AP Statistics Practice Test 1

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AP Statistics Practice Test 2

AP Statistics Practice Test 1

ANSWER SHEET FOR SECTION I

AP Statistics Practice Test 1

SECTION I

Time: 1 hour and 30 minutes

Number of questions: 40

Percentage of total grade: 50

Directions: Solve each of the following problems. Decide which is the best of the choices given and answer in the appropriate place on the answer sheet. No credit will be given for anything written on the exam. Do not spend too much time on any one problem.

A set of test scores has the following 5-number summary:

Which statement about outliers must be true?

There is exactly one outlier on the lower end.
There is at least one outlier on the lower end.
There is exactly one outlier on the higher end.
There is at least one outlier on the higher end.
There are no outliers.
The table above shows the regression output for predicting the number of calories from the number of milligrams of sodium for items at a fast-food chain. What percent of the variation in calories is explained by the regression model using sodium as a predictor?
7.36%
54.2%
53.8%
73.4%
73.6%
A drug company will conduct a randomized controlled study on the effectiveness of a new heart disease medication called Heartaid. Heartaid is more expensive than the currently used medication. The analysis will include a significance test with H ₀ : Heartaid and the current medication are equally effective at preventing heart disease and H _A: Heartaid is more effective than the current medication at preventing heart disease. Which of these would be a potential consequence of a Type II error?
Patients will spend more money on Heartaid, even though it is actually not any more effective than the current medication.
Doctors will begin to prescribe Heartaid to patients, even though it is actually not any more effective than the current medication.
Patients will continue to use the current medication, even though Heartaid is actually more effective.
Researchers will calculate the wrongP -value, making their advice to doctors invalid.
Researchers will calculate the correctP- value, but will misinterpret it in their conclusion.
Researchers in the Southwest are studying tortoises—a species of animal that is affected by habitat loss due to human development of the desert. A total of 78 tortoises are being studied by researchers at several sites. The data on gender and species of all these tortoises is organized in the table shown below.

If a tortoise from this study is to be selected at random, let A = the tortoise is female and B = the tortoise is a Morafka . Which of the following appropriately interprets the value of P(B|A)?

31.3% is the probability that a randomly selected Morafka tortoise is a female tortoise.
31.3% is the probability that a randomly selected female tortoise is a Morafka tortoise.
19.2% is the probability that a randomly selected Morafka tortoise is a female tortoise.
19.2% is the probability that a randomly selected female tortoise is a Morafka tortoise.
31.3% is the probability that a randomly selected tortoise is a Morafka tortoise.
The main purpose of blocking in an experiment is to:
reduce bias.
reduce confounding.
reduce variation within treatments.
reduce variation between treatments.
reduce the probability of a Type I error.
The midterm scores for a statistics course were approximately normally distributed with a mean of 52 points and a standard deviation of 4 points. The final exam scores were approximately normally distributed with a mean of 112 points and a standard deviation of 10 points. One student had a score of 58 points on the midterm. If she had the same standardized score (z -score) on the final exam, what must her score have been?
15 points
58 points
118 points
122 points
127 points
Which of the following is not an advantage of stratified random sampling over simple random sampling?
When done correctly, a stratified random sample is less biased than a simple random sample.
When done correctly, a stratified random sampling process has less variability from sample to sample than a simple random sample.
When done correctly, a stratified random sample can provide, with a smaller sample size, an estimate that is just as reliable as that of a simple random sample with a larger sample size.
A stratified random sample provides information about each stratum in the population as well as an estimate for the population as a whole, and a simple random sample does not.
When done correctly, a stratified random sample can provide a more reliable estimate than a simple random sample using the same sample size.
The correlation between height in inches and weight in pounds for a particular class is 0.65. If the heights are converted from inches to centimeters, what will the correlation be? (1 in. = 2.54 cm)
–0.65
–0.26
0.10
0.26
0.65
Breakfast cereals have a wide range of sugar content. Some cereals contain High Fructose Corn Syrup (HFCS) as a source of sugar and some do not. The boxplots above show the total sugar content of different types of cereal for those containing HFCS and for those that do not. Which statement is true based on the boxplots?
The number of cereals with HFCS is about the same as the number of cereals without HFCS.
The cereals with HFCS have a greater interquartile range than the cereals without HFCS.
The cereals without HFCS have a greater range than the cereals with HFCS.

D About half the cereals without HFCS have less sugar than about three-fourths of the cereals with HFCS.

About half the cereals with HFCS have more sugar than about three-fourths of the cereals without HFCS.
Semir rolls a six-sided die every morning to determine what he will have for breakfast. If he rolls a 1 or 2, he takes time to cook himself a big breakfast. If he rolls a 3 or larger he grabs a quick lighter breakfast. When he cooks himself a big breakfast, there is a 15% chance he will be late for school. If he has a lighter breakfast, there is a 6% chance he will be late for school. What is the probability Semir will be on time for school any given day?
0.09
0.21
0.80
0.91
0.94
Researchers were gathering data on alligators in an attempt to estimate an alligator”s weight from its length. They captured 29 alligators and measured their length and weight. They created three regression models. Each model, along with its residual plot, is shown below. y represents the weight in pounds and x represents the length in inches.

Which statement is true?

There is a linear relationship between weight and length, and model I is most appropriate.
There is a linear relationship between weight and length, and model II is most appropriate.
There is a nonlinear relationship between weight and length, and model I is most appropriate.
There is a nonlinear relationship between weight and length, and model II is most appropriate.
There is a nonlinear relationship between weight and length, and model III is most appropriate.
A polling company has been asked to do a poll to see if the proportion of people that would vote for their candidate has improved since the previous poll, as their analyst suspects. Using p _p = the proportion of all voters that would have supported their candidate in the previous poll, and p _c = the proportion of all voters that would support their candidate in the current poll, which is an appropriate pair of hypotheses for a significance test?
H₀ : p _c > P _P

H _a : p _c ≤ P _P

H₀ : p _c < P _P

H _a : p _c > P _P

H₀ : p _c > P _P

H _a : p _c = P _P

H₀ : p _c = P _P

H _a : p _c > P _P

H₀ : p _c = P _P

H _a : p _c ≠ P _P

A public health researcher suspected that there would be a relationship between the proportion of people in a state that engage in binge alcohol use and the proportion of people in the state that smoke cigarettes. A regression analysis was performed and some of the output is shown below.

What is the correlation coefficient for percent cigarette smokers and percent binge drinkers?

