AP STAT

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A survey was conducted to determine what percentage of college seniors would have chosen to attend a different college if they had known then what they know now. In a random sample of 100 seniors, 34 percent indicated that they would have attended a different college. A 90 percent confidence interval for the percentage of all seniors who would have attended a different college is

(.2621, .4179).34 +/- 1.645(sqrt((.34*.66)/100))

COMPLETE these formulas for combining random variables by inserting the right hand side of the formula. E(X +/- c) = ?E(aX) = ?E(X + Y) = ?E(aX+bY) = ?Var(X+/-c) = ?Var(aX) = ?Var (X+/-Y) = ?

E(X +/- c) = E(X)+cE(aX) = aE(X)E(X + Y) = E(X)+E(Y)E(aX+bY) = aE(X)+bE(Y)Var(X+/-c) = Var(X)Var(aX) = a2 v(x)Var (X+/-Y) = Var(X)+/-Var(Y)

Suppose X and Y are random variables with E(X) = 25, var (X) = 3, E(Y) = 30 and var (Y) = 4. What are the expected value and standard deviation of the random variable X + Y?Explain by showing substitutions.

E(X+Y) = E(X)+E(Y) = 55 V(X+Y) = V(X)+V(Y) = = 7 standard deviation (X+Y) = sqrt 7 approx 2.646 The problem gave variance so there is NO NEED to square them.

A tire manufacturer claims that their tires will last 40,000 miles with a standard deviation of 5,000 miles.Assuming that the claim is true, describe the sampling distribution of the mean lifetime of a random sample of 160 tires.(Use notation to describe the model...and include mean and standard deviation.)

The model would be normally distributed and have a mean of 40,000 miles and a standard deviation of 395.28 miles.

A city has two hospitals, with many more births recorded at the larger hospital than at the smaller one. Records indicate that in general babies are about equally likely to be boys or girls, but the actual gender ratio varies from week to week. Which hospital is more likely to report a week when over two-thirds of the babies born were girls? Explain.

The smaller hospital is more likely to report a weekn when over two-thirds of the babies born are girls. This is due to the fact that the smaller hospital has a smaller sample size which will result in more variability than the larger hospital that will have a more consistent ratio.

Your company is contracted by a car dealership to do a statistical study. The dealership is interested in testing to see if its percentage of satisfied customers is higher than the industry standard of 67%. The dealership informs your staff that it can only afford a study that surveys a random sample of 100 of their customers.Your staff completes the survey of 100 customers and the results state that 70% of its customers are highly satisfied. What is the value of your test statistic? (Standardize phat using this sample and the original info given)

The value of test statistics is .655

The parameter of interest, choice of test, and hypotheses all depend on only the population information. Identify these components for the scenario described.A local newspaper published a report that on the last state math tests, the average performance for a random sample of 15 urban districts was a score of 64. You believe that the performance in these districts was higher. a. Parameter of interestb. Choice of Testc. Hypotheses

parameter: mean performance of urban districts on the last state math testTest: one sample right tail t test for mean Hypotheses: H0: mu = 64 average score was 64HA: mu > 64 average score was greater than 64

If a player has 12 at bats and probability of getting a hit of .25, how many times would you expect him to get a hit?

Assume normal distribution.The mean score on a college placement exam is 500 with a standard deviation of 100. Ninety five percent of the test takers score above what?

336

A radio station has a contest in which contestants roll a regular 6-sided die. If he rolls a 1 or a 2, he wins $50. If he rolls a 3 or a 4, he wins $100. If he rolls a 5, he wins $1000. If he rolls a 6, he doesn't win anything.What is the probability that out of the first 5 contestants exactly 2 win $100?

binompdf.329(5, 1/3, 2) =.329

EPA is examining the relationship between ozone level (in parts per million) and the population (in millions) of U.S. Cities.Dependent variable: OzoneR-squared = 84.4% s = 5.454 with 16 - 2 = 14 dfVariable Coefficient SE(Coeff)Constant 18.892 2.395Population 6.650 1.910Given that the test statistic is 3.4817, determine the P-value to test the null hypothesis that slope = 0. Decide if the value is STATISTICALLY SIGNIFICANT.

.001833...statistically significant

An investigator indicates that the POWER of his test (at a significance of 1%) of a sample mean resulting from his research is 0.87. What is the probability that he made a Type II error?

1-.87=.13. The investigator has a higher probability of having a Type 1 error than he does of having a Type 2 error.

The city council has 6 men and 3 women. If we randomly choose two of them to co-chair a committee, what is the probability these chairpersons are the same gender? Select the correct fractional response. HINT: Consider there is NO REPLACEMENT of an individual who is already selected.

1/2

The independent random variables X and Y are defined by the following probability distribution tables. X 1 3 6 Y 2 3 5 7 P(X) .6 .3 .1 P(Y) .1 .2 .3 .4 Determine the standard deviation of 3Y + 5.

5.44

In a test for acid rain, an SRS of 49 water samples showed a mean pH level of 4.4 with a standard deviation of 0.35. Find a 90% confidence interval estimate for the mean pH level.

90% confidence interval: (4.3 , 4.5)(or (4.31775,4.48225) exactly) 90% CI=mean+/-1.645(standard deviation/square root of the sample size)=4.4+/-1.645(.35/square root of 49)

Define Type II error.

A type 2 error refers to a false negative. It means that there was failure in rejecting a false null hypothesis.

Tests for adverse reactions to a new drug yielded the results given in the table. Drug Placebo headaches 11 7No headaches 73 91The data will be analyzed to determine if there is sufficient evidence to conclude that an association exists between the treatment (drug or placebo) and the reaction (whether or not headaches were experienced. The results of the test are Χ2 = 1.798 P-value = .1799.Identify the correct conclusion.

Fail to reject the null hypothesis. Report that there insufficient evidence to conclude that the distribution of headaches is uniform for the drug and placebo.

