Ch.8 Comparing More Than Two Proportions Verified Test Bank

Chapter 8

Introduction to Statistical Investigations Test Bank

Note: TE = Text entry TE-N = Text entry - Numeric

Ma = Matching MS = Multiple select

MC = Multiple choice TF = True-False

DD = Drop-down

CHAPTER 8 LEARNING OBJECTIVES

8.1: Carry out a simulation-based analysis for comparing multiple population proportions using a Mean Group Diff statistic.

8.2: Carry out a theory-based analysis for comparing multiple population proportions using a
chi-square statistic.

8.3: Carry out both a simulation-based and theory-based analysis for a chi-square goodness-of-fit test.

Section 8.1: Comparing Multiple Proportions:

Simulation-Based Approach

8.1-1: Understand how multiple comparisons can increase the probability of a Type I error.

8.1-2: Compute the Mean Group Diff statistic from a dataset when comparing multiple proportions.

8.1-3: Understand that larger values of the Mean Group Diff statistic suggest stronger evidence against the null hypothesis.

8.1-4: Use the 3S strategy with the Mean Group Diff statistic.

8.1-5: Use the Multiple Proportions applet to carry out an analysis using the Mean Group Diff statistic to compare multiple proportions.

8.1-6: Explain why the simulated null distribution of the Mean Group Diff statistic looks different from other simulated null distributions.

8.1-7: Conduct a follow-up test after using the Mean Group Diff statistic.

Questions 1 through 7: Researchers analyzed eating behavior and obesity at Chinese buffets. They estimated people’s body mass indexes (BMI) as they entered the restaurant then categorized them into three groups---bottom third (lightest), middle third, and top third (heaviest). One variable they looked at was whether or not they browsed the buffet (looked it over) before serving themselves or served themselves immediately. Treating the BMI categories as the explanatory variable and whether or not they browsed first as the response, the researchers wanted to see if there was an association between BMI and whether or not they browsed the buffet before serving themselves. They found the following results:

• Bottom Third: 35 of the 50 people browsed first
• Middle Third: 24 of the 50 people browsed first
• Top Third: 17 of the 50 people browsed first

1. Fill in the following two-way table with the observed data.

LO: 8.1-2; Difficulty: Easy; Type: TE-N

1. For each BMI category, calculate the conditional proportion of people that browsed first and write these numbers below as decimals, not fractions.

Bottom Third___________ Middle Third____________ Top Third____________

LO: 8.1-2; Difficulty: Easy; Type: TE-N

1. What is the null hypothesis to determine if there is an association between BMI and if a person browses first?
2. Compute the Mean Group Diff statistic.

LO: 8.1-2; Difficulty: Medium; Type: TE-N

1. Use the Multiple Proportions applet to calculate an approximate p-value for this test. Use at least 1,000 shuffles.

Bottom Middle Top

Browse 35 24 17

Not 15 26 33

Then click “Use Table”. Select “Show Shuffle Options” and enter a large number (e.g., 1000) for Number of Shuffles. Then click “Shuffle. Count Samples greater than or equal to 0.24.

LO: 8.1-5; Difficulty: Hard; Type: TE-N

1. What is the shape of the resulting null distribution of MAD statistics shown in the output of the Multiple Proportions applet?
   1. Symmetric
   2. Normal
   3. Left skewed
   4. Right skewed
2. Based upon the p-value of 0.001, what is the appropriate conclusion for this test?
   1. We have strong evidence of an association between BMI and if a person browses first among all people who eat at Chinese buffets.
   2. We have strong evidence of an association between BMI and if a person browses first among people who eat at Chinese buffets similar to those in the study.
   3. We have strong evidence of no association between BMI and if a person browses first among all people who eat at Chinese buffets.
   4. We have strong evidence of no association between BMI and if a person browses first among people who eat at Chinese buffets similar to those in the study.
3. Which of the following is not one of the 3 “S”’s in the 3S strategy?
   1. Simulate
   2. Statistic
   3. Sample
   4. Strength of evidence
4. Suppose three students were interested in studying the association between a person’s political party (Democrat, Republican, Independent, or Other), and whether they voted in the last election (Yes, No). Each of the three students selects a random sample of 40 students and asks each student their political party and if they voted in the last election. Their three Mean Group Diff statistics are:

Mean Group Diff 1 = 0.12 Mean Group Diff 2 = 0.35 Mean Group Diff 3 = 0.27

Which Mean Group Diff statistic provides the strongest evidence against the null hypothesis?

• 1. 0.12
  2. 0.35
  3. 0.27

1. True or False: As the Mean Group Diff statistic increases, the p-value decreases.
2. True or False: The Mean Group Diff statistic can never be negative.
3. True or False: The Mean Group Diff statistic is the average distance sample proportions are from each other.
4. True or False: The Mean Group Diff statistic is the average of a group of sample proportions.