–0.1887
–0.0356
0.0160
0.0356
0.1887
Marine biologists want to determine whether a small increase in the acidity of sea water would adversely affect marine life. They randomly assigned several Entacmaea quadricolor , one species of sea anemone, to one of eight aquariums that had been prepared with environments as similar as possible to the natural habitat. Then they randomly selected four tanks, and gradually increased the acidity of the water. They monitored the health of the anemones for the duration of the study. Which of these statements is NOT true about this study?
An advantage of using only one variety of sea anemones is that there should be less variability in the response to each treatment.
A disadvantage of using only one variety of sea anemones is that the scope of inference is limited only to that type of anemone.
An advantage of using aquariums is that it is easier to maintain control to avoid confounding factors.
A disadvantage of using aquariums is that the anemones might respond differently in the ocean than they do in aquariums.
If the anemones in the aquariums with increased acidity are less healthy than those without increased acidity, it cannot be determined that the increased aciditycaused the response.
In a large high school, the student council wants to select a sample to survey the student body. They will first select 25 freshmen, then select 25 sophomores, then 25 juniors, and then finally 25 seniors. This is an example of a
Cluster sample
Stratified sample
Simple random sample
Convenience sample
Systematic sample
Listed above are the summary statistics for the daily high temperatures (in degrees F) for the month of September in a Midwestern U.S. city. One particular day has a z -score of 0.70. What was the likely temperature that day?
68
70
74
76
80
In American football, the running back carries the football forward in a quest to score points. Many people create fantasy football teams by selecting players as their “team.” These fans analyze data to judge which players are outstanding. The following regression output analyzes the linear relationship between the number of times a running back is given the football (carries) and the total number of yards that player gains in a season for a random sample of 18 running backs.

The appropriate calculation for a 99% confidence interval for the slope of the least squares regression line for the total yards gained versus number of carries in a season for all running backs is

4.3441 ± 2.576.(1.811)/
4.3441 ± 2.921.(1.811)
4.3441 ± 2.921.(1.811)/
4.3441 ± 2.898.(1.811)
4.3441 ± 2.898.(1.811)/
Above is a cumulative relative frequency plot for the scores of a large university class on a 90-point statistics exam. Which of the following observations is correct?
The median score is at least 60 points.
The distribution of scores is skewed to the right.
The distribution of scores is skewed to the left.
The distribution is roughly symmetric.
If a passing score is 60, most students passed the test.
Self-efficacy (the belief that one has control over one”s situation) as it related to job satisfaction was studied. When a group of teachers rated their ability to control their situation and their satisfaction with their job, the two variables had a correlation of 0.30. Which statement follows from this correlation?
If you want teachers to be happy with their job, give them more control over their situation.
If you want teachers to take more control over their situation, make them happier at their jobs.
Teachers in the study who were more satisfied with their job were less confident in their ability to control their situation.
There is evidence of a causal relationship.
9% of the variability in job satisfaction can be explained by the linear model with self-efficacy as a predictor.
A university sent out a survey to a random sample of 120 recent graduates to see how those graduates feel about the education they received. Thirty-two of the graduates did not respond to the survey. Which of these would be the best way to proceed?
Use the 88 who did respond, using 88 as the sample size in the analysis.
Use the 88 who did respond, using 120 as the sample size in the analysis.
Attempt to contact the 32 who did not respond and get the information from them.
Select 32 more graduates at random and use them to replace the nonresponders.
Start over, sending the survey to more graduates to increase the number of responses.
Some health professionals suspect that doctors are more likely to order cardiac tests for men than women, even when women describe exactly the same symptoms. Young doctors, training to work in the emergency room, were randomly assigned to two groups—Group A and Group B—and presented with the exact same description of common symptoms of heart disease. However, those in Group A were told that the patient was a 55-year-old female and those in Group B were told that the patient was a 55-year-old male. All other patient characteristics were exactly the same.

Which of the following is the appropriate statistic to test whether doctors are less likely to order cardiac tests when the patient is female?

A study looked at medical records of about 23,000 patients, mostly in Asia. They found that patients who drank at least three cups of green tea per day had a much lower incidence of depression. In an American newspaper article about this study, which of the following statements should not be made?
It is possible that people who drink green tea also tend to have healthier lifestyles than those who don”t.
It is possible that people who drink green tea also tend to have a more positive outlook on life than those who don”t.
Because this is observational data, the association between drinking green tea and a lower incidence of depression is not an indication of a causal relationship.
Because most of these patients lived in Asia, where tea drinking is much more common than in the United States, a similar association may not exist in the United States.
People who want to avoid depression should consider drinking green tea on a daily basis.
A polling company wants to estimate, using a 95% confidence interval, the proportion of likely voters that would vote for a particular candidate in the upcoming presidential election. Of the following choices , which is the smallest sample size that will ensure a margin of sampling error less than or equal to 2 percentage points?
250
950
1,150
2,250
2,450
The director of a local food bank asks for data on all donations given during the month of November. Of the 100 checks received, the average donation is $155 with a standard deviation of $32. Which of the following is the most appropriate statement?
This November, the average donation is $155.
50% of all donations this November are more than $155.
We are 95% confident that the average donation in November is between about $91 and $219.
We are 95% confident that the average donation in November is between about $149 and $161.
This November, about 95% of all donations are between $91 and $219.
A multiple-choice test has 30 questions, and each question has five options. A student has not studied and will guess on every question. Which calculation shows how to find the probability of getting between 15 and 20 (inclusive) questions correct?
0.2¹⁵0.8¹⁵ + 0.2¹⁶ 0.8¹⁴ + … + 0.2²⁰ 0.8¹⁰
Use the normal approximation to find .
Use the normal approximation to find .
Claire and Max each created a simulated sampling distribution for a test statistic. They each used samples of size 4 from a normal population with the same mean m and standard deviation s . Claire simulated the sampling distribution of and Max simulated the sampling distribution of , where s is the sample standard deviation.

Which statement makes a correct comparison of their simulated sampling distributions?

