Stats 155- Questions

What is your chance of being in a sample of 25, drawn randomly from a population of 25,000?

.001

If you roll a pair of fair dice, what is the probability of getting a sum of 5 (give the answer to 3 decimal places)?

.111

A report on the assests of American households says that the median net worth of US families is $77,300. The mean net worth of these families is $498,800. What explains the difference between these two measures of center?

Outliers: very few people make VERY large amounts of money (6)

A PC contains 8152 music and video files. The distribution of file size is highly skewed. Assume that the s.d. for this population is 0.82 megabytes. a. What is the s.d. of the average file size when you take an SRS of 16 files from this population? b. How many files would you need to sample if you wanted the s.d. of the sample mean to be no larger than 0.10 megabytes?

SD = SD/SQRT(n) = .82/SRT(16) = .205 .1 = .82/SQRT(n) = 67.24 = 68 (9)

Explain whether a significance test will answer the following questions: a) is the sample of the experiment properly designed? b) is the observed effect compatible with the null hypothesis? c) Is the observed effect important

only b (5)

A college wants 950 entering freshman. Experience shows that about 75% of admitted students accept. The college admits 1200 students. Hence assuming the decisions are made independently, the number who accept has a B(1200,0.75) distribution. What is the standard deviation of the number of students who accept?

sqrt(n*p*(1 - p)). 15 (7)

For the data: 0.8, 2.1, 2.6, 0.9, 2.2, 0.8, 2.2, 0.9 Make histograms using the bins: [0,1), [1,2), [2,3) [0.5,1.5), [1.5,2.5), [2.5,3.5) [0,1), [1,3) b) Why are bins [0,2), [1,3) inappropriate here? c) Why are bins [1,2), [2,5) inappropriate here?

(5)

Find the standard deviation of the number of aces Value of X/Probability 0/.8507 1/.1448 2/.0045

.373 (can't type this process, look at HW 6 written answer sheet) (6)

Calculate (and think about as "balance point") weighted average of 1,2,3,10 for the weights: a. 1/4,1/4,1/4,1/4 b. 0.1,0.1,0.1,0.7 c. 0.3,0.3,0.3,0.1 d. 1/3,1/3,1/3,0 e. 0,1,0,0

4 7.6 2.8 2 2 (6)

The following table gives the distribution of the number of servings of fruits and vegetables consumed per day in a population. Find the mean for this random variable. Number servings X/Probability 0/.3 1/.1 2/.1 3/.2 4/.2 5/.1

4.80 (6)

A study intentionally overfed 12 volunteers for eight weeks. The mean increase in fat was 2.32 kilograms and the standard deviation was 1.21 kilograms. A kilogram is 2.2 pounds. (Round to 2 decimal places) a. What is the mean increase in fat in pounds? b. What is the standard deviation in pounds?

5.10 2.66 (7)

A firm pays each of its six clerks $45,000, each of two accountants $70,000, and the owner gets $420,000. a. What is the mean salary at the firm? (Round to two decimal places) b. How many employees earn less than the mean? c. What is the median salary? d. The firm gives no raises to anyone but the owner, who now gets $500,000. How does this change affect the mean? Median?

=((45000*6)+(70000*2)+(420000))/9 = 92222.22 8 =MEDIAN(45000, 45000, 45000, 45000, 45000, 45000, 70000, 70000, 420000) = 45000 =((45000*6)+(70000*2)+(500000))/9 = 92222.22 Mean increases to $10,111.11. Median does not change (6)

You are planning an evaluation of an alcohol awareness campaign. Previous evaluation indicates that about 20% of the students surveyed will respond Yes to the question "Did the campaign alter your behavior toward alcohol consumption?" How large a sample of students should you take if you want the margin of error for 95% confidence to be about 0.08?

