviations). The first square root term is our pooled estimate for the value of this common population standard deviation term. What assumption about the two population variances is made when the following standard error estimate is used? In this case, we are assuming that the two populations have individual variances (standard deviations) that are NOT equal to each other. This is referred to as the separate variances estimate of the standard error of the difference of two sample means. Most of the time in two-population tests, we test the hypothesis H0: m1-m2 =0. What does it mean to test the hypothesis H0: m1-m2 = D0 where D0 does not equal to zero? The term D0 refers to the expected difference. When D0 does not equal zero, we are saying that we expect one of the means to differ from the other mean by this amount on average. So if we say m1-m2 =10, we are saying that on average we expect the difference between the means to be 10 units. What we are hoping to do is show (depending on the alternative hypothesis) that the true difference is different (less than or greater than) D0. Why is the Wilcoxon Rank Sum test use the words "Rank Sum" in its name? The test statistic used is constructed as the sum of ranks for the sorted data. In this case, the test statistic is the sum of the ranks for one of the populations. Use Table 5 in the Appendix to find the critical value of the Wilcoxon rank sum test for independent samples when n1=7 and n2 = 6 and the alternative hypothesis is "Population 1 is shifted to the right of Population 2" with Type I error probability of a=0.05. The alternative hypothesis tells us that we are in Case 1 and that the critical value will be TU[7,6,a=0.05,one-tailed]=54 How is the critical value for the Wilcoxon rank sum test found if one or both of the sample sizes are greater than 10? The answer is on page 292 in the book. Essentially we revert to a one-sample z-test where the mean we are comparing the observed rank sum T to is given by the sample sizes (see the mean value on page 298). The true variance for the z-statistics depends on the sample sizes as well as the numbers of tied ranks. Note that the theory behind the Wilcoxon rank sum test requires that the underlying distributions are continuous. What do we mean when we talk about "Paired Data"? Paired data refer to data collection (or experimental) conditions where the measurements for the two "factors" of interest are taken on the same unit or individual. Because of this, we expect some correlation or lack of independence between the two measurements, e.g. association between the paired measurements caused by their being taken on the same individual. The statistical analyis must take into account this structural aspect of the sampling. Another way of looking at the paired data case is as follows. In the two independent sample case we can conceptually envision randomly assigning study units to treatments. In the paired data case, the study units are created in pairs, each pair is defined on one individual. Each individual gets both treatments, treatments randomly assigned to the paired study units for that individual. Individuals are hence assumed random. Is a two-sample Paired Data t-test equivalent to a one-sample t-test performed on the differences in values for each sample unit? Why of course! Which of the two tests is for testing the difference in means from samples of two independent populations? The Wilcoxon Rank Sum Test or the Wilcoxon Signed Rank Test? So what does the other test? The Wilcoxon Rank Sum Test is the nonparametric equivalent of the two independent sample t-test. The Wilcoxon Signed Rank Test is essentially the equivalent of the one-sample t-test. It is used to for paired data situations as well as one-sample testing. The equation for estimating sample sizes for the two-sided hypothesis test of differences of means is given by:. Can you define each of the terms in this equation? See the discussions on pages 315 and 316 in the book. The D term is defined by how big of a difference between the population means we feel must occur if the populations are to be declared truly different. The a and b relate to the probabilities of Type I and Type II errors. The sd is the value of the true standard deviation of the differences. Do problem 6.83 as a paired samples t-test and using the Wilcoxon Signed Rank test, both with a Type I error probability of 0.05. Do you get different results? Using Excel we can easily compute the differences, the appropriate t-statistic and compute the associated t-critical value and p-value. ``` Sample Analyst_1 Analyst_2 Difference Signed_Rank 1 31.4 28.1 3.3 5 2 37.0 37.1 -0.1 -1 3 44.0 40.6 3.4 6 4 28.8 27.3 1.5 3.5 5 59.9 58.4 1.5 3.5 6 37.6 38.9 -1.3 -2 Mean Difference= 1.38 Variance Difference= 3.43 Standard Deviation= 1.85 sample size(n)= 6 t-statistic = 1.83 t-critical(0.05,5)= 2.02 =tinv(0.10,5) p-value= 0.06 =TDIST(D14,5,1)``` From this we can determine that the null hypothesis that Analyst_1 and Analyst_2 read the same cannot be rejected in favor of the alternative that Analyst_1 reads higher than Analyst_2. For the Wilcoxon Signed-Rank test we use the computed ranks of the differences and associated sign as given in the table above. Note that I have used the average rank for the two tied rankings. Since the differences were taken as Analyst_1 minus Analyst_2, the alternative hypothesis is that the median differences should tend to be larger than zero (Case 1). The test statistic is T-, the absolute value of the sum of the negative ranks. Here T- = |-3| = 3. The critical value for the test is obtained from Table 6 for n=5 and a = 0.05 one-tailed. T-critical = 2. Since T- = 3 > T-critical=2, we do NOT reject the null hypothesis and conclude that both analysts are reading similarly. We conclude that both tests lead us to the same decision. Again, using the scenario of problem 6.83 (page 334), how many water samples would we need if we wanted to be certain that the two Analysts did not differ by more than 2 ppm with Type I error probability of 0.05 and Power of 0.90 assuming the underlying variance in the differences were 1.0? Equation on page 316 was used and programmed into Excel. The following results were obtained. ``` P(Type I error)= 0.05 =C20 P(Type II error)= 0.10 =C21 Variance of difference= 1.00 =C22 z(alpha) = 1.644853 =NORMSINV(0.95) z(beta)= 1.281550794 =NORMSINV(0.90) Delta= 2.00 =C25 Expected Sample size = 3 =CEILING(C22*((C23+C24)^2)/(C25^2),1) ``` From this we conclude that only three samples are needed. Note that the assumed variances of the differences here is 1.0 which is less than the 3.43 we observed in the previous section. If we assumed the variance were 4.0 instead, the needed sample size would be 9 individuals. If you have Excel, you can download the simple spreadsheet on which these calculations were based from here. Review the Key Formulas on pages 317-318. For students in agriculture and environmental fields. Answers For students in engineering fields. Answers For students in toxicology and health science fields. Answers This answer set also has links to SAS, MINITAB and SPSS solutions to the problem. Look at them if you want an idea of alternative ways of doing this problem. For students in community development, education and social services fields. Answers