Both simulated sampling distributions are approximately normal.
Both simulated sampling distributions have the same spread.
Claire”s simulated sampling distribution is skewed to the right, and Max”s is roughly symmetric.
Claire”s simulated sampling distribution is roughly symmetric, and Max”s is skewed to the left.
Claire”s simulated sampling distribution is approximately normal, and Max”s is approximately at -distribution.
A spreadsheet contains a data set that consists of 100 values such that IQR > variance . However, these values must be converted using the following formula: NewValue = 2(OldValue ) + 10. Which of the following statements is true about the data set consisting of the NewValues?
The IQR has been increased by 10 units.
The IQR is now less than the variance.
The IQR is now double its previous size.
The IQR has been doubled and then increased by 10 units.
The IQR has been cut in half.
A test for heartworm in dogs shows a positive result in 96% of dogs that actually have heartworm, and shows a negative result in 98% of dogs with no heartworm. If heartworm actually occurs in 10% of dogs, what is the probability that a randomly selected dog that tested positive for heartworm actually has heartworm?
11%
18%
84%
88%
96%
A recent poll reported that 43% of Americans approve of the job the president is doing, with a margin of sampling error of ±3.2% at a 95% level of confidence. Which of these correctly interprets that margin of error?
There is a 3.2% percent chance that the proportion of all Americans who approve of the job the president is doing is not 43%.
There is a 95% chance that the proportion of all Americans who approve of the job the president is doing is 43%.
About 95% of polls conducted in this way will find that between 39.8% and 46.2% of those sampled approve of the job the president is doing.
About 95% of polls conducted in this way will give a sample proportion within 3.2 percentage points of the actual proportion of all Americans who approve of the job the president is doing.
About 3.2% of all Americans approve of 95% of what the president does.
Two events A and B each have a nonzero probability. If A and B are independent, which of the following statements is true?
P(A and B ) = P (A ) · P (B )
Aand B may or may not be mutually exclusive.
Aand B must be mutually exclusive.
P(A | B ) = P (B | A )
P(A and B ) = 0
Players in the National Football League weigh, on average, about 248 pounds with a standard deviation of about 47 pounds. If four players are to be selected at random, the expected value of the random variable W , the total combined weight of the four players, is 992 pounds. The standard deviation of W is approximately
47 pounds
67 pounds
94 pounds
141 pounds
188 pounds
When a patient complains to the doctor about a certain set of symptoms, the doctor diagnoses the patient with Condition A 15% of the time. If a patient with these symptoms is diagnosed with Condition A, he or she is diagnosed with Condition B 70% of the time. A patient with these symptoms that is not diagnosed with Condition A is diagnosed with Condition B 10% of the time. What is the probability that a patient with this set of symptoms will be diagnosed with at least one of these conditions?
0.235
0.250
0.765
0.850
0.950
A doctor hopes that a new surgery technique will shorten the recovery time compared to the standard technique. To test this, he designed an experiment in which patients who required this type of surgery were randomly assigned to the standard technique or the new technique. Then the mean recovery time for each treatment group was compared. Assuming conditions for inference were met, which analysis should be used?
At -test for a mean.
At -test for a difference in means.
Az- test for a mean.
Az -test for a difference in means.
Az -test for a difference in proportions.
For a class project, a student wants to see if boys and girls at their large high school differ in the number of contacts they have stored in their phone. The student conducts a survey of 50 randomly sampled boys and 40 randomly selected girls, and asks them to report the number of contacts. Which of the following is true about this situation?
Because the population standard deviations are not known and conditions are met, the studentshould use a two-sample t -test.
Because the sample sizes are different, the student shouldnot use a two-sample t -test.
Because the sample sizes are both greater than 30, the student shouldnot use a two-sample t -test.
Because the shape of the population distribution is not known, the student shouldnot use a two-sample t -test.
Becausenp and n (1 – p ) are both at least 10, the student should use a two-proportion z -test.
Which of these is the best description of a P -value?
The probability of making a Type I error.
The probability of making a Type II error.
The probability of rejecting the null hypothesis if it is, in fact, false.
The probability of getting a test statistic at least as extreme as the observed test statistic, if the null hypothesis is true.
The probability of getting a test statistic at least as extreme as the observed test statistic, if the null hypothesis is false.
A farmer who raises hens for egg production wants his eggs to have a mean mass of 56 grams. He is considering the purchase of a different type of hen, so he took a random sample of 18 eggs laid by this type of hen. The distribution of the masses is symmetric and mound-shaped with a mean of 54.1 grams and no outliers. The farmer conducted a t -test to see if there is evidence that the eggs from these hens have a mean mass that is different from 56 g and got a test statistic of t = –1.973. If he uses a 5% significance level, which is the correct conclusion and reason?
Becauset is more extreme than ±1.96, he should reject the null hypothesis. He has convincing evidence at the 5% significance level that the mean mass of eggs from these hens is different from 56 grams.
Becauset is less extreme than the critical value of t for 17 degrees of freedom, he should not reject the null hypothesis. He does not have convincing evidence at the 5% significance level that the mean mass of eggs from these hens is different from 56 grams.
Becauset is less extreme than the critical value of t for 18 degrees of freedom, he should not reject the null hypothesis. He does not have convincing evidence at the 5% significance level that the mean mass of eggs from these hens is different from 56 grams.
Becauset is more extreme than the critical value of t for 18 degrees of freedom, he should reject the null hypothesis. He has convincing evidence at the 5% significance level that the mean mass of eggs from these hens is different from 56 grams.
Because the sample mean was less than 56, he should use a one-sided alternative hypothesis. Thus,t is more extreme than the critical value of t for 17 degrees of freedom and he should reject the null hypothesis. He has convincing evidence at the 5% significance level that the mean mass of eggs from these hens is different from 56 grams.
An experiment is conducted in which the response variable is the average gain in participants” performance in the long jump. A two-sample t -test with a 5% level of significance will be used to analyze the results. If all else is kept the same, which of the following descriptions of a possible change in procedure is true?
Change from equal size treatment groups to very different size treatment groups would increase the power of the test.
Change from a 5% significance level to a 1% significance level would increase the power of the test.
Taking more careful measurements to reduce variability in the response would increase the power of the test.
Increasing the sample size would reduce the probability of a Type I error.
Increasing the sample size would increase the probability of a Type I error.
Researchers wish to test how rangelands growth is affected by wildfire. A very large open area is divided into 28 equally sized plots and subjected to a controlled burn. One year later, the researchers calculate the percentage of grass returned in each plot after the fire damage. The average percentage return for these 28 plots was 58.5% with a standard deviation of 6.5%. Which of the following represents a 95% confidence interval to estimate the average proportion of grass return one year after a fire?
0.585 ± 1.96(0.065/
0.585 ± 2.052(0.065/
0.585 ± 2.052(0.065/
0.585 ± 1.96(0.065/
Which of the following is NOT true of the χ² probability distribution function?
The area under the χ²curve is 1.
χ²is defined only for nonnegative values of the variable.
For small degrees of freedom, the curve displays strong right-skewness.
For the same α, as the number of degrees of freedom increases, the critical value for the rejection region decreases.
χ²is actually a family of probability distributions defined by its degrees of freedom.
Tucson, Arizona, hosts the world”s largest gem and mineral show every February. A government official claims that more than 50% of city residents have been to the Gem Show at least once. You decide to test this theory, take a simple random sample of 200 Tucson residents, and discover that 108 have attended. Which of the following is the appropriate calculation of the test statistic for the hypothesis proposed by the government official?

END OF SECTION I

AP Statistics Practice Test 1

SECTION II

Time: 1 hour and 30 minutes

Number of questions: 6

Percentage of total grade: 50

General Instructions

There are two parts to this section of the examination. Part A consists of five equally weighted problems that represent 75% of the total weight of this section. Spend about 65 minutes on this part of the exam. Part B consists of one longer problem that represents 25% of the total weight of this section. Spend about 25 minutes on this part of the exam. You are not necessarily expected to complete all parts of every question. Statistical tables and formulas are provided.

