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Chapter 11 Elementary Statistics Larson Farber Nonparametric Tests.

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Presentation on theme: "Chapter 11 Elementary Statistics Larson Farber Nonparametric Tests."— Presentation transcript:

1 Chapter 11 Elementary Statistics Larson Farber Nonparametric Tests

2 The Sign Test Section 11.1

3 Left-tailed test: H 0 : median k andH a : median < k Right-tailed test: H 0 : median k and H a : median > k Two-tailed test: H 0 : median = k andH a : median k Nonparametric Tests A nonparametric test is a hypothesis test that does not require any specific conditions about the shape of the populations or the value of any population parameters. Tests are often called distribution free tests. The Sign Test is a nonparametric test that can be used to test a population median against a hypothesized value, k. Hypotheses or

4 Sign Test To use the sign test, first compare each entry in the sample to the hypothesized median, k. If the entry is below the median, assign it a – sign. If the entry is above the median, assign it a + sign. If the entry is equal to the median, assign it a 0. Compare the number of + and – signs. (Ignore 0s.) If the number of + signs and the number of – signs are approximately equal, the null hypothesis is not likely to be rejected. If they are not approximately equal, however, it is likely that the null hypothesis will be rejected.

5 Sign Test Test Statistic : When n 25, the test statistic is the smaller number of + or – signs. When n > 25, the test statistic is: For n > 25, you are testing the binomial probability that = 0.50.

6 Application A meteorologist claims that the daily median temperature for the month of January in San Diego is 57º Fahrenheit. The temperatures (in degrees Fahrenheit) for 18 randomly selected January days are listed below. At = 0.01, can you support the meteorologists claim? Write the null and alternative hypothesis. H 0 : median = 57º and H a : median 57º 2. State the level of significance. = Determine the sampling distribution. Binomial with p = 0.5

7 Since H a contains the symbol, this is a two-tail test. There are 8 + signs and 9 – signs. So, n = = – – – – – – –

8 6. Find the test statistic. 5. Find the rejection region. 4. Find the critical value.With n = 17, use Table 8 Critical value is 2. Reject H 0 if the test statistic is less than or equal to 2. The test statistic is the smaller number of + or – signs, so the test statistic is 8.

9 7. Make your decision. 8. Interpret your decision. The test statistic, 8, does not fall in the critical region. Fail to reject the null hypothesis. There is not enough evidence to reject the meteorologists claim that the median daily temperature for January in San Diego is 57. The sign test can also be used with paired data (such as before and after). Find the difference between corresponding values and record the sign. Use the same procedure.

10 The Wilcoxon Test Section 11.2

11 Wilcoxon Signed-Rank Test The Wilcoxon signed-rank test is a nonparametric test that can be used to determine whether two dependent samples were selected from populations with the same distribution. Find the difference for each pair: Sample 1 value – Sample 2 value Find the absolute value of the difference. Rank order these differences. Affix a + or – sign to each of the rankings. Find the sum of the positive ranks. Find the sum of the negative ranks. Select the smaller of the absolute values of the sums. To find the test statistic, w s

12 Application The table shows the daily headache hours suffered by 12 patients before and after receiving a new drug for seven weeks. At = 0.01, is there enough evidence to conclude that the new drug helped to reduce daily headache hours? 1. Write the null and alternative hypothesis. 2. State the level of significance. = 0.01 H 0 : The headache hours after using the new drug are at least as long as before using the drug. H a : The new drug reduces headache hours. (Claim)

13 Before After – – Diff Abs Rank – – Sign Rank

14 The sum of the positive ranks is = 33. The sum of the negative ranks is –1.5 + (–1.5) = –3. The test statistic is the smaller of the absolute value of these sums, w s = 3. There are 8 + and – signs, so n = 8. The critical value is 2. Because w s = 3 is greater than the critical value, fail to reject the null hypothesis. There is not enough evidence to conclude the new drug reduces headache hours.

15 Wilcoxon Rank-Sum Test The Wilcoxon rank-sum test is a nonparametric test that can be used to determine whether two independent samples were selected from populations having the same distribution. Both samples must be at least 10. Then n 1 represents the size of the smaller sample and n 2 the size of the larger sample. When the samples are the same size, it does not matter which is n 1.

16 Wilcoxon Rank-Sum Test Test statistic: Combine the data from both samples and rank it. R = the sum of the ranks for the smaller sample. Find the z-score for the value of R. where

17 The Kruskal-Wallis Test Section 11.3

18 The Kruskal-Wallis Test The Kruskal-Wallis test is a nonparametric test that can be used to determine whether three or more independent samples were selected from populations having the same distribution. H 0: There is no difference in the population distributions. H a: There is a difference in the population distributions. Combine the data and rank the values. Then separate the data according to sample and find the sum of the ranks for each sample. R i = the sum of the ranks for sample i.

19 The sampling distribution is a chi-square distribution with k – 1 degrees of freedom (where k = the number of samples). Given three or more independent samples, the test statistic H for the Kruskal-Wallis test is: where k represents the number of samples, n i is the size of the i th sample, N is the sum of the sample sizes, and R i is the sum of the ranks of the i th sample. Reject the null hypothesis when H is greater than the critical number. (Always use a right-tail test.) The Kruskal-Wallis Test

20 Application You want to compare the hourly pay rates of accountants who work in Michigan, New York and Virginia. To do so, you randomly select 10 accountants in each state and record their hourly pay rate as shown below. At the.01 level, can you conclude that the distributions of accountants hourly pay rates in these three states are different?

21 = 0.01 H 0 : There is no difference in the hourly pay rate in the 3 states. H a : There is a difference in the hourly pay in the 3 states. 1. Write the null and alternative hypothesis. 2. State the level of significance. The sampling distribution is chi-square with d.f. = 3 – 1 = 2. From Table 6, the critical value is Find the rejection region. 4. Find the critical value. 3. Determine the sampling distribution. X2X2

22 Test Statistic Michigan salaries are in ranks: 2, 3, 4, 5, 6, 7, 13, 15, 17.5, 22 The sum is New York salaries are in ranks: 8, 14, 19, 21, 23, 24, 27, 28, 29, 30 The sum is 223. Virginia salaries are in ranks: 1, 9, 10, 11, 12, 16, 17.5, 20, 25, 26 The sum is

23 R 1 = 94.5, R 2 = 223, R 3 = n 1 = 10, n 2 = 10 and n 3 = 10, so N = 30 The test statistic falls in the rejection region, so reject the null hypothesis. There is a difference in the salaries of the 3 states. Find the test statistic. Make Your Decision Interpret your Decision

24 Rank Correlation Section 11.4

25 (There is a significant correlation between the variables.) Rank Correlation The Spearman rank correlation coefficient, r s, is a measure of the strength of the relationship between two variables. The Spearman rank correlation coefficient is calculated using the ranks of paired sample data entries. The formula for the Spearman rank correlation coefficient is where n is the number of paired data entries and d is the difference between the ranks of a paired data entry. The hypotheses: (There is no correlation between the variables.)

26 Rank Correlation Seven candidates applied for a nursing position. The seven candidates were placed in rank order first by x and then by y. The results of the rankings are listed below. Using a.05 level of significance, test the claim that there is a significant correlation between the variables. (There is no correlation between the variables.) (There is a significant correlation between the variables.) x y

27 Application Critical Value = Since the statistic does not fall in the rejection region, fail to reject H 0. There is not enough evidence to support the claim that there is a significant correlation. x y d = x – y d – –1 1 20


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