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slides 3-chapter 15part1

slides 3-chapter 15part1 - Chapter 15 fl Sections 15.2 15.3...

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Unformatted text preview: Chapter 15: fl Sections 15.2, 15.3, and 15.5 Nonparametric Statistical Inference: Comparing Two Populations i Outline . Sections 15.2, 15.3, and 15.5 will cover: - The Wilcoxon Rank Sum Test: Independent Random Samples - The Sign Test: Paired Experiment . The Wilcoxon Signed—Rank Test: Paired Experiment I Table used: . Tables 1, 7, and 8 i Nonparametric Statistical Inference . Nonparametric statistical inference methods... . ...do not require the normality assumption to be satisfied. . ...specify hypotheses in terms of population distributions rather than population parameters. . ...use the ranks or signs of observations rather than the actual observations themselves. - ...are usually more powerful than their parametric counterparts when normality is not satisfied. 3 The Wilcoxon Rank Sum Test: $ Independent Random Samples - The Wilcoxon Rank Sum Test is used when we wish to compare two populations and the following conditions have been met: . We have two independent SRSs, one from each population. . The normality assumption has been violated or cannot be validated, and the sample sizes are not large enough to apply the Central Limit Theorem. i Step 1: State The Hypotheses H0: The distributions for Population 1 and Population 2 are identical. versus one of the following alternative hypotheses: : The distributions for Population 1 and Population 2 are different. (two-tailed test) : The distribution for Population 1 lies to the left of that for Population 2. (left-tailed test) : The distribution for Population 1 lies to the right of that for Population 2. (right-tailed tests) i Step 2: Compute Test Statistic T - In order to compute the test statistic, we must a. Let n1 denote the smaller of the two sample sizes, and denote the population of the origin of this sample as Population 1. b. Rank all nl + n2 observations from smallest (1) to largest (111+ n2). c. Compute T1 = the sum of the ranks of the first sample. This is the test statistic Tfor a left tailed test, so if that is the test of interest, stop. Otherwise, proceed to d. i Step 2: Compute Test Statistic T d. Compute Tf = n,(nl + n2 +1)— T,. This is the sum of the ranks of the first sample if the assigned ranks had been reversed from large to small. This is the test statistic T for a right tailed test, so if that is the test of interest, stop. Otherwise, proceed to e. e. The test statistic T for a two-tailed test is the minimum of TI and T1 NOTE: When ranking the observations, tied observations are assigned the average of the ranks that the observations of differing values would have received. 7 i Step 3: Critical Value - The critical value for the test statistic T computed in Step 2 can be found in Table 7. - These critical values are given for one-tailed tests with a = 0.05, 0.025, 0.01, and 0.005. - NOTE: For a two-tailed test, the values of a are doubled. - The table gives the critical value c such that P(T S c) S a . i Step 4: Decision . If our observed test statistic T is less than or equal to the critical value c, reject H0 and accept HA. Exercise 1 . The weights of turtles caught in two different lakes were measured to compare the effects of the two lake environments on turtle growth. Lake Weight (in 02) At 05 = 0-05 do the data provide 1 14.1 15.2 13.9 14.5 14.7 sufficientevidence to 13'8 14'0 16'1 12'7 15'3 indicateadifference 2 12.2 13.0 14-1 13-6 12-4 in the distributionsof 11.9 12.5 13.8 turtle weights between the two lakes? The Wilcoxon Rank Sum Test: i Large Sample Approximation . If our samples sizes are large enough (say at least 10), the Wilcoxon rank sum test statistic T1 will follow an approximately normal distribution with 111(n1 +n2 +1) and 02 : I’lll’l2(l’ll +n2 +1) . 2 "' 12 T _ [UT approx ~ N(0, 1). #1' = . Thus, 2 = 0T . We can use this as the basis for a test statistic, and carry out a hypothesis test in the usual way, using either the p-value or critical value method. 