Be sure to write clearly and legibly. If you make an error, you may save time by crossing it out rather than trying to erase it. Erased or crossed-out work will not be graded.
Show all your work. Indicate clearly the methods you use because you will be graded on the correctness of your methods as well as the accuracy of your final answers. Correct answers without support work may not receive credit.

Statistics, Section II, Part A, Questions 1–5

Spend about 65 minutes on this part of the exam; percentage of Section II grade: 75.

Directions: Show all your work. Indicate clearly the methods you use because you will be graded on the correctness of your methods as well as on the accuracy of your results and explanation.

In response to drought, some plants react with reduced biological nitrogen fixation (BNF), hindering their ability to grow at a typical rate. Scientists are working to develop techniques to supplement crops so that the BNF rate is not as greatly reduced during drought. One promising treatment is to be compared to standard growing conditions on randomly selected acres of soybeans. The following summary data is for the bushels per acre of soybeans for each group:
The maximum yield of the treated acres was 51.50 bushels of soybeans. Is this an unusually high yield compared to the other treated acres? Justify your answer.
Researchers indicate that the average yield for the treatment group is statistically significantly higher than that of the control group. Explain what this means in simple terms.
Field workers report to the scientists that the machine to measure the yield in the treatment group was poorly calibrated—each of the bushel values was over-reported by 0.5 bushels. So each of the measures should be reduced (subtracted) by 0.5 bushels. Describe the impact doing so would have on the mean, median, standard deviation, and interquartile range of the treatment group.
A particular bottle-filling machine is supposed to put 16 ounces of water into a bottle. The amount actually dispensed is approximately normal with mean 16.1 ounces and a standard deviation of 0.12 ounces. A bottle is rejected by quality control if it contains less than 15.7 ounces or more than 16.3 ounces.
What is the probability that a randomly selected bottle gets rejected by quality control?
What is the probability that, out of 20 randomly selected bottles, at least 2 are rejected by quality control?
What is the probability that a set of 5 randomly selected bottles contains, on average, more than 16.2 ounces of water?
A student was curious about a news article that reported the percentage of Americans who believe that a “higher power” affects who wins sporting events. With the assumption that the percentage reported in the article is correct, she simulated selecting samples of 100 Americans and calculating the percentage of those that believe a higher power has a hand in who wins a sporting event. The results of 200 trials are shown below.
Based on the results of these trials, what appears to be the reported percentage of Americans that believes the results of a sporting event are influenced by a “higher power?” Explain.
Another student believes that the proportion reported in the article is too low. He takes a random sample of 100 Americans and finds that 44 percent believe in the influence of a “higher power” on the outcome of a sporting event. Based on the dotplot above, does it appear the percentage in the article is correct? Explain why or why not.
Below are the residual plots generated for the linear regression on three separate sets of bivariate data.
Is a linear model appropriate for Data Set A? Justify your answer using Residual Plot A.
What does Residual Plot B reveal about the reliability of the estimates from the regression line in Data Set B?
Explain what information is being conveyed about the linear regression for Data Set C in Residual Plot C.
New formulas of gel nail polish have recently been introduced. A local salon wished to test number of days of wear of the new formula against the old formula as a control. Identical bottles of both formulas are prepared and 14 clients have their nails polished on their first visit with one formula (chosen at random) and then the next visit with the other formula. The number of days of wear is then compared.

Do these data provide evidence that the mean number of days of wear is greater for the new formula of gel nail polish than for the old? Use α = 0.05.

Statistics, Section II, Part B, Question 6

Spend about 25 minutes on this part of the exam; percentage of Section II grade: 25.

Directions: Show all of your work. Indicate clearly the methods you use because you will be graded on the correctness of your methods as well as on the accuracy of your results and explanation.

Overuse of lawn chemicals causes environmental problems when runoff gets into waterways. Researchers tested three types of fertilizer spreaders typically used by homeowners: drop spreaders, hand-held broadcast rotary spreaders, and push rotary spreaders. The goal was to determine if the type of spreader influenced the ability of the user to apply the correct amount of fertilizer.

Fifteen volunteers who were homeowners with no particular training in the use of this equipment operated the spreaders. Each spreader was randomly assigned to a 1,000-square-foot plot of lawn. The researchers prepared the spreaders by correctly setting the dispensation amount and providing the same amount of fertilizer in each spreader. The spreader with fertilizer was weighed before and after the fertilizer was applied to determine the amount dispensed.

What were the experimental units? The treatments? The response variable?

The plot below shows the amounts dispensed by each type of spreader in kg/100 m² .

Based on the plot alone, comment on the strength of evidence of a difference in the amount of fertilizer dispensed by the three types of spreaders.

Volunteers with no particular training were used because regular homeowners are the population of interest. However, one researcher points out that there may be very different skill levels among the volunteers. He suggests using a randomized block design, with each volunteer as a block. Each volunteer will use all three spreaders in a randomized order. They ran the experiment with just five volunteers using this design. The results, along with the operator of the spreader, are displayed below.

Based on the plot above, comment on the association between the operator of the spreader and the amount of fertilizer dispensed.

To remove the effect of the operators” skill levels on the response variable, the researchers calculated the mean amount dispensed for each operator. Miya”s amounts dispensed are given in the table below.

To adjust the amount dispensed for Miya, calculate the mean amount of fertilizer she dispensed in her three trials, and subtract that mean from each of her amounts dispensed. Show all computations.

The plot below shows the Adjusted Amount Dispensed for the other four volunteers.

Add Miya”s Adjusted Amount Dispensed for each type of spreader to the plot above. Then comment on what this new plot reveals about the differences in amounts dispensed between the types of spreaders.