A teacher is wondering if 1st period students tend to do better on tests than 2nd period students. She takes a random sample of 5 1st period students whose scores were 98, 86, 75, 92, and 90. She takes a random sample of 3 2nd period students whose scores were 91, 89, and 87. Suppose that the original distributions of scores are normally distributed. Do these data give evidence that the 1st period students do better? Perform the appropriate hypothesis test. Show all work.

Graded Hypothesis: The null hypothesis is that the difference between the average scores of period 1 and period 2 are not differnet. The alternative hypothesis is that the average scores of period 1 and period 2 do not equal each other. Significance level: .05 The test of signifcance provided a p-value that is greater than that of the significance level. This mean that the null hypothesis is rejected. There is more evidence in favor of the alternative hypothesis. z=.37

Daniel believes that in a given conficence interval the researchers should change the confidence level from 98% to 99% so that they can get a more narrow margin with which to "pin down" the true population mean. You disagree with Daniel. What is the basis for your argument against Daniel's suggestion?

Increasing the confidence interval increases the margin of error also increases. You cannot increase the confidence interval because it will not change the margin of error because the population is still the same. The only way to narrow the margin of error is to take more sample.

An investigator indicates that the POWER of his test (at a significance of 1%) of a sample mean resulting from his research is 0.87. Interpret this statement in your own words.

It means that his investigation resulted in an 87% chance of detecting some effect (if there is one).

A example of a data set for a matched-pairs t-test might look like this: PretestPost-test831133722452528295031373549382542365269 ----------------------------------------------------------- Var.1: Mean = 29.5 Unbiased SD = 13.2Var. 2: Mean = 40.7 Unbiased SD = 13.0t-statistic = 2.69Degrees of freedom = 9Two-tailed probability = .025 State the hypotheses that you would use and then state the conclusion based on the P--value given.

Pretest scores will be higher than post-test scores. The conclusion rejects the hypotheses because the p-value is less than the significance level, meaning that the original hypothesis is rejected.

Which of the following is TRUE of chi-square distributions?

They take on only positive values.

The lengths of fish in a neighborhood pond are normally distributed with a mean of 5 inches and a standard deviation of 1.2 inches. What is the probability that a sample of 10 randomly selected fish will have a mean length of over 6 inches?

normalcdf(lb, ub,mean, sd) for sample size 10 the sd is 1.2/sqrt10 or .379 normalcdf(6, 10000000, 5, .379) = .00416

nference for regresssion on the population regression slope is based on which of the following distributions?

t-distribution (n - 2 degrees of freedom)

A recent survey concluded that the proportion of American teenagers who have a cell phone is 0.27. The true population proportion of American teenagers who have a cell phone is 0.29. For samples of size 1,000 that are selected at random from this population, what are the mean and standard deviation, respectively, for the sampling distribution of the sample proportion of American teenagers who have a cell phone.

D .29, square root of ((.29)(.71))/1000

A random sample of 100 traffic tickets given to motorists in a large city is examined. The tickets were classified according to the race of the driver. The results are summarized in the following table. White Black Hispanic Other Number of tickets 46 37 11 6 The proportion of the population of the city in each of the race categories above is the following. White Black Hispanic Other Proportion 0.65 0.30 0.03 0.02 We wish to test whether the racial distribution of traffic tickets in the city is the same as the racial distribution of the population of the city. To do so we use the X2 statistic.The component of this X2 statistic corresponding to the Hispanic category is ______________.

(O - E)2/E = 21.33.

Police report traffic accidents they investigated last year indicated 40% of the accidents involved speeding, 25% involved alcohol, and 10% involved both risk factors.What is the probability that an accident involved neither alcohol nor speed?HINT: draw a well-labeled venn diagram

.45

0.64 ± 0.094.

The distribution of actual weights of 8-ounce chocolate bars produced by a certain machine is normal with mean 8.1 ounces and standard deviation 0.1 ounces. If a sample of five of these chocolate bars is selected, there is only a 5% chance that the average weight of the sample of five of the chocolate bars will be below _______.(Use the formula backwards to solve for X...do NOT round until the very last step.) Select the closest answer.

7.95 ounces.

The sampling distribution of a statistic is A) the probability that we obtain the statistic in repeated random samples. B) the mechanism that determines whether randomization was effective. C) the distribution of values taken by a statistic in all possible samples of the same size from the same population. D) the extent to which the sample results differ systematically from the truth.

If P(A) = 0.2 and P(B) = 0.3, find P(A or B) if it is known that events A and B are not disjoint.

Incorrect answer: P(A or B)= 0.44

A recent study investigated the amount of drink that was put into the 2 liter bottles for Yummy Orange Soda. The results were statistically significant and found that the bottles were being over filled. Briefly explain the meaning of 'statistically significant' in this context.

It means that the null hypothesis was true. In the case, it means that there was no difference between the predicted amount (being over filled) and the actual amount.

The confidence interval created for the slope resulting from a linear regression analysis is (.079, .173).What is the slope of the LSRL?

Find the midpoint of the given interval ie.,average the endpoints

H0: There is no relation between candy choice and the age of a student. Ha: There is a relation between candy chocie and the age of a student.

A survey of local car dealers revealed that 64% of all cars sold last month had CD players, 28% had alarm systems, and 22% had both CD players and alarm systems.Are having a CD player and an alarm system disjoint events? Explain.

Incorrect answer: No they are not disjointed events. This is because they can both occur at the same time, if they were disjointed then the probability of having both should be 0.

A state's Department of Education reports that 12% of the high school students in that state attend private high schools. The State University wonders if the percentage is the same in their applicant pool. Admissions officers plan to check a random sample of the over 10,000 applications on file to estimate the percentage of students applying for admission who attend private schools.They actually select a random sample of 450 applications, and find that 46 of those students attend private schools. Create a 90% confidence interval.Should the admissions officers conclude that the percentage of private school students in their applicant pool is lower than the statewide enrollment rate of 12%? Explain.