Questions 14 and 15: Is there an association between a person's political affiliation (Democrat, Republican, or Independent) and whether or not they give to charity?

1. Describe the appropriate parameter(s) and assign symbols(s) to the parameters. Select all that apply
   1. = probability that a Democrat will give to charity
   2. = probability that a Republican will give to charity
   3. = probability that a Independent will give to charity
2. Assume that, in reality, there is no association between a person’s political affiliation and whether or not they give to charity. Which of the following has the largest probability?
   1. The p-value for a test of is greater than 0.05.
   2. The p-value for a test of is greater than 0.05 AND the p-value for a test of is greater than 0.05.
   3. The p-value for a test of is greater than 0.05 AND the p-value for a test of is greater than 0.05 AND the p-value for a test of is greater than 0.05.
   4. All of the above have equal probability.

Section 8.2: Comparing Multiple Proportions:

Theory-Based Approach

8.2-1: Find the value of the chi-square test statistic using the Multiple Proportions applet, recognize that larger values of the statistic mean more evidence against the null hypothesis and why the distribution of the chi-square statistic is non-negative and follows a right-skewed distribution.

8.2-2: Conduct a chi-square test of significance using the Multiple Proportions applet, including appropriate follow-up tests.

8.2-3: Identify whether a chi-square test meets appropriate validity conditions.

Questions 16 through 22: Hope student researchers want to determine if students yawning (when being yawned at) is associated with the time of day (morning, afternoon, and evening). They found the following results when they yawned in front of people around campus:

• Morning: 25 of the 39 people yawned
• Afternoon: 15 of the 38 people yawned
• Evening: 29 of the 44 people yawned

1. Fill in the following two-way table with the observed data.

LO: 8.2-2; Difficulty: Easy; Type: TE-N

1. Are validity conditions met to conduct a chi-square test of association?
   1. Yes, since 39, 38, and 44 are all larger than 20.
   2. Yes, since 121 is larger than 20.
   3. Yes, since each cell in the table has at least 10 observations.
   4. No, since several cells in the table have less than 20 observations.
2. For each time of day, calculate the conditional proportion of people that yawned and write these numbers as decimals not fractions.

Morning___________ Afternoon____________ Evening____________

LO: 8.2-1; Difficulty: Easy; Type: TE-N

1. Which applet would you use to conduct a theory-based test of association between time of day and whether a person yawns?
   1. One Proportion
   2. Two Proportions
   3. Multiple Proportions
   4. Theory-Based Inference
2. When this test is run, a p-value of 0.03 is found. What would be an appropriate conclusion?
   1. We have strong evidence that the probability a person yawns is different between all three times of day: morning, afternoon, and evening.
   2. We have strong evidence of an association between time of day and whether a person yawns.
   3. We have strong evidence of no association between time of day and whether a person yawns.
   4. We do not have strong evidence of an association between time of day and whether a person yawns.
3. The chi-square statistic for this test was 6.991. If instead, the chi-square statistic had been 5.332, would the p-value be smaller, larger, or remain the same?
   1. Smaller
   2. Larger
   3. Remain the same
4. Follow-up confidence intervals for all pairwise differences in proportions are as follows:

• 0.0299 to 0.4627: Morning – Afternoon
• -0.2237 to 0.1876: Morning – Evening
• -0.4736 to -0.0551: Afternoon – Evening

Based off of these intervals, which of the following conclusions are correct?

• 1. The probability of yawning in the morning differs significantly from the probability of yawning in the evening.
  2. The probability of yawning in the morning differs significantly from the probability of yawning the afternoon.
  3. The probability of yawning in the afternoon does not differ significantly from the probability of yawning in the evening.
  4. The probability of yawning in the morning does not differ significantly from the probability of yawning in the afternoon.

Questions 23 through 25: Can telling a joke affect whether or not a customer in a coffee bar leaves a tip for the waiter? A study investigated this question at a coffee bar in France. The waiter randomly assigned coffee-ordering customers into three groups: one received a card telling a joke with their bill, another group received a card containing an advertisement for a local restaurant, and the third received no card. The results are shown below.

1. Are validity conditions met to conduct a chi-square test of association?
   1. Yes, since 80 is larger than 20.
   2. Yes, since 240 is larger than 20.
   3. Yes, since each cell in the table has at least 10 observations.
   4. No, we did not collect a random sample of customers.
2. Use the Multiple Proportions applet to calculate the chi-square statistic and theory-based p-value for these data.

Chi-square statistic =

p-value =

Joke Ad No

Tip 32 16 24

No 48 64 56

Then click “Use Table”. Check the Show chi-square output box.

