AP Statistics Quiz a – Chapter 26 – Key 1. a Biology Professor Reports

AP Statistics Quiz A – Chapter 26 – Key

1. A biology professor reports that historically grades in her introductory biology course have been distributed as follows: 15% A’s, 30% B’s, 40% C’s, 10% D’s, and 5% F’s. Grades in her most recent course were distributed as follows: Grade A B C D F Frequency 89 121 78 25 12

a. Test an appropriate hypothesis to decide if the professor’s most recent grade distribution matches the historical distribution. Give statistical evidence to support your conclusion. We want to know if the most recent grade distribution matches the historical grade distribution. H : The most recent grade distribution matches the historical grade distribution. 0 H : The most recent grade distribution differs from the historical grade distribution. A Conditions: *Counted data: We have the counts of the number of students who earned each grade. *Randomization: We have a convenience sample of students, but no reason to suspect bias. *Expected cell frequency: There are a total of 325 students. The smallest percentage of expected grades are F’s, and we expect 325(0.05) = 16.25. Since the smallest expected count exceeds 5, all expected counts will exceed 5, so the condition is satisfied.

Under these conditions, the sampling distribution of the test statistic is F 2 with 5 – 1 = 4 degrees of freedom, and we will perform a chi-square goodness-of-fit test.

Grade A B C D F Observed Frequency 89 121 78 25 12 Expected Frequency 48.75 97.5 130 32.5 16.25 χ2 component 33.232 5.6641 20.8 1.7308 1.1115

2 2 Obs Exp F ¦ Exp 2 2 2 2 2 89 48.75 121 97.5 78 130 25 32.5 12 16.25 48.75 97.5 130 32.5 16.25 62.538

The P-value is the area in the upper tail of the F 2 model with 4 degrees of freedom above the computed F 2 value. P-value = P F 2!62.538 8.5 u 10 13 The P-value is very small, so we reject the null hypothesis. There is strong evidence that the most recent grade distribution differs from the historical grade distribution.

b. Which grade impacted your decision the most? Explain what this means in the context of the problem. The F 2 component for A’s is the largest, 33.232:. More students earned A’s than we would have expected based on the historical grade distribution.

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Copyright 2010 Pearson Education, Inc. 2. As part of a survey, students in a large statistics class were asked whether or not they ate breakfast that morning. The data appears in the following table: Breakfast Yes No Total Male 66 66 132 Sex Female 125 74 199 Total 191 140 331

Does it appear eating breakfast is independent of the student’s sex? Test an appropriate hypothesis. Give statistical evidence to support your conclusion.

We want to know whether the categorical variables “eating breakfast” and “student’s sex” are statistically independent. H : Eating breakfast and student’s sex are independent. 0 H : There is an association between eating breakfast and student’s sex. A Conditions: *Counted data: We have the counts of individuals in categories of two categorical variables. *Randomization: We have a convenience sample of students, but no reason to suspect bias. *Expected cell frequency: The expected values (shown in parenthesis in the table) are all greater than 5, so the condition is satisfied. Under these conditions, the sampling distribution of the test statistic is F 2 with r − 1 c − 1 = 2 − 1 2 − 1 = 1 degree of freedom, and we will perform a chi-square test of independence. ( ) ( ) ( ) ( ) Breakfast Yes No Total Male 66 (76.169) 66 (55.831) 132 Sex Female 125 (114.83) 74 (84.169) 199 Total 191 140 331

2 2 2 2 2 2 Obs Exp 66 76.169 66 55.831 125 114.83 74 84.169 F 5.339 ¦ Exp 76.169 55.831 114.83 84.169 The P-value is P F 2 !5.339 0.0209 .

The P-value of 0.0209 is pretty small, so we reject the null hypothesis. There is evidence of an association between the student’s sex and whether or not breakfast is eaten. It appears that females may be more likely to eat breakfast.

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AP Statistics Quiz B – Chapter 26 – Key

1. In a local school, vending machines offer a range of drinks from juices to sports drinks. The purchasing agent thinks each type of drink is equally favored among the students buying drinks from the machines. The recent purchasing choices from the vending machines are shown in the table. Lemon Lime Kiwi Tropical Grape Drink Type/Flavor Sports Drink Strawberry Punch Sports Drink Order Frequency 159 198 174 149 Expected Frequencies 170 170 170 170 F 2 components 0.712 4.612 0.094 2.594 a. Test an appropriate hypothesis to decide if the purchasing agent is correct. Give statistical evidence to support your conclusion. We want to know if the types of drinks are uniformly distributed (equally favored) among the students buying drinks from the machines. H : The types of drinks are uniformly distributed (equally favored) among the students buying drinks. 0 H : The types are not uniformly distributed (equally favored) among the students buying drinks. A Conditions: * Counted data: We have the counts from a sample of purchasing data. * Randomization: We don’t want to make inferences about other high schools, so we don’t need to check this condition. * Expected cell frequency: The null hypothesis expects 25% of the 680 drinks, or 170, should occur in each flavor. These expected values are all greater than 5, so the condition is satisfied. Under these conditions, the sampling distribution of the test statistic is F 2 with 4 – 1 = 3 degrees of freedom. We will perform a chi-square goodness of fit test. 2 2 2 2 2 2 Obs Exp 159 170 198 170 174 170 149 170 F 8.012 ¦ Exp 170 170 170 170 The P-value is the area in the upper tail of the F 2 model for 3 degrees of freedom above the computed F 2 value. P P F 2 !8.012 0.0458 The P-value of 0.0458 is low, so we reject the null hypothesis. These data show evidence that the drink flavors are not uniformly chosen (equally favored) by the students.