=((NORM.INV(0.975,0,1)/0.08)^2)*(0.2*(1-0.2)) = 96

"What do you think is the ideal number of children?" A Gallup poll asked this question on 1020 randomly chosen adults. Over half (53%) thought that a total of two children was idea. Suppose that p = .53 is exactly true for the population of adults. Gallup announced a margin of error of +/- 4% for this poll. What is the probability that the sample proportion p-hat for an SRS size n = 1020 falls between .49 and .57? You see that it is likely, but nor certain, that polls like this give results that are correct within their margin of error. We will say more about margins of error in Chapter 6.

=(BINOM.DIST(0.57*1020,1020,0.53,TRUE))-(BINOM.DIST(0.49*1020,1020,0.53,TRUE)) (7)

Each subject tastes two unmarked cups of coffee, one of each type, in random order and states which he or she prefers. Of the 50 subjects who participate in the study, 15 prefer the instant coffee. Let p be the probability that a randomly chosen subject prefers fresh-brewed coffee to instant coffee. (p is the proportion of the population who prefers fresh-brewed coffee) a) Test the claim that a majority of people prefer the taste of fresh-brewed coffee. Report p-value.

=1-BINOM.DIST(34,50,0.5,TRUE) = .0033 AND we are asked to rework this using normal approximation to binomial: (direcly from JSM the man himself) Following the steps above, you arrive at: p-value = P{X >= 35 | p = 0.5} where X ~ Bi(50,p). Now the normal approximation says we approximate: X ~ N(np, sqrt(np(1 - p))) so put in n and p, and you get: p-value = P{X >= 35} = 1 - P{X < 35} = 1-NORM.DIST(35,50*0.5,sqrt(50*0.5*0.5),true) (5)

The scores of the WAIS IQ test are approximately Normal with mean 100 and standard deviation 15. The high IQ society, MENSA requires a score of 130 or higher. What percent of adults qualify for membership? (Round the percent to 2 decimal places)

=1-NORM.DIST(130,100,15,TRUE) = 2.28% (8)

A patient is classified as having gestational diabetes if her glucose level is above 140 mg/dl. Sheila's measured glucose level varies according to the Normal distribution with mean 125 mg/dl and s.d. 10 mg/dl. (Round to 4 decimal places) a. If a single glucose measurement is made, what is the probability that Sheila is diagnosed as having gestational diabetes? b. If measurements are made instead on 3 separate days and the mean result is compared with the criterion 140 mg/dl, what is the probability that Sheila is diagnosed as having gestational diabetes? c. What is the level L such that there is probability only 0.05 that the mean glucose level of three test results falls above L for Sheila's glucose level distribution? (Round to 1 decimal place)

=1-NORM.DIST(140,125,10,TRUE) = .0668 X is about equal to (mean,SD/SQRT(n)) =1-NORM.DIST(140,125,5.77,TRUE) = .0047 =NORM.INV(0.95,125,5.77) (because above ) = 134.5 (9)

mean = 2403.7, SD = 880, H0 = miu = 2403.7, H1 = miu >2403.7 a) 100 n b) 500 n c) 2500 n

=1-NORM.DIST(2453.7,2403.7,88,TRUE) = .2843 =1-NORM.DIST(2453.7,2403.7,39.35,TRUE) = .1020 =1-NORM.DIST(2453.7,2403.7,17.6,TRUE) = .0023 (10)

The Federal Aviation Administration assumes that passengers average 190 pounds. However, passengers vary: the FAA gave a mean but not an s.d. A reasonable s.d. is 35 pounds. Weights are not Normally distributed, but they are not very non-Normal (why even say that tho ew i hate this **** class). A commuter plane carries 25 passengers. What is the approximate probability that the total weight of the passengers exceeds 5200 pounds? (Round to 4 decimal places)

=1-NORM.DIST(5200,4750,175,TRUE) = .0051 (9)

A TV ad claims that 30% of people prefer Brand X. Should we dispute this claim if a random sample of 10 people show: 2 people who prefer Brand X 3 people who prefer Brand X 6 people who prefer Brand X 10 people who prefer Brand X A manager asks 12 workers, of whom 7 say they are satisfied with working conditions. Does this contradict the CEO's claim that ¾ of the workers are satisfied?