11 i Exercise 2 - Suppose you wish to detect a shift in distribution 1 to the right of distribution 2 based on sample sizes HI = 12 and n2 = 14. If T1 = 193, what do you conclude? Use a = 0.05. i The Sign Test: Paired Experiment - The Sign Test is a non-parametric counterpart of the paired-difference test in Section 10.5. It should be used when any of the paired-difference assumptions have been violated or cannot be validated. These assumptions are: - Sample differences are at least approximately normally distributed. . Equal variances across pairs. i The Sign Test: Paired Experiment . Recall that our data consists of ordered pairs (A, B), and that we take the difference between these pairs to obtain A , B. When the two populations distributions are identical, the probability that A exceeds B would bep = 0.5. LetX represent the number of times A exceeds B. Then under the assumption that the two population distributions are identical, it follows that X~ B(n, 0.5), where n is the number of observed ordered pairs. 14 i Step 1: State The Hypotheses H0: The population distributions are identical and P(A exceeds B) =p = 0.5. versus one of the following alternative hypotheses: HA: The population distributions different and p i 0.5. HA: The population ofA measurements is shifted to the right of the population of B measurements andp > 0.5. HA: The population ofA measurements is shifted to the left of the population of B measurements and p < 0.5. i Step 2: Compute Test Statistic x - x is simply the number of times that A exceeds B. - In other words, x is the number of times the difference A , B is positive. . NOTE: Delete tied pairs — that is, pairs with A = B — from the data set. Reduce the sample size n accordingly. i Step 3: p-value . While it is possible to use a critical value method here, it’s much easier to compute the p-value. The way in which this is done depends on the direction of HA, and on the assumption thatX~ B(n, 0.5): HA : p>0.5:> p—value=P(X2x) HA :p<0.5:>p—value= P(XSx) 2P(X2x) if x>n/2 2P(XSx) if x<n/2 HAzp¢0.5:>p—value={ 17 i Step 4: Decision . We make a decision based on the p-value in the usual way. i Exercise 3 . An investment analyst wants to test whether differences exist between the returns on two mutual funds. Paired data of annualized rates of return (in 0/o) for the two mutual funds during 16 randomly chosen months are given on the next slide. Is there evidence at a = 0.10 to suggest whether or not the distribution of returns are equal? Exercise 3 The Sign Test: i Large Sample Approximation - If our samples size is large enough (say at least 25), we can approximate our binomial random variable/test statistichith a normal distribution. - Assuming H0 is true, this means approx X N N(O.5n,0.25n). x — 0.5n WW ~ N 0,1). 0.5% ( . We can use this as the basis for a test statistic, and carry out a hypothesis test in the usual way. 21 I Thus, 2 = i Exercise 4 - A clinic is interested in comparing the popularity of two doctors on staff, Dr. Pepper and Dr. Feelgood. They select 30 patients at random to see both doctors about the same problem and then ask them to rate each doctor on a scale from 1 to 10. . Three of the patients rated the doctors the same. Of the remaining patients, only 8 preferred Dr. Pepper. At a = 0.05, is there evidence to suggest that Dr. Pepper is less popular than Dr. Feelgood? 22 The Wilcoxon Signed-Rank Test: i Paired Experiment . The Wilcoxon Signed-Rank Test is another non- parametric counterpart of the paired-difference test in Section 10.5. Like the Sign Test, it too should be used when any of the usual paired-difference assumptions have been violated or cannot be validated. The Wilcoxon Signed-Rank Test: Paired Experiment - Recall that our data consists of ordered pairs (X, Y), and that we take the difference between these pairs to obtain X— Y. - When the two populations distributions are identical, we would expect half, or n/2, of these differences to be negative and half to be positive. - We would also expect that positive and negative difference of equal magnitude to occur with equal probability. i Step 1: State The Hypotheses H0: The two population distributions are identical. versus one of the following alternative hypotheses: H A: The two population distributions