b. Which type of drink impacted your decision the most? Explain what this means in the context of the problem. Obs Exp 198 170 Kiwi Strawberry; its standardized residual is the largest 2.15 , and its Exp 170 F 2 component was 4.612. More students preferred the Kiwi Strawberry compared to what was expected by the purchasing agent. The purchasing agent should order more Kiwi Strawberry than other flavors.

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Copyright 2010 Pearson Education, Inc. 2. A manufacturing plant for recreational vehicles receives shipments from three different parts vendors. There has been a defect issue with some of the electrical wiring in the recreational vehicles manufactured at the plant. The plant manager wonders if all of the vendors might be contributing equally to the defect issue. The plant manager reviews a sample of quality assurance inspections from the last six months. The data are shown in the table below.

Purrfect Parts Co. Made-4-U Co. 25 Hour Parts Co. Rejected 53 (57.69) 48 (51.51) 70 (61.81) Perfect 93 (85.01) 71 (75.90) 88 (91.08) Not perfect but acceptable 22 (25.30) 31 (22.59) 22 (27.11)

Test an appropriate hypothesis to decide if the plant manager is correct. Give statistical evidence to support your conclusion.

We want to know whether if all of the vendors might be contributing equally to the defect issue. H : The type of defects in vehicles made by the three vendors has the same distribution (are homogeneous). 0 H : The type of defects in vehicles made by the three vendors does not have the same distribution (are not A homogeneous).

Conditions: * Counted data: We have the counts from a sample of quality assurance inspections from the last six months. * Randomization: The data are from a sample of quality assurance inspections from the last six months. * Expected cell frequency: The expected values (shown in parenthesis in the table) are all greater than five. Under these conditions, the sampling distribution of the test statistic is F 2 with (3 – 1)(3 – 1) = 4 degrees of freedom.

We will perform a chi-square test of homogeneity. 2 2 2 2 Obs Exp 53 57.69 48 51.51 F ¦ ... 7.40 Exp 57.69 51.51

2 P P (F ! 7.40) 0.1161

The P-value of 0.1161 is rather high, so we fail to reject the null hypothesis. There is little statistical evidence to indicate that the types of defects vary by vendor.

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AP Statistics Quiz C – Chapter 26 – Key

1. When two competing teams are equally matched, the probability that each team wins any game is 0.5. The NBA championship goes to the team that wins four games in a best-of-seven series. If the teams were equally matched, the probability that the final series ends with one of the teams sweeping four straight games would be 2(0.5) 4 = 0.125 . Further probability calculations indicate that 25% of these series should last five games, 31.25% should last six games, and the other 31.25% should last the full seven games. The table shows the number of games it took to decide each of the last 57 NBA champs. Do you think the teams are usually equally matched? Give statistical evidence to support your conclusion.

Length of series 4 games 5 games 6 games 7 games NBA finals 7 13 22 15 Expected counts 7.125 14.25 17.8125 17.8125

H : The distribution of length of series is consistent with a 50-50 chance for each team to win each game. 0 H : The distribution of lengths of series isn’t consistent with a 50-50 chance for each team to win each A game. Conditions: * Counted data: These are counts of categorical data. * Randomization: We will assume that these 57 series are typical of past and future NBA championships. * Expected cell frequency: The expected values are all greater than five. Under these conditions, we can perform a F 2 goodness-of-fit test with 3 degrees of freedom. 222 2 2 2 Obs Exp 7 7.125 13 14.25 22 17.8125 15 17.8125 F ¦ 1.54 Exp 7.125 14.25 17.8125 17.8125 P = 0.67. The high P-value means we do not reject the null hypothesis; there is no evidence that the NBA championship series are inconsistent with the conjecture that the teams are evenly matched.

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Copyright 2010 Pearson Education, Inc. 2. Could eye color be a warning signal for hearing loss in patients suffering from meningitis? British researcher Helen Cullington recorded the eye color of 130 deaf patients, and noted whether the patient’s deafness had developed following treatment for meningitis. Her data are summarized in the table below. Test an appropriate hypothesis and state your conclusion.

Deafness related to… Eye color meningitis other Light 30 72 Dark 2 26

H : Deafness and eye color are independent. 0 H : There is an association between deafness and eye color. A These are counts of categorical data, assumed to be representative of deaf patients in Britain, with expected counts (25.1, 76.9, 6.9, and 21.1) all at least 5. OK to do a F 2 test for independence, with df = 1 2 2 22 2 2 Obs Exp 30 25.1 72 76.9 2 6.9 26 21.1 F ¦ 5.87 Exp 25.1 76.9 6.9 21.1 P = 0.015

Since P = 0.015 is low, I reject the null hypothesis. There is strong evidence that hearing loss is associated with eye color. It appears that people with dark-colored eyes are at less risk of suffering deafness from meningitis.

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