=BINOM.DIST(2,10,.3,TRUE) + =1-BINOM.DIST(3,10,.3,TRUE) = .733 Same for all these 1 .076 5.9E-6 Manager question--TRICKY TRICKY =BINOM.DIST(7,12,0.75,TRUE) + =1-BINOM.DIST(10,12,0.75,TRUE) = .316 Because gotta use timeline: 9 is 75% of 12, 7 is two away, 11 is two above, use them numbers yo (5)

Suppose that all adults who think that having two children is idea is p = .53. What is the probability that a sample proportion p falls between .49 and .57 (+/- 4%) if the sample is an SRS of size n = 300? n = 5000? Combine these results with your work in exercise 5.60 to make a general statement about the effect of larger samples in a sample survey.

=BINOM.DIST(300*0.57,300,0.53,TRUE)-BINOM.DIST(300*0.49,300,0.53,TRUE) = .817 =BINOM.DIST(5000*0.57,5000,0.53,TRUE)-BINOM.DIST(5000*0.49,5000,0.53,TRUE) = .999 The larger the sample size, the larger the probability that the sample survey falls within 8% of sample proportion. (7)

What's the excel functions for the 5-number summary?

=MINIMUM, =QUARTILE.INC(x,Q1), =MEDIAN, =QUARTILE,INC(x,Q3), =MAXIMUM (7)

Find the proportion of observations from a standard Normal distribution for each of the following events. (Round to 4 decimal places) (a) Z <= -1.7 (b) Z => -1.7 (c) Z > 1.9 (d) -1.7 < Z < 1.9

=NORM.DIST(1.7,0,1,TRUE) = .0446 =1-NORM.DIST(-1.7,0,1,TRUE) = .9554 =1-NORM.DIST(1.9,0,1,TRUE) = .0287 =NORM.DIST(1.9,0,1,TRUE)-NORM.DIST(-1.7,0,1,TRUE) = .927 (8)

Averages are less variable than individual observations. It is reasonable to assume that the can volumes in exercise 5.17 (mean = 250, sd = .5) vary according to a Normal distribution. In that case ,the mean xbar of an SRS of cans also has a normal distribution. b) What is the probability that the volume of a single randomly chosen can differs from the target value by 1 ml or more? c) What is the probability that the mean volume of an SRS of 4 cans differs from the target value by 1 ml or more?

=NORM.DIST(249,250,0.5,TRUE)+1-NORM.DIST(251,250,0.5,TRUE) = .0456 .5/SQRT(4) = 2 is sample SD I dontknow this one, but book says: P = 2P(X<249) = 2P(z<-4) is about equal to 0 *law of averages: averaging increases accuracy by factory of 1/SQRT(n) (9)

U.S. women aged 20 and over have a mean HDL (good cholesterol) of 55 mg/dl with a standard deviation of 15.5 mg/dl. Assume the distribution is Normal. (Round the percents to 2 decimal places) a. What percent of women have 40 mg/dl or less? b. What percent have HDL levels of 60 mg/dl or higher? c. What percent are more than 40 mg/dl but less than 60 mg/dl?

=NORM.DIST(40,55,15.5,TRUE) = 16.6% =1-NORM.DIST(60,55,15.5,TRUE) = 37.35% =NORM.DIST(60,55,15.5,TRUE) - NORM.DIST(40,55,15.5,TRUE) = 45.99% (8)

For the standard normal distribution N(0,1), what are the two values farthest from 0 that would not be considered outliers according to the 1.5*IQR criterion? Note that one value is negative and one is positive. (Give all answers to two decimal places.) Positive value: z= Negative value: z=

=NORM.INV(.25,0,1) = -.67 =NORM.INV(.75,0,1) = .67 IQR = 1.34 Outliers: Q1 - (IQR*1.5) Q3 + (IQR*1.5) +/-2.7 (8)

For the standard normal distribution, find the values (to 2 decimal places) of z such that (a) 26% of values are below z b) 33% of values are above z

=NORM.INV(.26,0,1) = -.64 =NORM.INV(.67,0,1) = .44 (for above, take 1-p) idk why (8)