LO: 8.2-1; Difficulty: Hard; Type: TE-N

1. Use the Multiple Proportions applet to calculate the follow-up 95% confidence intervals for difference in proportions. Based on the output, which of the following statements are correct? Select all that apply.
   1. The probability a customer leaves a tip is statistically significantly larger if a joke card is left than if an ad card is left.
   2. The probability a customer leaves a tip is statistically significantly larger if a joke card is left than if no card is left.
   3. The probability a customer leaves a tip is not statistically significantly different between the ad card and no card conditions.
   4. The probability a customer leaves a tip is not statistically significantly different between the joke card and ad card conditions.

1. True or False: As the chi-square statistic increases, the p-value increases.
2. True or False: As the chi-square statistic increases, we have more evidence against the null hypothesis.
3. True or False: The chi-square statistic can be negative.
4. True or False: The chi-square statistic is the ratio of variability between the groups and the variability within the groups.
5. What shape is the distribution of the chi-square statistic?
   1. Symmetric
   2. Normal
   3. Right skewed
   4. Left skewed
6. Why do we do overall tests when comparing multiple proportions and not just do the follow-up confidence intervals?
   1. Doing an overall test will more likely lead to significant results than just doing the follow-up confidence intervals.
   2. Doing an overall test allows us to see exactly which group is significantly different from which other groups.
   3. Doing an overall test allows us to quickly get our results, while the follow-up confidence intervals are very time-consuming.
   4. Doing an overall test allows us to keep the probability of a type I error at 5% or whatever significance level we would like.

Section 8.3: Chi-Square Goodness-of-Fit Test

8.3-1: Calculate expected counts based on the hypothesized model in a chi-square goodness-of-fit test.

8.3-2: Conduct a simulation-based chi-square goodness-of-fit test using the MAD (mean absolute difference) statistic in the Goodness of Fit applet.

8.3-3: Conduct a simulation-based chi-square goodness-of-fit test using the chi-square statistic in the Goodness of Fit applet.

8.3-4: Identify whether or not validity conditions are met for a theory-based chi-square goodness-of-fit test.

8.3-5: Conduct a theory-based chi-square goodness-of-fit test.

1. Suppose we conduct a chi-square goodness-of-fit test to determine if a 6-sided die is fair. We roll the die 10 times, and the resulting p-value is very large. What may we conclude? Select all that apply.
   1. We have proven that the die is fair.
   2. The hypothesis that the die is fair is plausible.
   3. There is evidence that the die is fair.
   4. There is evidence against the hypothesis that the die is not fair.
   5. There is little to no evidence that the die is not fair.
   6. We have proven that the hypothesis that the die is not fair is false.

Questions 33 through 39: Are people equally likely to be born on any day of the seven days of the week? Or are some days more likely to be a person’s birthday than other days? To investigate this question, days of birth were recorded for the 147 “noted writers of the present” listed in The World Almanac and Book of Facts 2000. The counts for the seven days of the week are given in the following table.

1. Let be the probability that a person is born on a particular day. What is the null hypothesis?
2. Fill in the table below with the expected counts based on the hypothesized model that people are equally likely to be born on any day of the seven days of the week.

Tol: +/- 0

LO: 8.3-1; Difficulty: Medium; Type: TE-N

1. Use the Goodness of Fit applet to conduct a simulation-based chi-square goodness-of-fit test using the MAD (mean absolute difference) statistic, and report the statistic and the p-value. Use at least 1000 shuffles.

MAD statistic =

p-value =

Outcome Count

M 17

T 26

W 22

H 23

F 18

Sat 15

Sun 25

Choose MAD for the Statistic. Select Show Sampling Options. Enter hypothesized probabilities: 1/7, 1/7, 1/7, 1/7, 1/7, 1/7, 1/7. Enter a large number of samples (e.g., 10000); click Sample; then count samples greater than or equal to 3.589.

Note: Currently, the applet doesn’t accept fractions, so one option is to enter the following hypothesized probabilities: 0.1429, 0.1429, 0.1429, 0.1429, 0.1428, 0.1428, 0.1428.

LO: 8.3-2; Difficulty: Hard; Type: TE-N

1. Use the Goodness of Fit applet to conduct a simulation-based chi-square goodness-of-fit test using the chi-square statistic, and report the statistic and the p-value. Use at least 1000 shuffles.

chi-square statistic =

p-value =

Outcome Count

M 17

T 26

W 22

H 23

F 18

Sat 15

Sun 25

Choose chi-square for the Statistic. Select Show Sampling Options. Enter hypothesized probabilities: 1/7, 1/7, 1/7, 1/7, 1/7, 1/7, 1/7. Enter a large number of samples (e.g., 10000); click Sample; then count samples greater than or equal to 5.117.

Note: Currently, the applet doesn’t accept fractions, so one option is to enter the following hypothesized probabilities: 0.1429, 0.1429, 0.1429, 0.1429, 0.1428, 0.1428, 0.1428.

