Wk6 DQ - Data Analysis & Business Intelligence

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Nonparametric Methods: Analysis of Ordinal Data

Chapter 16

16-1

This chapter continues our discussion of hypothesis tests designed especially for nonparametric data. We do not need to assume anything about the shape of the population distribution; these are distribution-free tests. The tests do require that the variables can be sorted and ranked; in other words, the variables must be measured with an ordinal, interval, or ratio level scale. We’ll consider 5 distribution-free tests and the Spearman coefficient of rank correlation.

Learning Objectives

LO16-1 Use the sign test to compare two dependent populations

LO16-2 Test a hypothesis about a median

LO16-3 Test a hypothesis of dependent populations using the Wilcoxon signed-rank test

LO16-4 Test a hypothesis of independent populations using the Wilcoxon rank-sum test

LO16-5 Test a hypothesis of several independent populations using the Kruskal-Wallis test

LO16-6 Test and interpret a nonparametric hypothesis test of correlation

16-2

The Sign Test

The sign test is based on the sign difference between two related observations

No assumptions need to be made about the shape of the two populations

For small samples, find the number of + or – signs and refer to the binomial distribution for the critical value

Example

A dietitian wishes to see if by taking a certain mineral, a person’s cholesterol level decreases

She measures the individuals before and after

If there has been an decrease “+,” an increase “−”

16-3

We are not interested in the magnitude of the difference, just the direction of the difference, a + or −. If there has been no change, a zero, that person is dropped from the study.

The Sign Test Example

The director of information systems at Samuelson Chemicals recommended that an in-plant training program be instituted for certain managers in Payroll, Accounting, and Production Planning. A sample of 15 managers is randomly selected from the three departments and rated on their technology knowledge. Then, after a 3 month training program, the same assessment rated the managers knowledge again. A “+” sign indicates an improvement, and a “−” sign indicates a decline in technology competence.

Did the in-plant training program increase the managers technical knowledge?

16-4

Step 1: State the null and the alternate

H0: ≤ .50 There has been no change in the technology knowledge base of the managers as a result of the training program

H1: > .50

There has been an increase in the technology knowledge base of the managers as a result of the training program

The managers are rated on his or her technology knowledge based on an assessment of how they use technology to solve problems. Based on the results, they were rated as outstanding, excellent, good, fair, or poor. Are the managers more knowledgeable after the training program than before?

We’ll use the six-step hypothesis-testing procedure. The symbol refers to the proportion in the population with a particular characteristic. The test statistic follows the binomial probability distribution because the sign test meets all the binomial assumptions; there are only two outcomes, the probability of success is the same for all trials, the total number of trials is fixed, and each trial is independent.

The Sign Test Example Continued

Step 2: Select the level of significance; we select .10

Step 3: Decide on the test statistic; it is the number of “+” signs

Step 4: Formulate a decision rule; if the number of pluses in the sample is 10 or more, the null hypothesis is rejected and the alternate hypothesis accepted

Step 5: Make a decision; reject H0, the number of pluses is 11

Step 6: Interpret, we conclude the three-month training course was effective. It increased the technology knowledge of the managers

16-5

To formulate the decision rule, we begin by eliminating Andy Love since he had a 0 change and could not be assigned to either group. That leaves us with a n=14. The binomial probability table from Appendix B.1, for an n of 14 and a probability of .50, has been copied here. The number of successes are in column 1, the probabilities of success in column 2, and the cumulative probabilities in column 3; the cumulative probabilities are arrived at by adding the probabilities in column 2 up, not down. We add up because the region of rejection is in the right tail (we would have added down, if this had been a left-tailed test). For illustration, to get the cumulative probability of 11, add .000+.001+.006+.022=.029. Now, go to the cumulative probability column and reading up from the bottom until we reach the cumulative probability nearest to, but not exceeding the level of significance of .10; this is .090 which corresponds to a number of successes of 10. Therefore, since the decision rule is if the number of pluses is 10 or more, we reject the null hypothesis.

11 out of the 14 managers increased their technology knowledge as a result of the training program.

Testing a Hypothesis about a Median

The median test is used to test a hypothesis about a population median

Find μ and σ for a binomial distribution

The z distribution is used as the test statistic

The value of z is computed from formula 16-1, where x is the number of observations above or below the median

A value above the median is assigned a “+”

A value below the median is assigned a “−”

A value that is the same as the median is dropped from the analysis

16-6

The sign test is one of the few tests that can be used to test the value of a median.