No they should, this is due to the fact that 12% is within the interval of 90% confidence.

Normally distributed normalcdf(lb, ub,mean, sd) for sample size 10 the sd is 1.2/sqrt10 or .379 normalcdf(6, 10000000, 5, .379) = .00416

A example of a data set for a matched-pairs t-test might look like this: PretestPost-test831133722452528295031373549382542365269 Write the hypotheses, find the P-value, and state the appropriate conclusion.

Null hypothesis: pretest score averages are equal to post-test score averages. Alternative hypothesis: the difference between pretest score averages and post-test score averages is greater than 0. p-value: .02807 Conclusion: The p value is less than the signifance level (.05) which means that the null hypothesis is rejected. This mean that there is more evidence towards its alternative hypothesis.

TRUE-FALSEIn normal distributions, the mean and median are equal.

True

Suppose the probability of a particular baseball player hitting a homerun is 0.19, based on the player's prior history. Assume the probability of the player hitting a homerun is the same for each at bat and assume that the player has 50 at bats during the season.Suppose we are interested in the number of at bats before the player gets a Homerun. Is this situation binomial or geometric?

by definition...awaiting first success...waiting problem

Tests for adverse reactions to a new drug yielded the results given in the table. The data will be analyzed to determine if there is sufficient evidence to conclude that an association exists between the treatment (drug or placebo) and the reaction (whether or not headaches were experienced.Drug Placeboheadaches 11 7No headaches 73 91 Which chi-square test would be appropriate for this situation?

chi-square for independence

An investigator indicates that the power of his test (at a significance of 1%) of a sample mean resulting from his research is 0.87. If n increases, then the power of the test...

increases.

A probability experiment involves a series of identical, independent trials with two outcomes (success/failure) per trial and the probability of a success on each trial is 0.1. Determine the number of trials, n, in a binomial experiment such that the expected number of successes in that binomial experiment will be equal to the EXPECTED number of trials in a geometric experiment.HINT: manipulate formulas

letting the two formulas for expected values for binom and geom equal each other....we getnp = 1/p then we substitute the .1 in for the p and solve for n so 100 (???)

Given the random variable described as A with mean = 80 and standard deviation = 12 , find the mean and standard deviation of the random variable 3X.

mean 3X = 240standard deviation 3X = 36

A consumer organization inspecting new cars found that many had appearance defects (dents, scratches, paint chips, etc.). While none had more than three of these defects, 7% had three, 11% had two, and 21% had one defect. Find the expected number of appearance defects in a new car, and the standard deviation.

mean = .64standard deviation = .93

The null hypothesis for a linear regression significance test for slope is always ________________.(state you answer in both symbols and words)

0 (zero)

The average outstanding bill for delinquent customer accounts for a national department store chain is $187.50 with a standard deviation of $54.50. In a simple random sample of 50 delinquent accounts, what is the probability that the mean outstanding bill is over $200?

0.0524

Suppose that 35% of all business executives are willing to switch companies if offered a higher salary. If a headhunter randomly contacts an SRS of 100 executives, what is the probability that over 40% will be willing to switch companies if offered a higher salary? Select the closest answer.

0.1470

Family size can be represented by the random variable X. Determine the average family size. X 2 3 4 5 P(X) .17 .47 .26 .10

3.29

No they are not disjointed events. This is because there is a probabality of .22 of having both CD player and an alarm sustem. If they were disjointed there would be no probability of both exisiting in a car.

Suppose X and Y are random variables with E(X) =10, std dev (X) = 3, E(Y) =15, and std dev (Y)= 4. Given that X and Y are independent, what are the mean and standard deviation of the random variable X - Y?

mean: -5 ; standard deviation: 5

The amount of cereal that can be poured into a small bowl varies with a mean of 1.5 ounces and a standard deviation of .3 ounces. A large bowl holds a mean of 2.5 ounces with a standard deviation of .4 ounces. You open a new box of cereal and pour one large and one small bowl.What is the standard deviation of this difference?

Assume that a school district has 10,000 sixth graders. In this district, the average weight of a sixth grader is 80 pounds, with a standard deviation of 20 pounds. Suppose you draw a random sample of 50 students. What is the probability that the average weight of a sampled student will be less than 75 pounds? Give an approximate answer.

3.9%

Forty-five percent of the WA upper school student body are male. 80% of the females love math, while only 60% of the males love math. What percentage of the TOTAL student body love math? (Stated another way...what is the probability that a randomly selected student will love math?)

71%

Which is true about a 99% confidence interval based on a given sample? READ CAREFULLYI. The interval contains 99% of the population.II. Results from 99% of all samples will lie in this interval.III. The interval is wider than a 95% confidence interval would be.

II and III only

In general, how does doubling the sample size change the confidence interval size?

Divides the interval size by the square root of 2.

A researcher found that day care providers are more likely to get the flu than people who are not typically around children with a p-value of .004. Briefly explain what this p-value means.

The p-value means that there null hypothesis was true, because it has greater statistical significance.

From time to time police set up roadblocks to check cars to see if the safety inspection is up to date. At one such roadblock they issued tickets for expired inspection stickers to 22 of 628 cars they stopped.Based on the results at this roadblock, construct and INTERPRET a 95% confidence interval for the proportion of autos in that region whose safety inspections have expired.

We have a confidence of 95% that between 2.1% and 4.9% of cars in the region have an expired safety inspection.