Calculate the standard deviation for the following lists, and compare qualitatively in terms of spread: 1, 3, 3, 1 -6, -4, -4, -6 1, 5, 5, 1 1, 1, 1, 1

=STDEV.S(1, 3, 3, 1) = 1.15 =STDEV.S(-6, -4, -4, -6 = 1.15 =STDEV.S(1, 5, 5, 1) = 2.31 =STDEV.S(1, 1, 1, 1) = 0

A company makes 40% of its cars at factory A, and the rest at factory B. Factory A produces 10% lemons, and Factory B produces 5% lemons. A car is chosen at random. What is the probability that: It came from Factory A? It is a lemon, if it came from Factory A? It is a lemon, from Factory A? It is a lemon? It came from Factory A, if it is a lemon (give the answer to 2 decimal places)?

A: .4 of the cars come from factory A; therefore, .4. A: We definitely know that it came from Factory A: that part has already been presented to us. Therefore, we will not use this as a given, since to visualize this, we are already in Factory A; Factory B is not even an option. Now that we are in Factory A, what is the chance that it's a lemon? 10%, or .1, as was given. A: We do not know that we are in factory A. Therefore, this needs to be considered as part of the problem. The question is an AND question: P(lemon AND factory A). P(lemon AND factory A)*P(Factory A) = .1*.4 = .04 A: We must find the different probabilities of it being a lemon for either choice and add them together. First, for Factory A, we found in the last question that lemon AND factory A = .04. If we do the same for Factory B, P(lemon AND Factory B) = P(Lemon and Factory B)*P(Factory B) = .05 * .6 = .03. .04 + .03 = .07. A: We know that the probability that getting a lemon from Factory A is .04. The probability of getting a lemon is .07+ .4/.07= .57. (2)

Is a binomial distribution reasonable for the random variable X? Check all that apply. 1.Toss a fair coin until you get a head. X is the number of tosses you made 2.A poll of students asks if they are irritable in the mornings. X is the number who do feel irritable in the mornings. 3.One-twelfth of sample survey calls made at random in NYC succeed in talking to the person. A survey calls 500 randomly selected numbers. X is the number of times a person is reached. 4.Deal 10 cards from a standard, shuffled deck. X is the number of black cards

A: 2, 3 (3)

Is a binomial distribution reasonable for the random variable X? Check all that apply. 1. A fitness study chooses a random number of students. X is the mean daily exercise time. 2. A nutritionist chooses a SRS of students and asks whether or not they eat at least 3 servings of vegetables a day. X is the number who say they do. 3. X is the number of days you skip a class during the school year. 4. A manufacturer of shoes randomly picks 20 shoes from the production of shoes each day for inspection. X is the number of shoes with a defect.

A: 2,4 (3)

The 2010 American Community Survey estimates that 14.6% of Americans live below the poverty line, 20.7% speak a language other than English at home, and 4.2% fall into both categories. Answer the following questions (give all percents to the nearest whole percent). a) Are living below the poverty level and speaking a language other than English at home disjoint events? b) Just in case you were guessing in part a), if the events were disjoint, what percent would fall into both categories? (Give your answer to the nearest whole percent) c) Are living below the poverty level and speaking a language other than English at home independent events? d) Just in case you were guessing in part c), if the events were independent, what percent would fall into both categories? (Give your answer to the nearest whole percent) e) If I choose two Americans at random, what is the probability that both live below the poverty line? (Give your answer to the nearest whole percent) f) If I choose two Americans at random, what is the probability that both speak English at home? (Give your answer to the nearest whole percent) g) If I choose two Americans at random, what is the probability that one speaks English at home and the other does not? (Give your answer to the nearest whole percent)

A: 62%; (1-.207)=.793^2=.628 A: 3% (For independent, multiplying both: .146*.207 = .03) A: Independent = not reliant on another. No, they are not independent. A: 0% would fall into both categories; disjoint = separate events A: 2%; .146 * .146 A: No, there is an overlap between the two. (3)

A factory makes 10% defective items & items are independently defective. Find P{9 or more good items in 10} a. Using X = # good items, and Binomial probability distribution function.