Testing a Hypothesis about a Median Example

The U.S. Bureau of Labor Statistics reported in 2018 that the median amount spent eating out by American families is about $3,000 annually. The food editor of the Portland Tribune wishes to know if the citizens of Portland differ. She selected a random sample of 22 families and found 15 spent more than $3,000 last year eating out, 5 spent less than that, and 2 spent exactly $3,000. Is it reasonable to conclude that the median amount spent this year in Portland, Oregon is not equal to $3,000?

Step 1: State the null and the alternate hypothesis

H0: Median = $3,000

H1: Median ≠ $3,000

Step 2: Select the level of significance; we select 0.05

Step 3: Decide on the test statistic; it is based on count

16-7

The strategy here is similar to the strategy used for the sign test.

1. We use the binomial distribution as the test statistic.

2. The number of trials is 20.

3. The hypothesized probability of a success is .50.

If the population median is $3,000, then we expect about half of the sampled couples spent more than $3,000 last year and about half spent less than $3,000. After discarding the two couples that spent exactly $3,000, we would expect 10 to be above the median and 10 to be below the median.

Testing a Hypothesis about a Median Example Continued

16-8

Step 4: Formulate the decision rule; if the count is 5 or less, or 15 or more, reject the null hypothesis.

Step 5: Make decision; reject the null hypothesis, 15 families spent more than $3,000

Step 6: Interpret; the food editor should conclude that there is a difference in the median amount spent in Portland from that reported by the BLS in 2018

In the food editor’s sample 15 families spent more than $3,000. Fifteen is in the rejection region, so the null hypothesis is rejected. The sample evidence indicates that the median amount spent annually is not equal to $3,000. The conclusion is that for families in Portland, there is a difference in the annual median amount spent eating out when compared to the U.S. Bureau of Labor Statistics information.

The Wilcoxon Signed-Rank Test for Dependent Populations

The Wilcoxon sign-rank test is a nonparametric test for differences between two dependent populations

The assumption of normally distributed populations is not required

The steps to conduct the test are

Rank absolute differences between the related observations

Apply the sign of the differences to the ranks

Sum negative ranks and positive ranks

The smaller of the two sums is the computed T value

Refer to Appendix B.8 for the critical value, and make a decision regarding H0

16-9

There are instances when we want to study the differences between dependent observations and we cannot assume that the distribution of the differences approximates a normal distribution; this is frequently a problem when working with ordinal level data.

The Wilcoxon Signed-Rank Test for Dependent Populations Example

Fricker’s is a family restaurant chain located primarily in the southeastern part of the United States. It offers a full menu, but its specialty is chicken. Recently, the owner, Bernie Frick, developed a new spicy flavor for the batter in which the chicken is cooked. Before replacing the current flavor, he wants to be sure that patrons will like the spicy flavor. To begin the taste test, he selects a random sample of 15 customers. Each customer is given a piece of the current chicken and asked to rate it on a scale of 1 to 20 and then the customer is given a piece of spicy chicken and asked to rate it.

Is it reasonable to conclude that the spicy flavor is preferred?

16-10

In the taste test, a value near 20 means the participant liked the flavor; a value near 0 indicates they did not like the flavor. Each participant is asked to rate both flavors of chicken, so the ratings are dependent or related (later we’ll find the differences between the ratings). Because of the subjective nature of the rankings, we are not sure that the differences follow the normal distribution so we decide to use the Wilcoxon signed-rank test.

The Wilcoxon Signed-Rank Test for Dependent Populations Example Continued

Step 1: State the null and the alternate hypothesis

H0: There is no difference in the ratings of the two flavors

H1: The spicy ratings are higher

Step 2: Select the level of significance; we select .05

Step 3: Select the test statistic; we’ll use T

Step 4: Formulate the decision rule; reject the null hypothesis if the smaller of the rank sums is 25 or less

16-11

We’ll conduct the six-step hypothesis-testing procedure. This is a one-tailed test because the owner will want to change the chicken flavor only if the sample participants show that the population of customers like the new flavor better. The level of significance is .05. We’ll use the Wilcoxon signed rank test, T, since we are working with ordinal level data and cannot assume the distribution of differences is normal. We’ll use Appendix B.8, the level of significance row, , is used for one-tailed tests. We find the .05 significance level and move down to the row where n=14. One person rated the chicken flavors the same and was dropped from the study, making the usable sample size 14. The value at that intersection is 25, so the critical value is 25.