END OF SECTION II

Answers to Practice Test 1, Section I

Solutions to Practice Test 1, Section I

(B) Using the outlier guideline, any value above Q3 + 1.5 ⋅IQR = 28.5 + 1.5 ⋅ 10.5 = 44.25, or below Q1 − 1.5 ⋅ IQR = 18 − 1.5 ⋅ 10.5 = 2.25, is considered an outlier. Because the maximum is 33, there are no values above 44.25. The minimum of 2 is an outlier. There may be more than one value below 2.25, so all we can say is that there is at least one outlier on the lower end.
(B) You should recognize that this is the interpretation ofr ² . According to the printout, that value is 0.542313, or about 54.2%.
(C) A Type II error means that the null hypothesis is false, but we fail to reject it. In this case, a false null hypothesis means that Heartaid actually does work better. Failing to reject that null hypothesis means they don”t have convincing evidence that Heartaid is better, so they won”t use it.
(B) This is asking for a conditional probability.P (Morafka|Female). Out of the 48 female tortoises, 15 are Morafka. 15/48 = 0.3125, or about 31.3%.
(C) Randomization is what fights against confounding by making the groups as alike as possible. With blocking, similar units are grouped together so the effect of being in that group can be accounted for in the analysis.
(E) On the midterm, thez -score is So she scored 1.5 standard deviations above the mean. To achieve the same standardized score on the final exam, she must score 1.5 standard deviations, or 1.5 ⋅ 10 = 15 points, above the mean on that exam. 112 + 15 = 127 points.
(A) A simple random sample is already unbiased. A stratified random sample cannot improve on that.
(E) The correlation coefficient remains unchanged when converting units.
(E) The median of the HFCS group is the same as Q3 of the No HFCS group.
(D) . 0.06 = 0.09. 1 – 0.09 = 0.91.
(D) Because there is a curve in residual plot I, which is for weight vs. length, the relationship between those two variables is not linear. Because residual plot II shows a no curve and plot III does, plot II represents a better model.
(D) The null hypothesis always contains equality. Because they are interested in whether the candidate”s support has improved, the alternative hypothesis is one-sided.
(A) The correlation,r , is the square root of r ² , which is given in the table. So r could be ±0.1887. Since the slope is negative, r is negative.
(E) Because this is a properly conducted randomized experiment, a causal relationship can be determined. E is the choice that is not true.
(B) The population is divided into homogenous subgroups, and a random sample is selected from each.
(E)
(B) The formula for the confidence interval for the slope isb ₁ ± t ^* · s_b ₁ · From the printout b ₁ = 4.3441 and s_b ₁ = 1.811. t ^* for 18 – 2 = 16 degrees of freedom and 99% confidence is 2.921.
(B) The steeper part of the graph corresponds to the higher bars on a histogram. So the higher bars would be on the left with a tail stretching toward the right.
(E) No causal relationship is established simply because of a correlation. Butr ² is 0.09.
(C) Nonresponse bias is a problem because the kind of people who don”t respond might feel differently from those who do. Getting responses from more of the people that do respond won”t fix the problem.
(B)Under the null hypothesis, the proportions are the same, so the standard error formula is based on the pooled proportion .
(E) Choice E implies a causal relationship, which is not justified because this is an observational study.
(E) The margin of error is largest for = 0.5, so if ≠ 0.5, the margin of error will be even smaller. So use . Solve for n and round up.
(A) They have all the data for November, so there is no inference to do. 50% of the donations are above the median, not the mean. A is the only conclusion we can make.
(A)n is not large enough to use a normal approximation. This is a binomial situation, and choice A correctly applies the binomial calculation.
(E) Usings in the denominator makes Max”s an approximate t -distribution. That has heavier tails and more spread than Claire”s approximately normal distribution.
(C) Multiplying affects both measures of location and spread. Adding affects only measures of location.
(C) This is easiest to see in a two-way table.

The required probability is 0.096/0.114 = 0.842.

(D) The confidence interval is based on one sample of many possible samples. The process has a 95% chance of capturing the true proportion of all Americans within the interval created. But there is nothing special about that one interval.
(A) That is the multiplication rule for independent events.
(C) Since these are four independent occurrences ofW , the variances add. So the standard deviation is
(A) The bottom path in the tree diagram shows that the probability that a patient has neither condition is 0.85 ⋅ 0.90 = 0.765. 1 – 0.765 = 0.235.
(B) They are comparing means without a known population standard deviation.
(A)They are comparing means, so a t -test would be the appropriate test. The students were randomly selected. With such large sample sizes, population shapes are not a concern. Conditions are met.
(D) That is the definition of aP -value.
(B) The critical value of t forn – 1 = 17 degrees of freedom is ±2.11. His test statistic was not outside those values.
(C) Reducing measurement variability would give a lower standard error of the mean and, therefore, a greater test statistic (in absolute value). Thus, the probability of rejecting the null hypothesis increases.
(C) This is a confidence interval for a mean with unknown s, sot for 27 degrees of freedom is used rather than z . The standard error of the mean is .
(D) As the number of degrees of freedom increases, the entire distribution is moved to the right, so the critical value for the rejection region increases.
(B) The official should use the hypothesized value ofp , which is 0.5, in the calculation. Also, n = 200, not 108.

Solutions to Practice Test 1, Section II, Part A

a. The 1.5IQRguideline can be used. IQR = 46.46 – 40.52 = 5.94. 1.5IQR = 8.91 bushels. The largest possible nonoutlier would be 46.46 + 8.91 = 55.37 bushels. The maximum of 51.5 bushels is below this value, so it is not an unusually high soybean yield.

(It would be acceptable to use the mean + 2 standard deviations, which is 43.555+2.3.937 = 51.429. Using this guideline, the maximum is more than 2 standard deviations from the mean, so one might consider it unusual.)

Even if the treatment has no effect on soybean yield, the means of the groups will probably be different due to random variability. If the treatment group”s yield was statistically significantly higher than that of the control group, that means the difference is too large to be reasonably attributed to chance.
The mean and median would both decrease by 0.5 bushels. The standard deviation and interquartile range would be unchanged.
a. Using a normal model withμ = 16.1 and σ = 0.12, the z -score for . The z -score for . P (15.7 < x < 16.3) = P (–3.33 < z < 1.67) = 0.048.
This is a binomial situation withn = 20 and p = 0.048 (from part a).
This is the sampling distribution of , which will be approximately normal with μ = μ = 16.1, and .
a. The mean of the distribution appears to be about 26%. This is the best estimate of the proportion used in the simulation.
A sample proportion of 44% or more appeared only 3 times out of 100. Because that is less than 5% of the trials, this is convincing evidence that the proportion that appears in the article is too low.
a. Yes. The residual plot shows no pattern and the points show random scatter. A linear model is appropriate.
The reliability of the estimates decreases as thex variable increases. This is evident from the fan shape in the residual plot.
There is a definite curve in the residual plot, which shows that a linear model is not appropriate for this relationship.
This is a matched pairs design.

Hypotheses :

Check conditions :

Treatments are assigned to subjects in a random order.

The plot of the differences is reasonably symmetric with no outliers.

Conditions are met for a matched pairs t -test.

Computations :

Comparing to a t distribution with 13 degrees of freedom, the P -value is 0.112.

Conclusion:

Because the P -value = 0.112 > α = 0.05, we fail to reject the null hypothesis. We do not have convincing evidence that there is a difference in the number of days of wear between the new formula and the old formula.