A Chi-square distribution has 11 degrees of freedom. Find the χ2 value corresponding to a right-hand tail area of .01

24.725

Determine the probability distribution's missing value. The probability that a tutor sees 0, 1, 2, 3,or 4 students on a given day. x 0 1 2 3 4 P(x) 5/19 6/19 4/19 _____ 1/19

3/19

.08

Chattahoochee Battery Company has developed a new laptop computer battery. On average, the battery lasts 240 minutes on a single charge. The standard deviation is 16 minutes. Suppose the company randomly samples production during each day at a specified time and selects 12 laptop batteries for testing. The standard deviation of the battery life for the selected batteries is 20 minutes. What is the chi-square statistic represented by this test?Χ2 = [ ( n - 1 ) * s2 ] / σ2

17.188

The P-value of a test of significance is the probability that:

???

A reporter believed that police officers were required to write a specific quota of traffic tickets during a month. In order to meet the alleged quota, he believed officers would need to write more tickets during the last week of the month. To investigate the CLAIM, the reporter collected the number of tickets written by the local police force in a month and organized them by weeks as show in the table below.State the null and alternate hypotheses. Week # 1st 2nd 3rd 4th Total # tickets 133 112 154 165 564 Expected # tickets 141 141 141 141 564

H0: Police officers do not need to write more tickets during the last week of the month in order to meet the alleged quota. Ha: Police officers do need to write more tickets during the last week of the month in order to meet their alleged quota.

Your company is contracted by a car dealership to do a statistical study. The dealership is interested in testing to see if its percentage of satisfied customers is higher than the industry standard of 67%. The dealership informs your staff that it can only afford a study that surveys a random sample of 100 of their customers.Give the appropriate hypotheses for a test of this information.

HO: p = .67, Ha: p > .67

In the first eight games of this season, LeRoy, a starting player for a major college basketball team, made 25 free throws in 40 attempts. You want to investigate how successful his team should expect him to be at the free throw line this season.Construct a 90% confidence interval for the proportion of free throws LeRoy will make this season.Interpret this confidence interval for Leroy's coaches and teammates.

The confidence interval indicates that his coach and teammates should except him to be relatively successful since his rate of free throws is greater than 50%.

Binomial and geometric probability situations share all of the following conditions except one. Identify the choice that is not shared.Read the answer choices carefully.

The focus of the problem is the number of successes in a given number of trials. (??)

Which of the following is not true concerning discrete probability distribution?

The standard deviation of the distribution is between -1 and 1.

If the P-value of a test is less than the level of significance, then which of the following is a correct conclusion?

The value of the test statistic is in the rejection region for this test.

A company supplying candy to school vending machines needs to know about candy preferences of middle school and high school age students. A random sample of 400 students was taken. The company wants to know if there is any relationship between the candy choices and the age of students.After performing a chi-square test of independence, the chi-square test statistic = 24.68. The appropriate critical value for df and desired alpha level = 11.34.Is there sufficient evidence to reject the null hypothesis? Explain.

There is sufficient evidence to reject the null hypothesis. This is due to the fact that ther test statistic falls to the right of the critical value, meanign that we reject the null hypothesis.

Police report that traffic accidents they investigated last year indicated 40% of the accidents involved speeding, 25% involved alcohol, and 10% involved both risk factors.Do these two risk factors appear to be independent?Use the numbers given in the problem to justify your answer.

They are dependent.

When a police officer responds to a call for help in a case of spousal abuse, what should the officer do? A randomized controlled experiment in Charlotte, North Carolina, studied three police responses to spousal abuse: advise and possibly separate the couple, issue a citation to the offender, and arrest the offender. The effectiveness of the three responses was determined by re-arrest rates. The table below shows these rates. Assigned Treatment # of Re-arrests Arrest Citation Advise/Separate 0 175 181 187 1 36 33 24 2 2 7 1 3 1 1 0 4 0 2 0 Suppose we wish to test the null hypothesis that the proportion of subsequent arrests is the same regardless of the treatment assigned. Which of the following statements is true?

We cannot test this hypothesis because the expected cell counts are less than five in too many of the cells.

A random variable X has mean mX and standard deviation sX. Suppose n independent observations of X are taken and the average of these n observations is computed. We can assert that if n is very large, the sampling distribution of is approximately normal. This assertion follows from......

the central limit theorem.

Suppose the Normal model describes the length of the fish in a neighborhood pond. One particular fish has a z-score of 3.1. This means that ...

the length of the fish was 3.1 standard deviations longer than the average fish in the pond.

It is generally believed that nearsightedness affects about 12% of children. A school district gives vision tests to 133 incoming kindergarten children.Describe the sampling distribution model for the sample proportion by identifying its shape and telling its mean and standard deviation.Using your result find the the probability that in this group over 15% of the children will be found to be nearsighted?

Incorrect answer: This sample has a standard deviation of .028 which is calculated by taking the sqaure root of ((.12)(.88))/(133). The mean is 15.96 of the students will be near sighted. The probability of this group having over 15% of the children being nearsighted is .142 ir 14.2%. This is calculated by first figuring out the Z score; ((.15-.12)/.028) which is 1.07. The probability of this z-score is .8577, but this is only for up to 15%, in order to get above 15% of children you need to subtract the value from 1, 1-.8577=.1423.

An investigator indicates that the POWER of his test (at a significance of 1%) of a sample mean resulting from his research is 0.87. What is the probability that he made a Type I error?

There is a 1% chance that the investigator made a type 1 error. This means that the possibility of getting a false positive (detecting something when there really was not) is 1 in a 100.

A survey of some AP Stats students recorded gender and whether the student was left or right-handed. Results were summarized in a table like the one shown. If it turned out that handedness was independent of gender, how may of the AP Stats students were lefty girls? LeftyRightyTotalBoy??66Girl??54Total20100120

set up a proportion x/66 = (20-x)/54 and solve for x (LEFTY BOYS) x = 11question asked for LEFT GIRLS so 20 - x = 9

The scores of individual students on the American College Testing (ACT) Program composite college entrance examination have a normal distribution with mean 18.6 and standard deviation 6.0. At Peppermill High School, 36 seniors take the test. If the scores at this school have the same distribution as national scores, what is the standard deviation of the sampling distribution of the average (sample mean) score for the 36 students?