The Wilcoxon Signed-Rank Test for Dependent Populations Example Concluded

Step 5: Make decision; in this case the smaller rank sum is 30, do not reject H0

Step 6: Interpret; we cannot conclude there is a difference in the flavor ratings between the current and the spicy.

Mr. Frick should stay with the current flavor of chicken!

16-12

Find the differences between the scores and report in column D. Then, add a column for the absolute value of the difference; this is column E. Next rank the absolute differences from smallest to largest, column F. In the table, Arquette, rated the spicy chicken a 14 and the current a 12. The difference of 2 is the smallest absolute difference, so it is given a ranking of 1. The next largest is Miller, with an absolute difference of 3, so it is given a rank of 2, and so on. Then each assigned rank is given the same sign of its original difference and these results are reported in columns G and H. Finally, the signed differences are summed. The sum of the positive ranks is 75 and the sum of the negative ranks is 30. The smaller of the two rank sums is referred to as T. 30 is not 25 or less, so we reject the null hypothesis.

The Wilcoxon Rank-Sum Test

The Wilcoxon rank-sum test is used to test whether two independent samples came from equivalent populations

The assumption of normally distributed populations is not required

The population variances need not be equal either

The data must be at least ordinal scale

If each sample contains at least 8 observations, use the standard normal distribution as the test statistic

16-13

If the null hypothesis is true, then the ranks will be about evenly distributed between the two samples and the sum of the ranks for the two samples will be about the same. To determine the value of the test statistic W, the sample observations are ranked from low to high as if they were from a single group and the sum of the ranks for each of the two samples is determined. W is used to compute z, where W is the sum of the ranks for population 1. The standard normal distribution, found in Appendix B.3 is the test statistic. In the formula, n1 is number of observations from the first sample, n2 is the number of observations from the second sample, and W is the sum of the ranks from the first population.

The Wilcoxon Rank-Sum Test Example

Dan Thompson, the president of OTG airlines, recently noted an increase in the number of bags that were checked in at the gate in Atlanta. He is interested in determining whether there are more gate-checked bags from Atlanta compared with flights leaving Chicago. A sample of nine flights from Atlanta and eight from Chicago are reported in the table.

Can we conclude that there are more checked bags for flights originating in Atlanta?

Ranked number of gate checked bags.

16-14

Mr. Thompson does not believe that the distributions are normal or have equal variances so a nonparametric test, the Wilcoxon rank-sum test, is appropriate. We begin by ranking the observations from both samples as if they were one (17 observations total). The Chicago flight with 8 checked bags is the least so it gets a rank of 1; the Atlanta flight with 25 checked bags is the highest, so it gets a rank of 17. There is a flight with 10 on each flight so these are tied for third place; to manage a tie, average the two ranks and report the average for each. So the flights with 10 are each ranked with 3.5. After ranking, sum the ranks. The sum of the Atlanta flights (population 1) is 96.5 and is the value of W in the formula on the next slide.

The Wilcoxon Rank-Sum Test Example Continued

Step 1: State the null and alternate hypothesis

H0: The number of gate-checked bags for Atlanta is the same or less than the number of gate checked bags for Chicago

H1: The number of gate-checked bags for Atlanta is more than the number of gate-checked bags for Chicago

Step 2: Select the level of significance; we select .05

Step 3: Select the test statistic; we use z

Step 4: Formulate the decision rule;reject H0 if z > 1.645

16-15

Mr. Thompson believes there are more checked bags for Atlanta flights, so a one-tailed, “right” tailed, test is appropriate. The test statistic follows the normal distribution and with a .05 significance level, the critical value found in the last row of Appendix B.5 is 1.645. Recall, in the formula, n1 is number of observations from the first sample, n2 is the number of observations from the second sample, and W is the sum of the ranks from the first population.

The Wilcoxon Rank-Sum Test Example Continued

16-16

Step 5: Compute the test statistic; make decision, do not reject H0

Step 6: Interpret; it appears that the number of checked bags in Atlanta are the same as those in Chicago

MegaStat output:

Because the computed z value (1.49) is less than 1.645, the null hypothesis is not rejected. The evidence does not show a difference in the distributions of the number of gate-checked bags. That is, it appears that the number of gate-checked bags is the same in Atlanta as in Chicago.