Solutions to Practice Test 1, Section II, Part B

a. The experimental units are the plots of land. The treatments are the three types of spreaders. The response variable is the amount of fertilizer dispensed.
There seems to be a slight difference in the amounts dispensed, but with the large amount of overlap between the groups it is not very convincing evidence of a difference.
There seems to be a strong association between the operator and the amount dispensed. Jackson and Sara dispensed smaller amounts with all three spreaders than the other operators. Bill and Miya dispensed larger amounts.
Miya”s mean amount dispensed is

There is now much less overlap between the treatments. With the variability of the operator accounted for, the evidence of a difference is more convincing. With drop spreaders, users tend to dispense the least fertilizer, and with push rotary spreaders, they tend to dispense the most.

AP Statistics Practice Test 2

ANSWER SHEET FOR SECTION I

AP Statistics Practice Test 2

SECTION I

Time: 1 hour and 30 minutes

Number of questions: 40

Percentage of total grade: 50

Sixty-two students were asked to complete a number puzzle. The number of minutes each student needed to complete the puzzle was recorded and the times are displayed in the histogram above. Which of the following describes the distribution?
The distribution is approximately normal.
The distribution is roughly symmetric.
The distribution is bimodal.
The distribution is skewed left.
The distribution is skewed right.
The table above shows the regression output for predicting the number of calories from the number of milligrams of sodium for items at a fast-food chain. Which is a correct interpretation of the slope of the regression line?
Every milligram of sodium contains 0.2457 calories.
A food item with one more milligram of sodium than another tends to have 0.2457 more calories.
A food item with one more milligram of sodium than another tends to have 136.69 more calories.
A food item with one more milligram of sodium than another tends to have 147.1427 more calories.
A food item with one more calorie than another tends to have 147.1427 more milligrams of sodium.
A new fertilizer is to be tested on tomato plants to determine if it increases the production of the plants. The available land is divided into eight plots. Four of the plots are on a hill and four are in a valley as shown. The plots in the valley are expected to get more water due to the influence of gravity, and those on the hill will get more direct sunlight. A randomized block design will be used, using two blocks of four plots each. Which of the following blocking schemes would be most appropriate?
Plots 1, 2, 3, and 4 in one block and plots 5, 6, 7, and 8 in the other
Plots 1, 2, 5, and 6 in one block and plots 3, 4, 7, and 8 in the other
Plots 1, 4, 5, and 8 in one block and plots 2, 3, 6, and 7 in the other
Plots 1, 3, 6, and 8 in one block and plots 2, 4, 5, and 7 in the other
The plots should be randomly assigned to blocks.
At a large hospital, the durations of emergency room visits, from check-in to discharge, are approximately normally distributed with a mean of 167 minutes and a standard deviation of 76 minutes. Which interval, to the nearest minute, contains the middle 80% of durations?
15 minutes to 319 minutes
53 minutes to 281 minutes
70 minutes to 264 minutes
91 minutes to 243 minutes
103 minutes to 231 minutes
Aliana works 6-hour weekend shifts at an area chain restaurant as a hostess, and she wants to explore whether customer preferences are influenced by how you ask them about seating. Aliana”s questioning strategy is as follows:

Aliana collected data over the course of several weekends, and her results are in the table below:

Which of the following is a true statement?

The number of degrees of freedom for the χ²is 102.
The sample size is a large enough size because it is bigger than 30.
We do not meet the conditions for a χ²test because the Table/Question A cell is less than 5.
We do not meet the conditions for a χ²because Aliana did not randomly assign treatments to subjects.
We do not meet the conditions for a χ²test because Aliana did not ask the same number of customers each of the three questions.
Biologists have two different methods for estimating the growth rate of a population of animals. One is based on counting the number of animals in a photographic image, and the other is based on a computer-calculated surface area of the animals in the image. (Larger animals are generally older and contribute more to reproduction.) The scatterplot above shows the estimates of the Population Growth Rate (PGR) for each method along with the line y = x . Which statement is true based on the scatterplot?
Thesurface area method overestimates the Population Growth Rate.
Thecount method underestimates the Population Growth Rate.
Both methods overestimate the Population Growth Rate.
Both methods give accurate estimates of the Population Growth Rate.
There is not enough information to tell how well either method estimates the actual Population Growth Rate.
An infrared thermometer measures the surface temperature of an object from a distance. The measurement errors for one particular model are approximately normally distributed with a mean of 0 degrees C and a standard deviation of 1.3 degrees C. If the temperature of an object is 20 degrees C, what is the probability that the thermometer will report the temperature as less than 18 degrees C?
0.023
0.062
0.097
0.903
0.938
A study was conducted to test a new style of keyboard in preventing repetitive stress disorders. Volunteers who have had problems with such injuries were randomly assigned to use either a traditional keyboard or the new design. A significance test was conducted with the alternative hypothesis that a smaller proportion of those using the new keyboard will suffer injuries than those using the traditional keyboard. The resulting P -value was 0.07. Which is a correct interpretation of this P -value?
The null hypothesis should be rejected.
The null hypothesis should be accepted.
There is a 7% chance that the null hypothesis is correct.
There is a 7% chance of getting a difference between the two groups at least as large as the observed difference if the new keyboard is really no better at preventing injuries.
There is a 7% chance that the observed difference is due to chance.
For a class project, Charlotte recorded the heights of all 28 students in her class and calculated several statistics. She then realized she made an error recording the height of the tallest person in the class. She correctly had him listed as the tallest, but needed to add two inches to his recorded height to correct it. Which of these measures of spread must remain unchanged?
Mean absolute deviation
Standard deviation
Variance
Interquartile range
Range
A large company grants its hourly employees pay raises after their six-month performance review. The amounts of the raises average $0.25 per hour, with a standard deviation of $0.05 per hour. Let represent the average pay raise of 50 randomly selected hourly employees. Which calculation would give the approximate probability that is at least $0.30 per hour?
P(z ≥ 0.30)
This probability cannot be calculated because no information is given about the shape of the distribution of all pay raises.
The table below shows the daily high temperatures for the month of October for two cities in different regions of the country.

Which of the following statements is true?

The range of the October temperatures for Hendersonville is greater than the range of October temperatures for Sheboygan.
The median October temperature for Hendersonville is less than the median October temperature for Sheboygan.
The median of the October temperatures for Hendersonville is greater than the third quartile of the October temperatures for Sheboygan.
The mean of the October temperatures for Hendersonville is less than the mean of the October temperatures for Sheboygan.
The interquartile range of the October temperatures for Hendersonville is about the same as the interquartile range of the October temperatures for Sheboygan.
Biologists around the world have increased efforts at conservation. Monitoring wildlife populations is important so that appropriate management efforts can be implemented, but such monitoring is often difficult. One study found a cheap and easy way to estimate the number of nesting sites of terns (a type of seabird) by monitoring the number of calls heard per minute. More calls happen when there are more birds, as one would expect. In fact, it turned out that the number of calls explained 71% of the variation in the abundance of nests between breeding sites. Which of the following statements is correct about the correlation between the number of calls and the abundance of nests?
The correlation coefficient is –0.71.
The correlation coefficient is 0.71.
The correlation coefficient is –0.84.
The correlation coefficient is 0.84.
The slope of the regression line is needed to determine the correlation.
A woman who works for a department store wants to sue the company for discrimination based on gender. Her reasoning is that women seem less likely to get promoted than men. She presented this table showing the employees eligible for promotion divided by both gender and whether they were promoted or not. Does the table provide some evidence to support her claim?
No, because more women than men were promoted.
No, because the proportion of women who were promoted is the same as the proportion of men who were promoted.
No, because the proportion of women who were promoted is less than the proportion of men who were promoted.
Yes, because the proportion of women who were promoted is less than the proportion of men who were promoted.
Yes, because the proportion of women who were promoted is less than the proportion of women who were not promoted.
Students in a statistics class, as part of an activity, measured the length of their hair in inches. Summary statistics are shown below.