1.0

Buffalo Battery Company has developed a new laptop computer battery. On average, the battery lasts 240 minutes on a single charge. The standard deviation is 16 minutes. Suppose the company randomly samples production during each day at a specified time and selects 8 laptop batteries for testing. The standard deviation of the battery life for the selected batteries is 20 minutes. What is the chi-square statistic represented by this test?Χ2 = [ ( n - 1 ) * s2 ] / σ2

10.938

The independent random variables X and Y are defined by the following probability distribution tables. X 1 3 6 Y 2 3 5 7 P(X) .6 .3 .1 P(Y) .1 .2 .3 .4 Determine the mean of X+Y.

7.2

The renowned soccer player, Levi Gupta scores a goal on 30% of his attempts. The random variable X is defined as the number of goals scored on 50 attempts. The renowned gambler, Mohammed Smith, wins at Blackjack 25% of the time. The random variable Y is defined as the number of games needed to win his first game.Define the random variable Z as the total number of soccer goals scored and blackjack games played. Determine the mean and standard deviation of the random variable Z.SHOW ALL WORK...will be graded using AP Scoring guidelines

ANSWER: Mean(Z)= 19; SD(Z)=4.74 (or square root of 22.5) Work: P(X)=.3 ; n(X)=50P(Y)=.25Z=X+Y; therefore Mean(Z)= (Mean(X))+(Mean(Y)) and SD(Z)=(SD(X))+(SD(Y)) mean and SD of X:Mean(X)=50*.30=15SD(X)=square root of (n(x)(1-P(X))=square root of (50*.3(1-.3))=square root of (50*.3*.7)= square root of 10.5. mean and SD of Y: Mean(Y)= 1/P(Y)=1/.25=4SD(Y)= square root of ((1-P(Y))/P(Y) squared)= square root of ((1-.25)/(.25)squared)=square root of ((.75)/.0625)= square root of 12. mean and SD of Z: Mean(Z)=Mean(X)+Mean(Y)=15+4=19.SD(Z)=SD(X)+SD(Y)=(square root of 10.5)+(square root of 12)= square root of 22.5 or 4.74

A grass seed company conducts a study to determine the relationship between the density of seeds planted (in pounds per 5000 sq ft) and the quality of the resulting lawn. Eight similar plots of land are selected and each is planted with a particular density of seed. One month later the quality of each lawn is rated on a scale of 1 to 100. The regression analysis is given below.Create a 95% confidence interval to determine if there is a relationship between seed density and lawn quality. Show all of your work...and explain our reasoning.Dependent variable : Lawn qualityR-square = 36.0%s = 9.073602 with 8 - 2 = 6 degrees of freedomVariable Coefficient SE(Coeff) t-ratio P-valueConstant 33.14815 7.510757 4.413423 .004503Seed Density 4.537037 2.469522 1.837213 .115825

CI = (-1.51, 10.58)from4.537037 +/- t*(df=6) (2.469522)4.537037 +/- (2.447)(2.469522)(-1.51, 10.58)zero is part of interval so it is possible that there is NO relationship between the two variables..seed density and lawn quality

Incorrect answer: If you were to take a sample size then you would have confidence that 95% of the results would fall within the expected parameters.

All airline passengers must pass through security screenings, but some are subjected to additional searches as well. Some travelers who carry laptops wonder if that makes them more likely to be searched.Data for 420 passengers aboard a cross-country flight are summarized in the table shown. Does it appear that being subjected to an additional search is independent of carrying a laptop computer? EXPLAIN. SearchedNot SearchedTotalLaptop 30 42 72No Laptop 145 203 348Total 175 245 420

It does appear that being searched and carrying a laptop are independent events. This is because the conditional probability of being searched while having a laptop is equal to the single event probability of carrying a laptop.

A grass seed company conducts a study to determine the relationship between the density of seeds planted (in pounds per 5000 sq ft) and the quality of the resulting lawn. Eight similar plots of land are selected and each is planted with a particular density of seed. One month later the quality of each lawn is rated on a scale of 1 to 100. The regression analysis is given below. Is there EVIDENCE of an association between seed density and lawn quality? Explain your thinkingDependent variable : Lawn qualityR-square = 36.0%s = 9.073602 with 8 - 2 = 6 degrees of freedomVariable Coefficient SE(Coeff) t-ratio P-valueConstant 33.14815 7.510757 4.413423 .004503Seed Density 4.537037 2.469522 1.837213 .115825

There is not enough evidence in order to reject the null hypothesis (that the population regression line is 0). This is due to the fact that the p-value (.115825) is greater than .05, which means we fail to reject the null hypothesis.

TRUE-FALSEThe area under the normal curve is always equal to 1 no matter what the mean and standard deviation are.

True

We will test the hypothesis that p = 60% versus p > 60%. We don't know it, but actually p is 70%. With which sample size and significance level will our test have the greatest power?

a = 0.05, n = 500

A reporter believed that police officers were required to write a specific quota of traffic tickets during a month. In order to meet the alleged quota, he believed officers would need to write more tickets during the last week of the month. To investigate the CLAIM, the reporter collected the number of tickets written by the local police force in a month and organized them by weeks as show in the table below. Week # 1st 2nd 3rd 4th Total # tickets 133 112 154 165 564 Expected # tickets 141 141 141 141 564 How many degrees of freedom should be used for this chi-square test of independence?

df = 3

Which two events are most likely to be independent?

having a driver's license, and having blue eyes

The parameter of interest (mean or proportion), choice of test, and hypotheses all depend on only the population information. Identify these components for the scenario described. BE SPECIFIC.A bookbag company published a report that in a recent comparison test 55% of middle school children preferred their bookbag to their closest competitor's. You want to determine if this is true.a. Parameter of interestb. Choice of Testc. Hypotheses

parameter: proportion of middle school children who prefer this particular bookbagTest: two tail one proportion z test Hypotheses:H0: p = .55 % of middle school children preferring certain bookbag is 55%HA: p not = .55 % of middle school children preferring certain bookbag is not 55%