The Kruskal-Wallis Test: ANOVA by Ranks

The Kruskal-Wallis one-way ANOVA by ranks is used to test whether several populations are the same

The assumption of normally distributed populations is not required

There are three requirements for this test:

The samples are from independent populations

The population variances must be equal

The samples are from normal populations

The test statistic follows the chi-square distribution, if there are at least five observations in each sample

16-17

The sample observations are combined and ranked from smallest to largest as though they were a single group and then the ordered values are replaced by ranks starting with 1 for the smallest value. In the formula, ∑R1, ∑R2 and so on is the sum of the ranks of samples; n1, n2, and so on are the sizes of the samples.

The Kruskal-Wallis Test: ANOVA by Ranks

16-18

To compute the Kruskal-Wallis test statistic

All the samples are combined

The combined values are ordered from low to high

The ordered values are replaced by ranks, starting with 1 for the smallest value

Compute the value of the test statistic from the following

The Kruskal-Wallis Test: ANOVA by Ranks Example

The Hospital Systems of the Carolinas operate three hospitals in the Greater Charlotte area: St. Luke’s Memorial, Swedish Medical Center, and Piedmont Hospital. The director of administration is concerned about the waiting time of patients with non-life-threatening injuries that arrive during weekday evenings at the three hospitals.

Is there a difference in the waiting times at the three hospitals?

Waiting times ranked and summed

16-19

The director selected random samples of patients at the three locations and determined the time, in minutes, between entering the particular facility and when treatment was completed. The samples are from independent populations, but we do not assume equal variances or that they follow a normal distribution. So we’ll use the Kruskal-Wallis test where the assumptions are not required. After ranking the wait times as if they were all three in one group and resolving any ties by finding the average of the two tied values—there are two 38s tied for 2nd place so they each get ranked with 2.5 (2 + 3 ÷ 2 = 2.5) —he sums the ranks. Use the sum of the ranks in the formula on the following slide.

The Kruskal-Wallis Test: ANOVA by Ranks Example

Step 1: State the null and alternate hypothesis

H0: The population distributions of waiting times are the same for the three hospitals

H1: The population distributions of waiting times are not all the same for the three hospitals

Step 2: Select the level of significance; he selects .05

Step 3: Select the test statistic; he’ll use chi-square

Step 4: Formulate the decision rule, reject H0 if H > 5.991

Step 5: Calculate H; make decision, do not reject H0

Step 6: Interpret; there is not enough evidence to conclude that there are differences in wait times

16-20

We use chi-square to formulate the decision rule. There are 2 degrees of freedom, there are three populations, so k = 3 and k − 1 = 2.Go to Appendix B.7, the critical value with 2 degrees of freedom and .05 significance level is 5.991. Use the sum of the ranks obtained from the table on the prior slide for the calculation of H. We do not reject the null hypothesis.

Rank-Order Correlation

Spearman’s coefficient of rank correlation is a measure of the association between two ordinal-scale variables

It can range from −1 up to 1

A value of −1 indicates perfect negative correlation

A value 1 indicates perfect positive correlation

A value of 0 indicates there is no association between the variables

Example

Two university staff members are asked to rank 10 faculty research proposals for funding purposes. Do the staff members rank the same proposals in the same way?

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This is a measure for ranked variables and allows us to describe the relationship between sets of ranked data.

Rank-Order Correlation Continued

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There are several conditions when it is not appropriate or can be misleading.

Those conditions include:

When the scale of measurement of one of the two variables is ordinal (ranked).

When the relationship between the variables is not linear.

When one or more of the data points are quite different from the others.

Spearman’s Coefficient Continued

The value of rs is computed from the following formula

Provided the sample size is at least 10, we can conduct a test of hypothesis using the following formula

The test statistic follows the t distribution

There are n-2 degrees of freedom

16-23

In the formula, d is the difference between the ranks for each pair and n is the number of paired observations.

Spearman’s Coefficient Example

Recent studies focus on the relationship between the age of online shoppers and the number of minutes spent browsing on the Internet. The table below shows a sample of 15 online shoppers who actually made a purchase last week. Included is their age and time, in minutes, spent browsing on the Internet last week.

Draw a scatter diagram.

What type of association do the sample data suggest?

Do you see any issues with the relationship between variables?

16-24

Here is the age and browsing minutes of a sample of 15 online shoppers. We begin the analysis by drawing a scatter diagram by plotting age on the horizontal axis and browsing minutes on the vertical axis. There appears to be a fairly strong inverse relationship but there are a couple of data points that seem different from the others. Because of these points we decide to use the coefficient of rank correlation.

Spearman’s Coefficient Example Continued

Find the coefficient of rank correlation.