A: = BINOM.DIST(9,10,.9,FALSE) + =BINOM.DIST(10,10,.9,FALSE) = .736 (4)

Answer from both gray-level and yes-no viewpoints: (a) A TV ad claims that less than 40% of people prefer Brand X. Suppose 7 out of 10 randomly selected people prefer Brand X. Should we dispute the claim? (b) 80% of the sheet metal we buy from supplier A meets our specs. Supplier B sends us 12 shipments, and 11 meet our specs. Is it safe to say the quality of B is higher?

A: =1-BINOM.DIST(6,10,.4,TRUE) P[ X ≥ 7 | p = 0.4] = p = 0.055 Yes/No-No, it is less than .05 A: =1-BINOM.DIST(10,12,.8,TRUE) = p = .275 No—this p-value is greater than .05 (5)

In a test for extrasensory perception, the experimenter looks at cards containing either a star, circle, wave, or square. The subject can not see the cards. As the experimenter looks at each of 20 cards, the subject names the shape on the card. What is the probability a subject correctly guesses at least 10 of the 20 shapes? (Round to 3 decimal places)

A: =1-BINOM.DIST(9,20,.25,TRUE) = .014 (4)

The CDC finds that 24.2% of women 18 to 24 years old binge drink. The study sampled 11,000 women in the above age category. Hence the population proportion is very close to p=0.24. Your college surveys a SRS of 200 female students and finds that 56 binge drink. If the proportion of women on your campus who binge drink is the same as the national proportion, what is the probability that the proportion in a SRS of 200 students is as large or larger than the result of the college's sample? (Round to 3 decimal places)

A: =BINOM.DIST(56,200,.24,TRUE) BUT excel can't do greater than, only less than, therefore: =1-BINOM.DIST(55,200,.24,TRUE) = .108 (4)

A survey of over 20,000 US HS students revealed that 20% of the students say tha they stole something from a store in the past year. This is down 7% from the last survey, which was performed 2 years earlier: you decide to take a random sample of 10 high school students from you city and ask them this question. a) If the high school students in your city match this 20% rate, what is the distribution of the number of students who say that they stole something in the past year? What is the distribution of students who do not say that they stole something from a store in the past year? b) What is the probability that 4 or more of the 10 students in your sample say that they stole something from a store in the past year?

A: Bi(10,.2) (because Bi(n,p) ; Bi(10,.8) A: =1-BINOM.DIST(3,10,.2,TRUE) (4)

75% Canadian teens use a fee-based website to download music. You decide to interview a random sample of 15 US teenagers. For now, assume that they behave similarly to the Canadian teenagers. a) What is the distribution of the number X who used a fee-based website to download music? Explain your answer. b) What is the probability that at least 12 of the 15 teenagers in your sample used a fee-based website to download music?

A: Bi(15,.75). 15 is n, .75 is p A: 1-BINOM.DIST(11,15,.75,TRUE) = .461 (4)

Decide whether the random variable is discrete or continuous, and give reason for your answer: A) Your web page has five different links, and a user can click on one of the links or leave the page. You record the length of time that a user spends on the web page before clicking on one of the links or leaving the page. B) The number of hits on your web page C) The yearly income of a visitor to your web page.

A: Continuous; lengths of time can be non-counting numbers A: Discrete; counting numbers, no fractions of people A: Discrete; only counting numbers (3) Continuous = measuring; discrete = counting

Since 2008, chain restaurants in California have been required to display calorie counts of each menu item. Prior to menus displaying calorie counts, the average calorie intake of diners at a restaurant was 1100 calories. After calorie counts started to be displayed on menus, a nutritionist collected data on the number of calories consumed at this restaurant from a random sample of diners. Do these data provide convincing evidence of a difference in the average calorie intake of diners at this restaurant? (H0, H1)

A: H0 = p = 1100, H1 = p does not equal 1100 (4)

New York is known as the city that never sleeps. A random sample of 25 New Yorkers were asked how much sleep they get per night. Do these data provide convincing evidence that New Yorkers on average sleep less than 8 hours a night? (H0, H1)

A: H0 = p = 8, H1 = p < 8 (4)

A building company bids on two large projects. The CEO believes the chance of winning the 1st is 0.6, the chance of winning the 2nd is 0.5, and the chance of winning both is 0.3. What is the chance of winning at least one of the jobs?