A newspaper article reported that a poll based on a sample of 800 voters showed the President's job approval rating stood at 62%. They claimed a margin of error of \pm±3%. What level of confidence were the pollsters

.03 = Z*(sqrt. (.62x.38)/(800)) .03/.017=Z* 1.75=Z* normalcdf(-100000,1.75)=.96 .96-.04 Confidence level of 92%

.03125

A survey of local car dealers revealed that 64% of all cars sold last month had CD players, 28% had alarm systems, and 22% had both CD players and alarm systems.What is the probability that a car had a CD player unprotected by an alarm system?Express your answer as a decimal.

.42

If performance on AP Statistics tests are independent and the probability of passing an AP Statistics test is 0.2, then the probability of passing three AP Statistics tests is:

0.008

A trucking firm determines that its fleet of trucks averages a mean of 12.4 miles per gallon with a standard deviation of 1.2 miles per gallon on cross country hauls and is approximately normally distributed. What is the probaility that one of the trucks averages fewer than 10 miles per gallon?

0.0228

The yearly mortality rate for American men from prostate cancer has been constant for decades at about 25 of every 100,000 men. (This rate has not changed in spite of new diagnostic techniques and new treatments.) In a group of 100 American men, what is the probability that at least 1 will die from prostate cancer?

0.0247

0.1472

GREAT QUESTIONWe have calculated a confidence interval based on a sample of n = 180. Now we want to get a better estimate with a margin of error only one third as large. We need a new sample with n at least ...

1620

Suppose you administer a certain aptitude test to a random sample of 9 students in your school and that the average score is 105. We want to try to determine the mean of the population of all students in the school. Assume a population standard deviation of 15 for the test. What is the z* for a 98% confidence interval?

2.33

The number of accidents per day at a large factory is noted for each of 64 days with a mean of 3.58 and a standard deviation of 1.52. With what degree of confidence can we assert that the mean number of accidents per day at the factory is between 3.20 and 3.96?

95%

A marketing company claims that 75% of adults prefer a certain toothpaste over all other competitors. A simple random sample of 50 adults indicated only 66% preferred that brand of toothpaste. Do these data indicate that the claim of the marketing company is too high at the 5% level? For full credit, you must show the essential pieces of your test:Assumptions, Hypotheses, Test Statistic, P-value, and make a conclusion in context of the problem.

Assumptions: That the data has a normal distribution and that it accurately represents the population. Hypothesis: Null hypothesis is that is that the p^hat is equal to .75. The alternative hypothesis is that it is less than .75. Test Statistic: .66-.75/sqrt((.75)(.25)/50)=-.09/sqrt(.00375)=-1.469 which is a probability of .0708. P-value:.0708 Conclusion: The p-value is greater the the alpha level which means that you fail to reject the null hypothesis. This means that there is not sufficient evidence to reach the conclusion that the claim of the marketing company is too high

Which has the larger mean and which has the larger standard deviation?

Based on answering incorrectly The mean of a graph is located along the horizontal axis...the higher the number on that number line...the larger the mean....we all know about spread as the standard deviation.graph b has BOTH the larger mean AND the larger standard deviation General Feedback partial credit:The mean of a graph is located along the horizontal axis...the higher the number on that number line...the larger the mean....we all know about spread as the standard deviation.graph b has BOTH the larger mean AND the larger standard deviation

An ice cream stand reports that 12% of the cones they sell are "jumbo" size. You want to see what a "jumbo" cone looks like, so you stand and watch the sales for a while. What is the probabilty that the first jumbo cone is the fourth cone you see them sell?Select the correct answer rounded to the nearest whole %.

First 4 cones are NOT jumbo but the fourth one is..(.88)^3 (.12) = .0817 (8%)

For purposes of making on-campus housing assignments, a college classifies its students as Priority A (seniors), Priority B (juniors), and Priority C (freshmen and sophomores). Of the students who choose to live on campus, 10% are seniors, 20% are juniors, and the rest are underclassmen. The most desirable dorm is the newly constructed Gold dorm, and 60% of the seniors elect to live there. 15% of the juniors also live there, along with only 5% of the freshmen and sophomores.What is the probability that a randomly selected resident of the Gold dorm is a senior?Explain your procedure CLEARLY and make sure to state your answer in a complete sentence.(Make sure you read this carefully enough to truly understand what is being asked.)

Graded The probability of randomly selected resident of the Gold dorm being a senior is .48. I calculated this number by adding the percentages of the Gold dorm residents, (.6 seniors + .15 juniors + .05 underclassmen = .8 ) I then multiplied the probability of being a senior (.6 ) by .8 and got .48 as the probability of a randomly selected resident of the Gold dorm being a senior.

For full credit, you must show the essential pieces of your test:Hypotheses, Test Statistic, P-value, and make a conclusion in context of the problem.An association of college bookstores reported that the average amount of money spent by students on textbooks for the Fall 1999 semester was $325.16 with a standard deviation of $76.42. A random sample of 75 students at the local campus of the state university indicated an average bill for textbooks for the semester in question to be $312.34. Do these data provide significant evidence at the 5% level that the actual average bill is different from the $325.16 that was reported?