differ. HA: The Population 1 distribution is shifted to the right (>) of the Population 2 distribution. HA: The Population 1 distribution is shifted to the left (<) of the Population 2 distribution. Step 2: Compute Test Statistic T . In order to compute the test statistic, we must a. Compute the differences X— onr each of the n pairs. Tied pairs are eliminated and n is reduced. b. Rank the absolute values of the differences from smallest (1) to largest. Average tied ranks. c. For the right tailed test, compute T’ = the sum of the (absolute value of the) ranks of the negative differences. For the left tailed test, compute T = the sum of the ranks of the positive differences. For the two-tailed test, take the minimum of TT and T . i Step 3: Critical Value . The critical value for the test statistic T computed in Step 2 can be found in Table 8. - These critical values are given for one-tailed tests with a = 0.05, 0.025, 0.01, and 0.005, or, equivalently, for two-tailed tests with a = 0.10, 0.05, 0.02, and 0.01. - The table gives the critical value T0 such that P(T 3 To) 3 a . i Step 4: Decision . If our observed test statistic T is less than or equal to the critical value To, reject H0 and accept HA. i Exercise 5 . Eight people were asked to perform a simple puzzle-assembly task under normal conditions and under stressful conditions. During the stressful time, mild shocks were given every 30 seconds. The highest blood pressure readings were recorded under both conditions, as given on the next slide. Do the data present sufficient evidence to indicate higher blood pressure readings under stressful conditions? Use a = 0.05. i Exercise 5 Stressful 130 118 125 120 121 125 130 120 10 The Wilcoxon Signed-Rank Test: Large Sample Approximation - If our samples sizes are large enough (say at least 25), the Wilcoxon signed-rank test statistic T‘ will follow an approximately normal distribution with x = ”W” and 0,2 w. 4 "* 24 T+ _ #T approx - ThUS,z= ~ N(0,1) UT - We can use this as the basis for a test statistic, and carry out a hypothesis test in the usual way, using either the p-value or critical value method. 31 i Exercise 6 . Suppose you wish to detect a shift in distribution 1 to the right of distribution 2 based on a paired- difference experiment consisting of n = 30 pairs. - You are given T ‘= 249. Use a 20.05 . Appendix Normal Probability Plots 11 1| Normal Probability Plots . We want to check our assumption that observed values are coming from a normal distribution. - We could check this assumption in a rough way by graphing a histogram or a stem-and-leaf plot of our sample data and checking to see if its shape is more or less “bell shaped”. - However, a more accurate/standard tool is a normal probability plot. 1| Normal Probability Plots . To start, observed values are first ranked from smallest to largest. - The ranked values are then plotted against their observed cumulative frequency. . If the observations are coming from a normal distribution, the plotted points will fall approximately along a straight line. - The determination of how well the points fall along a straight line is usually subjective. Normal Probability Plots 1| in MINITAB . Enter your data in one of the spreadsheet columns. Go to the top of the screen and click on Graph, then Probability Plot. Then, select Single. . Under Variables, simply indicate which column you are graphing data from. The default name is C1. - Click Distribution and under the Distribution tab, ensure “Normal” is selected. Under the Data Display tab, you can uncheck the option to “Show Confidence Interval”. - Click OK. Click OK again. 12 Normal Probability Plots 1| in MINITAB . In each of the three examples that follow, 500 observations were randomly generated (simulated) from some distribution. . We will look at the corresponding histogram and normal probability plot in each case. 1| Example 1: N(70,144) 50 p O o) O M O Frequency _x O j K—r—l l I 100 110 38 l l Percentile lllllll l 13 1| Example 2: eXp(5) 90 80 70 60 50 40 30 20 10 O Frequency 2 '5 C <1) 0 L a.) a. 1| Example 3: U(0,1) 30 N O Frequency 14 3| Example 3: U(0,1) 2 ‘E (1) U L d) o. 15 ...
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