LO: 8.3-3; Difficulty: Hard; Type: TE-N

1. Use the Goodness of Fit applet to conduct a theory-based chi-square goodness-of-fit test, and report the chi-square statistic and the p-value.

chi-square statistic =

p-value =

Outcome Count

M 17

T 26

W 22

H 23

F 18

Sat 15

Sun 25

Choose chi-square for the Statistic. Select Show Sampling Options. Enter hypothesized probabilities: 1/7, 1/7, 1/7, 1/7, 1/7, 1/7, 1/7. Enter a large number of samples (e.g., 10000); click Sample; then count samples greater than or equal to 5.117. Now, check the box next to “Overlay Chi-square distribution”.

LO: 8.3-5; Difficulty: Hard; Type: TE-N

1. Are validity conditions met to conduct a theory-based chi-square goodness-of-fit test?
   1. Yes, since all observed counts are at least 10.
   2. Yes, since we have at least 10 successes and 10 failures.
   3. No, since the p-values in questions 35 and 37 were different.
   4. No, since this was not a random sample.
2. Based upon a p-value of 0.53, what is the appropriate conclusion for the test?
   1. We have strong evidence that people are equally likely to be born on any day of the seven days of the week.
   2. We do not have strong evidence that people are equally likely to be born on any day of the seven days of the week.
   3. We have strong evidence that people are not equally likely to be born on any day of the seven days of the week.
   4. We do not have strong evidence that people are not equally likely to be born on any day of the seven days of the week.

Questions 40 through 46: According to the 2019 U.S. Census Bureau estimates, 63.4% of the U.S. self-identifies as Non-Hispanic white, 15.3% as Hispanic and Latino, 13.4% as Black or African American, 5.9% as Asian, and 2.0% as Other. A random sample of 1000 students graduating from California colleges and universities resulted in the following data on self-identified race:

You would like to determine if the distribution of self-identified race among California college and university students differs from the distribution of self-identified race in the U.S. overall.

1. What is the null hypothesis? Fill in the value for each parameter.

0.634

0.134

0.153

0.059

0.020

LO: 8.3-1; Difficulty: Easy; Type: TE-N

1. In the null hypothesis, what does the symbol represent?
   1. The proportion of the U.S. that self-identify as a certain Race
   2. The proportion of Californians that self-identify as a certain Race
   3. The proportion of all students graduating from California colleges and universities that self-identify as a certain Race
   4. The proportion of the sample of 1000 students that self-identified as a certain Race
2. What is the expected count based on the null model for the “Hispanic or Latino” cell?
   1. 77
   2. 15.3
   3. 11.8
   4. 153
3. Use the Goodness of Fit applet to conduct a simulation-based chi-square goodness-of-fit test using the chi-square statistic, and report the statistic and the p-value. Use at least 1000 shuffles.

chi-square statistic =

p-value =

Outcome Count

W 679

B 51

H 77

A 190

O 3

Choose chi-square for the Statistic. Select Show Sampling Options. Enter hypothesized probabilities: 0.634, 0.134, 0.153, 0.059, 0.02. Enter a large number of samples (e.g., 10000); click Sample; then count samples greater than or equal to 397.67.

LO: 8.3-3; Difficulty: Hard; Type: TE-N

1. Use the Goodness of Fit applet to conduct a simulation-based chi-square goodness-of-fit test using the MAD (mean absolute difference) statistic, and report the statistic and the p-value. Use at least 1000 shuffles.

MAD statistic =

p-value =

Outcome Count

W 679

B 51

H 77

A 190

O 3

Choose MAD for the Statistic. Select Show Sampling Options. Enter hypothesized probabilities: 0.634, 0.134, 0.153, 0.059, 0.02. Enter a large number of samples (e.g., 10000); click Sample; then count samples greater than or equal to 70.4.

LO: 8.3-2; Difficulty: Hard; Type: TE-N

1. Based upon a p-value of less than 0.001, what is an appropriate conclusion for the goodness-of-fit test?
   1. We do not have strong evidence that the distribution of self-identified race among California college and university students differs from the distribution of self-identified race in the U.S. overall.
   2. We have strong evidence that the distribution of self-identified race among California college and university students differs from the distribution of self-identified race in the U.S. overall.
   3. We have strong evidence that the distribution of self-identified race among California college and university students is the same as the distribution of self-identified race in the U.S. overall.
   4. We have strong evidence that the probabilities of each self-identified race among California college and university students are not all equal.
2. Are the validity conditions met for a theory-based chi-square goodness-of-fit test?
   1. Yes, since 1000 is larger than 20.
   2. Yes, since all of the expected counts are at least 10.
   3. No, since the observed count for Other is less than 10.
   4. Yes, since the p-values from both the simulation-based and theory-based tests are similar.