H0: mu = 325.16 avg money spent on textbooks was $325.16HA: mu not = 325.16 avg money spent on textbooks is different from 325.16one sample two tail t testt = -1.45 with p-value = .1505NOT statistically significantFail to reject H0not enough evidence to conclude that the avg $ spent on textbooks is different from 325.16

A total of 23 Gossett High School students were admitted to State University. Of those students, 7 were offered athletic scholarships. The school's guidance conselor looked at their componsite ACT scores (shown in the table), wondering if State U. might admit people with lower scores if they also were athletes. Assuming that this group of students is representative of students throughout the state, what do you think?Composite ACT ScoresNon-athletes: 25, 22, 19, 25, 24, 25, 24, 23, 21, 27, 29, 26, 30, 27, 26, 23Athletes: 22, 21, 24, 27, 19, 23, 17Create and interpret a 90% confidence interval. (For the difference in the mean ACT scores - athletes vs non-athletes. Use technology to assist.

If we compared the results on a pretend number line we would observe overlap, but the athletes have an interval that is below both of the other groups...you can also use the calculator and do a 2 sample t interval (meaning on the DIFFERE NCE) and get (.302, 5.48) showing exactly how different those ACT averages are.

A state's Department of Education reports that 12% of the high school students in that state attend private high schools. The State University wonders if the percentage is the same in their applicant pool. Admissions officers plan to check a random sample of the over 10,000 applications on file to estimate the percentage of students applying for admission who attend private schools.They actually select a random sample of 450 applications, and find that 46 of those students attend private schools. Create a 90% confidence interval.Explain what "90% confidence" means in this context.

If you took a sample, you would have a 95% confidence that between 7% and 14% of students attend private schools.

In order to have a margin of error of 4 points, then the sample size needs to be 77 students. This is calculated from first getting the z score which is 2.33, then multiplying it by the standard deviation (15) divided by the square root of the sample size. Once you rearrange the equation the square root of the sample size is about 8.7375, once this is squared you get a sample size of 76.3439 which you have to round up to 77.

Suppose the probability of a particular baseball player hitting a homerun is 0.19, based on the player's prior history. Assume the probability of the player hitting a homerun is the same for each at bat and assume that the player has 50 at bats during the season.Suppose we are interested in how many Homeruns this batter gets in a game. Is this situation binomial or geometric?

Incorrect answer: The situation is binomal, this is because each observaation is independent of each other. There is also a fixed number of observations.

Incorrect answer: We can calculate this based on the assumption that the scores are normally distributed. This leads to calculating the confidence interval with the x amount being 105, the z being +/- 2.33, a standard deviation of 15 and a sample size of 9. When this is calculated you get (93.35,116.65). This means that we are 98% confident that the scores fall between 93.35 and 116.65.

In a board game you determine the number of spaces you may move by spinning a spinner AND rolling a die. Each has its own oddities. The spinner has three regions: Half of the spinner is marked "5", and the other half is equally divided between "10" and "20." (Perhaps a sketch would be helpful) The six faces of the die show 0, 0, 1, 2, 3, and 4. When it is your turn, you spin AND roll, adding the numbers together to detmine how far you may move. The question is...HOW MANY SPACES COULD YOU EXPECT TO MOVE ON ANY GIVEN ROLL.Create a probablity model for each of the variables, the spinner variable and the die variable.Find the expected values for each and then combine them using the rules for combining random variables. Show the values that would be present in the tables and explain exactly how you calculated your final answer.

Incorrect answer: You could expect to move 11 2/3 spaces on any given roll. The table for the spinner would look like this. Number on spinner: 5,10,20 and probability: 1/2, 1/4, 1/4 (each respectively). The mean for the spinner is calculated by multiplying each number value by its probability and then adding them together, which equals 10 in this case. The variance for the spinner is found by square-rooting the addition of the squares of each number on the spinner by its probability. Now the probability model for the die looks as follows: number rolled: 0,1,2,3,4 and probability: 1/3,1/6,1/6,1/6,1/6,1/6 (each respectively). The mean is found by adding the product of each number rolled by its probability, which totals 5/3. The variance is then found by square rooting the sum of all the numbers rolled multiplied by their probability and squared, which equals 1.49. The final probability is calculated by adding the means of both the die and the spinner (10+5/3) which equals 11 2/3 spaces.

Increasing the confidence interval does not make it easier to pin down the true population mean but rather it increases room for error. When you decrease your confidence interval, you are decreasing the room for error. If you do not have the correct parameter of interest then it will only cause greater error.

Suppose the average height of a policeman is 71 inches with a standard deviation of 4 inches, while the average for a policewoman is 66 inches with a standard deviation of 3 inches. If a committee looks at all ways of pairing up one male with one female officer, what will be the mean and standard deviation for the difference in heights for the set of possible partners? You will actually be combining random variables...so be careful when determining the standard deviation.

Mean of 5 inches with a standard deviation of 5 inches.

Which of the following lead to binomial distributions?I. An inspection procedure at an automobile manufacturing plant involves selecting a sample of cars from the assembly line and noting for each car whether there are no defects, at least one major defect, or only minor defects.II. As students study more and more during their AP Statistics class, their chances of getting an A on any given test continue to improve. The teacher is interested in the probability of any given student receiving various numbers of A's on the class exams.III. A committee of two is to be selected from among the five teachers and ten students attending a meeting. What are the probabilities that the committee will consist of two teachers, of two students, or of exactly one teacher and one student?

None of the above gives the complete set of true responses.

A survey of local car dealers revealed that 64% of all cars sold last month had CD players, 28% had alarm systems, and 22% had both CD players and alarm systems.What is the probability one of these cars selected at random had neither a CD player nor an alarm system?Express your answer as a decimal.

P(neither CD nor alarm) = 1 - P(CD OR alarm) = 1 - .7 = .3

A sporting goods store announces a "Wheel of Savings" sale. Customers select the merchandise they want to purchase, then at the cash register they spin a wheel to determine the size of the discount they will receive. The wheel is divided into 12 regions, like a clock. Six of those regions are red, and award a 10% discount. The three white regions award a 20% discount and two blue regions a 40% discount. The remaining region is gold, and a customer whose lucky spin lands there gets a 100% discount - the merchandise is free!What is the probability that there is at least one gold winner among the first six customers?Express your answer rounded to 3 decimal places.