Which of the following statements is true?

Seventy-five percent of the students had hair at least 18.5 inches long.
Fifty percent of the students had hair between 9.18 and 29 inches long.
Fifty percent of the students had hair between 2 and 18.5 inches long.
The distribution of hair lengths can be appropriately modeled by a normal distribution.
The distribution of hair lengths has at least one outlier.
The scatterplot above shows data points and the regression line for predicting y from x . Which statement is true about the effect of removing point A or B on the regression model?
RemovingA would decrease the slope but removing B would increase the slope.
RemovingA would increase the slope but removing B would have little effect on the slope.
RemovingA would decrease the correlation but removing B would increase the correlation.
Removing pointA would increase the correlation and removing point B would not change the correlation.
Removing pointA would increase the y -intercept and removing point B would decrease the y -intercept.
Which of these statements correctly explains bias?
It describes a process that creates estimates that are too high on average or too low on average.
It describes a process that results in a sample that does not represent the population.
It describes a lack of randomization in the sampling process.
It describes a sample that is not representative of the population.
It describes a sample that was selected by convenience.
After graduation, you have the opportunity to open a local pizza restaurant. But to cover all your expenses as well as pay yourself a salary, you need an average revenue of about $9,000 per month. Anything less and you risk losing your business. You take a random sample of 30 different pizza restaurants similar to the one you hope to open and get data on their monthly sales. For the hypotheses

H _o : μ = $9,000 H_a: μ > $9,000

describe a Type I error and its consequence.

The consequences of a Type I error cannot be determined without an α-level.
You believe the average sales of pizza restaurants like yours is more than $9,000, so you open a pizza restaurant and have a high risk of losing your business.
You believe the average sales of pizza restaurants like yours is more than $9,000, so you open a pizza restaurant and have a high probability of a successful business.
You believe the average sales of pizza restaurants like yours is $9,000 or less, so you do not open a pizza restaurant yourself and will miss out on an opportunity to own a successful business.
You believe the average sales of pizza restaurants like yours is $9,000 or less, so you do not open a pizza restaurant because other pizza places are not successful.
A particular crop of one variety of onion has weights that are approximately normally distributed with mean 9.8 oz. and standard deviation 2.1 oz. How does an onion in the 28th percentile for weight compare to the mean?
1.22 ounces below the mean
0.59 ounces below the mean
0.59 ounces above the mean
1.22 ounces above the mean
2.26 ounces above the mean
A pharmaceutical company wants to test a new cholesterol-reducing drug against the previous drug. It does not anticipate much association between cholesterol level and gender, but it does anticipate an association between cholesterol level and the amount of exercise a person gets. For a randomized block design, it should:
Block on gender because it is not associated with cholesterol level.
Block on gender because males and females are different.
Block on the type of drug because it may be associated with cholesterol level.
Block on exercise level because it is associated with cholesterol level.
Block on both gender and exercise because blocking on more variables gives more reliable estimates.
A high school sent a survey to a randomly selected sample of 75 of last year”s graduating class. 27 of those selected did not return the survey. The best plan of action would be to:
Use the surveys that were returned and change the sample size to 48 for the analysis.
Use the surveys that were returned and leave the sample size at 75 for the analysis.
Randomly select 27 additional class members and send the survey to them.
Start over with a new sample but use a larger sample size.
Follow up with those that did not return the survey to encourage them to respond.
Which of these is a correct description of the term?
Afactor is a response variable.
Replicationmeans the experiment should be repeated several times.
Levelsare the same as treatments .
Experimental unitsare the same as subjects .
Theresponse variable is the variable directly controlled by the researchers.
The president of an online music streaming service whose customers pay a fee wants to gather additional information about customers who have joined in the past 12 months. The company plans to send out an e-mail survey to a sample of current customers with a link that gives participants a month of streaming service for free once the survey has been completed. They know that musical tastes vary by geographical region. Which of the following sample plans would produce the most representative sample of its customers?
Choose all of the customers who joined in the last month.
Make a list of all the customers who joined in the last 12 months and choose a random sample of customers on this list.
From the list of all customers who joined in the last 12 months, classify customers by the state in which they live, then choose 10 customers from each state.
From the list of all customers who joined in the last 12 months, classify customers by the state in which they live, then choose 3% of the customers from each state.
Make a list of all customers who joined in the last 12 months and choose every 10th customer on this list.
The reason we check that np and n (1 – p ) are at least 10 when calculating probabilities in a binomial situation is to:
Ensure that has a binomial distribution.
Be sure the sample is random enough.
Make sure the distribution of is symmetric enough to use a normal approximation.
Get an unbiased estimate forp .
Be able to use as the standard deviation of .
A maker of breakfast cereals plans to run a promotion in which an action figure is included in each box of cereal. The more popular figures will appear in fewer boxes. The probability of getting a box with each kind of toy is shown in the table below.

What is the probability that the first two randomly selected boxes contain Serenity and Blackstar?

0.0216
0.0432
0.0468
0.2784
0.3000
A department store has a promotion in which it hands out a “scratchcard” at the checkout register, with a percent discount concealed by an opaque covering. The customer scratches off the covering and reveals the amount of the discount. The table below shows the probability that a randomly selected card contains each percent discount.

What is the expected value of the percentage of discount?

5%
12%
15%
21%
40%
In basketball, an offensive rebound occurs when a player shoots and misses, and a player from the same team recovers the ball. For the 176 players on the roster for one season of professional men”s basketball, the third quartile for the total number of offensive rebounds for one season was 143. If five players are selected at random (with replacement) from that season, what is the approximate probability that at least three of them had more than 143 rebounds that season?
0.0127
0.0879
0.1035
0.8965
0.9121
A new smartwatch is manufactured in one part of a factory, then secured for shipping in another, independent part of the factory. The weight of the smartwatch has a mean of 62 grams and a standard deviation of 1.0 grams. The weight of the packaging (box, user”s guide, bubble wrap, etc.) has a mean of 456 grams and a standard deviation of 6 grams. Together, the distribution of the weight of the smartwatch and its packaging would have the following mean and standard deviation:
Mean 518 grams; standard deviation 7.0 grams
Mean 518 grams; standard deviation 3.5 grams
Mean 518 grams; standard deviation 6.1 grams
Mean 394 grams; standard deviation 6.1 grams
Mean 394 grams; standard deviation 3.5 grams
Students in a small high school were asked to vote for a meal for a school event. The results of the survey and the grades of the students are summarized below.