16-25

We continue the analysis by finding rs, the coefficient of rank correlation. Beginning by ranking the age of the shoppers; the youngest is 17 and she is given a rank of 1, the oldest is 60 and is given a rank of 15. When there is a tie, like for third place where both are 21 years of age, use the average of (3rd + 4th place) ÷ 2 = 3.5 and rank both at 3.5. The same ranking strategy is used for browsing minutes. Then calculate the coefficient of rank correlation. It is −0.724 and indicates a fairly strong negative association between the age of the Internet shopper and minutes spent browsing. Next, we conduct a hypothesis test.

Spearman’s Coefficient Example Concluded

Step 1: State the null and alternate hypothesis

H0: The rank correlation in the population is zero

H1: There is a negative association among the ranks

Step 2: Select the level of significance; we select .05

Step 3: Select the test statistic; we use t

Step 4: Formulate the decision rule; reject H0 if t < −1.771

Step 5: Make decision; reject the null hypothesis, t = −3.784

Step 6: Interpret; there is evidence of a negative association between the age of the Internet shopper and the time spent browsing the Internet

Conduct a test of hypothesis to determine if there is a negative association between the ranks.

16-26

Here, we conduct a test of hypothesis to investigate the significance of rs which was −0.724. This will be a one-tailed, left-tailed test. We go to Appendix B.5 and find that with a one-tail test and degrees of freedom, n − 2=13, the value is 1.771; so the critical value is −1.771. Since we have a sample of 10 or more, we use formula 16-7. We reject the null hypothesis. Younger shoppers spend more time browsing the Internet while shopping.

Chapter 16 Practice Problems

16-27

Question 3

16-28

Calorie Watchers has low-calorie breakfasts, lunches, and dinners. If you join the club, you receive two packaged meals a day. Calorie Watchers claims that you can eat anything you want for the third meal and still lose at least 5 pounds the first month. Members of the club are weighed before commencing the program and again at the end of the first month. The experiences of a random sample of 11 enrollees are:

We are interested in whether there has been a weight loss as a result of the Calorie Watchers program.

State H0 and H1.

Using the .05 level of significance, what is the decision rule?

What is your conclusion about the Calorie Watchers program?

LO16-1

Question 5

16-29

According to the U.S. Department of Labor, the median salary for a chiropractor in the United States is $81,500 per year. A group of recent graduates employed in the state of Colorado believe this amount is too low. In a random sample of 18 chiropractors who recently graduated, 13 began with a salary of more than $81,500. Is it reasonable to conclude that the starting salary in Colorado is more than $81,500?

State the null and alternate hypotheses.

State the decision rule. Use the .05 significance level.

Test the hypothesis and interpret the results.

LO16-2

Question 7

16-30

An industrial psychologist selected a random sample of seven young urban professional couples who own their homes. The size of their home (square feet) is compared with that of their parents. At the .05 significance level, can we conclude that the professional couples live in larger homes than their parents?

LO16-3

Question 13

16-31

Tucson State University offers two MBA programs. In the first program, the students meet two nights per week at the university’s main campus in downtown Tucson. In the second program, students only communicate online with the instructor. The director of the MBA experience at Tucson wishes to compare the number of hours studied last week by the two groups of students. A sample of 10 on-campus students and 12 online students revealed the following information.

Do not assume the two distributions of study times (in hours) follow a normal distribution. At the .05 significance level, can we conclude the online students spend more time studying?

LO16-4

Question 19

16-32

Davis Outboard Motors Inc. recently developed an epoxy painting process to protect exhaust components from corrosion. Bill Davis, the owner, wishes to determine whether the durability of the paint was equal for three different conditions: saltwater, freshwater without weeds, and freshwater with a heavy concentration of weeds. Accelerated-life tests were conducted in the laboratory, and the number of hours the paint lasted before peeling was recorded. Five boats were tested for each condition.

Use the Kruskal-Wallis test and the .01 level to determine whether the number of hours the paint lasted is the same for the three water conditions.

LO16-5

Question 23

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Ten new sales representatives for Clark Sprocket and Chain, Inc. were required to attend a training program before being assigned to a regional sales office. At the end of the program, the representatives took a series of tests and the scores were ranked. For example, Arden had the lowest test score and is ranked 1; Arbuckle had the highest test score and is ranked 10. At the end of the first sales year, the representatives’ ranks based on test scores were paired with their first year sales.

Compute and interpret the coefficient of rank correlation between first-year sales and class rank after the training program.

At the .05 significance level, can we conclude that there is a positive association between first-year sales dollars and ranking in the training program?

LO16-6