P(not none) = 1 - P(none) = 1 - (11/12)^6 =.4067

After once again losing a football game to the college's arch rival, the alumni association conducted a survey to see if alumni were in favor of firing the coach. An SRS of 100 alumni from the population of all living alumni was taken. Sixty-four of the alumni in the sample were in favor of firing the coach. Let p represent the proportion of all living alumni who favor firing the coach.Suppose you wish to see if the majority of alumni are in favor of firing the coach.To do this you test the hypothesesH0: p = 0.50, Ha: p > 0.50The P-value of your test is ________Make a statement about the p-value obtained and form a conclusion regarding the coach's firing.

T p-value is .00175. The p-value is less than the null hypothesis which can help us conclude that the null hypothesis is rejected. This means there is a greater favoring for the alternative hypothesis.

Helen scores 680 on the math part of the SAT. The distribution of SAT scores in a reference population is normally distributed with mean 500 and standard deviation 100. Juan takes the ACT math test and scores 27. ACT scores are normally distributed with mean 18 and standard deviation 6.Clearly explain the procedure for deciding which student performed better on his/her respective test. Include all formulas and substitutions and state your conclusion in the context of the problem.

The best way to create a basis of comparison for Helen and Juan is to first standardize the scores by converting them into z scores. Z score is equal to x minus mean divided by standard deviation, in which x is equal to the score they each got. Helen's Z score is equal to 680 minus 500 divided by 100 which is equal to 1.8. Juan's Z score is equal to 27 minus 18 divided by 6 which equals 1.5. Helen performed better on her respective test because her Z score was greater than Juan's.

TRUE-FALSEThe area under the standard normal curve between 0 and 2 is half the area between -2 and 2.

True

Textbook authors must be careful that the reading level of their book is appropriate for the target audience. Some methods of assessing reading level require estimating the average word length. We've randomlly chosen 20 words from a randomly selected page in Stats: Modeling the World and counted the number of letters in each word:5, 5, 2, 11, 1, 5, 3, 8, 5, 4, 7, 2, 9, 4, 8, 10, 4, 5, 6, 6For a more definitive evaluation of reading level the editor wants to estimate the text's mean word length to within 0.5 letters with 98% confidence. How many randomly selected words does she need to use? (Since we are missing "n" for this problem and we are dealing with a t-distribution we must use undefined for our df.)Explain a little about how your calculations will look.

Using the undefined t* value is our only option since the n (and therefore the df) is unknown. .5 = (2.326)(2.685)/sqrtn or sqrt n = (2.326)(2.685)/.5 or 12.49...however this is sqrt n and we need n so approx 156 would be the best answer.

Rebecca's doctor is concerned that she may suffer from gestational diabetes (high blood glucose levels associated with pregnancy). There is variation both in the actual glucose level and in the blood test that measures the level. A patient is diagnosed with gestational diabetes if their glucose level is above 140 milligrams per deciliter (mg/dl) one hour following a sugary drink being ingested. Rebecca's measured glucose level one hour after ingesting the sugary drink varies according to the normal distribution with mu = 125 mg/dl and standard deviation = 10 mg/dl.Answer the following questions. Show all work. State reasons for your procedure.a) If a single glucose measurement is made, what is the probability that Rebecca is diagnosed with gestational diabetes?b) If measurements are made instead on four separate days and the mean result is computed, describe the center, shape and spread of the resulting sampling distribution. Simply use correct notation and explain your reasons.c) If the sample mean of Rebecca's four readings is compared with the criterion 140 mg/dl, what is the probability that Rebecca is diagnosed as having gestational diabetes?

a) There is a possibility of .0668 that Rebecca is diagnosed with gestational diabetes.First you have to get the z score which is 140-125 divided by 10 which equals 1.5. Then you have to refer to a z-score possibility table to get the probability which is equal to .9332. We want the possibility of her glucose levels to be above 140 and not up until 140 so this probabililty needs to be subtracted from 1 to get .0668. b) There is a probability of .0013 that Rebecca is diagnosed with gestational diabetes. The distribution will be normal.The probabillity is calculated the same as part a, except that the standard deviation needs to be divided by the square root of 4 in order to adjust to the sample size. c) Over a 7 days sample period, there is a probability of .00003 that Rebecca will be diagnosed with gestational diabetes. First you have to have the z score accomodate for the sample size so the standard deviation needs to be divided by the square root of the sample size (7) which is equal to about 3.78. Now the z-score can be found by dividing 140-125 by 3.78 to get the new z-score of 3.97. This has a probability of .99997. This then has to be subtracted from 1 meaning that there is a probability of .00003 that Rebecca will be diagnosed with gestational diabetes.

In a random sample of 300 elderly men, 65% were married, while in a similar sample of 400 elderly women, 48% were married. Determine a 99% confidence interval estimate for the DIFFERENCE between the percentages of elderly men and women who were married.

phat for difference = .65 -.48 or .17z* for 99% CI = 2.575SE for difference =sqrt (.0014) or .037 .17 +/- 2.575(.037) = (.075, .265) or 7.5% and 26.5% We are 99% confident that the true difference in proportion of married elderly men and women is between 7.5% and 26.5%. This large interval is partially due to the 99% CI level.Anything close will be considered correct. .17 +/- 2.575(.037) = (.075, .265) or 7.5% and 26.5% We are 99% confident that the true difference in proportion of married elderly men and women is between 7.5% and 26.5%. This large interval is partially due to the 99% CI level.

In a hypothesis test, the decision between a one sided test and a two sided alternative hypothesis is based on:

the alternative hypothesis appropriate for the context of the problem.

If a statistic used to estimate a parameter is such that the mean of its sampling distribution is equal to the true value of the parameter being estimated, the statistic is said to be ___________.

AP STAT

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