Are the events Student is a junior and Student chose pizza independent?

Yes, because the proportion of juniors who chose pizza is the same as the proportion of all students who chose pizza.
Yes, because the proportion of students that preferred pizza who are juniors is different from the proportion of all students who are juniors.
No, because the proportion of juniors who chose pizza is the same as the proportion of all students who chose pizza.
No, because the proportion of students who are juniors that preferred pizza is different from the proportion of all students who are juniors.
No, because the proportion of juniors who chose pizza is different from the proportions of other grades who chose pizza.
The total cholesterol level in a large population of people is strongly skewed right with a mean of 210 mg/dL and a standard deviation of 15 mg/dL. If random samples of size 16 are repeatedly drawn from this population, which of the following appropriately describes the sampling distribution of these sample means?
The shape is unknown with a mean of 210 and a standard deviation of 15.
The shape is somewhat skewed right with a mean of 210 and a standard deviation of 3.75.
The shape is approximately normal with a mean of 210 and a standard deviation of 15.
The shape is approximately normal with a mean of 210 and a standard deviation of 3.75.
The shape is now skewed left with a mean of 210 and a standard deviation of 3.75.
In one metropolitan region, technical writers have an annual salary that is approximately normally distributed with a mean of $55,800. The first quartile of salaries is $48,815. What is the standard deviation?
$6,984
$10,356
$10,476
$13,968
$20,709
Which of these explains why t should be used instead of z for inference procedures for means.
The Central Limit Theorem applies to proportions but not means.
We usez for proportions because proportions are approximately normal. Means are not.
We usez for proportions because the sampling distribution of sample proportions is approximately normal, but that is not the case for the sampling distribution of sample means.
The sampling distribution of sample means is at -distribution, not a z -distribution.
When using the sample standard deviation to estimate the population standard deviation, more variability is introduced into the sampling distribution of the statistic.
Which of the following is least likely to reduce bias in a sample survey?
Following up with those who did not respond to the survey the first time
Asking questions in a neutral manner to avoid influencing the responses
Using stratified random sampling rather than simple random sampling
Selecting samples randomly
Allowing responses to be anonymous
In a study to determine whether there is a difference in color preferences for young children based on their gender, researchers had a random sample of children choose from a selection of toys that were identical except for their color. The table below shows the number of each color that was chosen by boys and by girls.

How many degrees of freedom should be used in a test of independence of gender and color preference?

2
4
9
10
137
A local library has a scanner to detect library materials that have not been checked out. Each item has a chip somewhere inside. Upon checkout, the chip is deactivated so the scanner will not set off the alarm. The scanner has a 98% chance of detecting an active chip (meaning the material has not been checked out) and setting off the alarm. The scanner also has a 3% chance of sounding the alarm when someone passes through without an active chip. It is estimated that 0.5% of library customers actually try to leave the library with an active chip. What is the probability that, if the alarm sounds, the patron leaving the library has an item with an active chip?
0.0049
0.0348
0.1410
0.9700
0.9800
Which of the following is the best description of the power of a significance test?
The probability that the null hypothesis is true.
The probability of getting a Type I error.
The probability of getting a Type II error.
The probability of getting a test statistic at least as extreme as the actual test statistic, if the null hypothesis is true.
The probability of rejecting the null hypothesis if it is, in fact, false.
A school board of a large school district is proposing a new dress code for students. Some students feel that this dress code unfairly targets female students. To see if there is a difference between boys and girls in their opposition to the new dress code, they conduct a poll of 60 randomly selected male and 70 randomly selected female high school students in the district. They find that 66 females oppose the dress code and 50 males oppose the dress code. Which of the following explains why a two-proportion z -test is not appropriate?
The sample sizes are different.
The sample sizes are too large.
The number of successes and the number of failures for the two groups are not all large enough.
The shapes of the population distributions are not known.
The population standard deviations are not known.
Researchers are conducting an experiment using a significance level of 0.05. The null hypothesis is, in fact, false. If they modify their experiment to use twice as many experimental units for each treatment, which of the following would be true?
The probability of a Type I error and the probability of a Type II error would both decrease.
The probability of a Type I error and the power would both increase.
The probability of a Type II error and the power would both increase.
The probability of a Type I error would stay the same and the power would increase.
The probability of a Type II error would stay the same and the power would increase.
A producer of skin care products has created a new formula for its cream to cure acne. To compare the effectiveness of the new cream to that of the old cream, it conducted a double-blind randomized experiment. Volunteers with acne tried the old formula on one side of their face and the new formula on the other, and which side got which formula was determined randomly. The response variable was the difference in the number of pimples (old formula – new formula). Which is the correct significance test to perform?
A two-proportionz -test
A two-samplet -test
A matched pairst -test
A chi-square test of independence
A chi-square goodness of fit test
Smartphones with larger screens tend to be more expensive than smartphones with smaller screens. A random sample of currently available smartphones was selected. A 95% confidence interval for the slope of the regression line to predict price from screen size is (61, 542). Which of the following statements concerning this interval is correct?
If many samples of this size were taken, about 95% of confidence intervals for the slope based on those samples would contain the slope of the line for all smartphones that are currently available.
If many samples of this size were taken, about 95% of those samples would have a regression line with a slope between 61 and 542.
There is convincing evidence that the size of the screen is the reason for the difference in price.
Because this interval is so large, there is not convincing evidence of a relationship between screen size and price.
There is not enough information to make a statement about the relationship between screen size and price because memory capacity and other features must also be taken into account.
In the 50 states of the United States, the financial climate for people living in retirement varies. Some states are very tax friendly to retirees. Other states are much less tax friendly to retirees. As a member of the Oregon state legislature, you wonder if Oregon”s tax policies are driving older, retired people to move to other states. You have a cousin in Florida with the opposite idea—that the tax-friendly nature of Florida attracts more retirees. Each of you takes a random sample of the adults in your respective states and finds the statistics in the table below.

The 99% confidence interval for the difference in the mean age of adults in Oregon and adults in Florida is (–16.73, 8.69). Does this interval provide convincing evidence that the mean ages of adults in the two states differ?

Yes, because 0 is in the interval.
No, because 0 is in the interval.
Yes, because the mean ages are different in the two samples.
No, because the mean ages of the samples are less than one standard deviation apart.
Yes, because this is at -interval for a difference in means.