A: P(1) = .6, P(2) = .5, P(1 and 2) = .3 P(1 or 2) = P(1) + P(2) - P(1 and 2) .6 + .5 - .3 = 0.8

Choose an American household randomly, let A be the event that the selected household is prosperous and B the event that it is educated. As the current survey, P(A)=0.138, P(B)=0.261, while the probability that a household is both prosperous and educated is P(A and B)=0.082. What is the probability P(A or B) that the household selected is either prosperous or educated?(give the answer to 3 decimal places)

A: P(A or B) = P(A) + P(B) - P(A and B) = .138 + .261 - .082 = 0.317 (2)

Using data from 4.122: A = Connecticut Office of the Chief Medical Officer B = the New Jersey Division of Criminal Justice C = the federal Disaster Mortuary Operations Response Team P(A) = .7, P(B) = .5, P(C) = .3, P(A and B) = .3, P(A and C) = .1, P(B and C) = .1, P(A and B and C) = 0 If Julie is offered the federal job, what is the conditional probability that she is also offered the New Jersey job? If Julie is offered the New Jersey job, what is the conditional probability that she is also offered the federal job?

A: P(B, given C) = P(C and B)/P(C) = .1/.3 = 1/3 A: P(C, given B) = P(B and C)/P(B) = .1/.5 = 1/5 = .2 (2)

The workforce in a town has (20%, 50%, 30%) workers with (no HS, HS-no C, C) education. Past experience indicates that (10%, 30%, 90%) of workers with (no HS, HS-no C, C) education can perform a given task. Find the probability that a randomly chosen worker (give all answers with 2 decimal places): a. Can perform the task b. Is College educated if (s)he can perform the task

A: P(Workers AND Can perform task) = disjoint so P(Workers) * P(Task) = SUM of last column on chart = .44 A: Total who can perform the task = .44; Total who can perform the task who have been to college = .27. P(College Educated if can perform task) = P(college educated, given can perform task) = .27/.44 = .61 (2)

The "random walk" theory of securities prices holds that price movement is disjoint time periods which are independent of each other. Suppose that we only record whether the price is up and down per year. The probability that our portfolio rises in price in any one year is 0.65. Give all answers with 2 decimal places. a. What is the probability that our portfolio goes up for three consecutive years? b. Knowing that the portfolio has risen in price for two years in a row, what probability do you assign to the event that it will go down next year? c. What is the probability that the portfolio's value moves in the same direction in both of the next two years?

A: Since disjoint, P(up AND up AND up) = P(up) * P(up) * P(up) = .653= .27 A: Disjoint, so previous years have no effect on this year. 1-.65 = .35 A: Same direction; two different probabilities. P(up AND up) = .652 = .42. P(down AND down) = .352= .12. .42 + .12 = .55

Suppose events A, B, C all have probability 0.4, A & B are independent, and A & C are mutually exclusive (give all answers in decimal form). (a) Find P{A or B} (b) Find P{A or C} (c) Find P{A and B} (d) Find P{A and C}

A: They are independent; they don't rely on each other. Therefore, P(A) + P(B) - P(A and B) = .4 + .4 -(.16) = .64 A: A and C are mutually exclusive. That means one cannot happen while the other happens; one will be chosen. Therefore, .4 + .4 = .8. A: A and B are independent; one does not affect the other. Therefore, .4 * .4 = .16. A: A and C are mutually exclusive; they cannot happen together. Therefore, 0.

Muscular dystrophy is an incurable muscle-wasting disease. The most common and serious type, called DMD, is caused by a sex-linked recessive mutation. Specifically, women can be carriers but do not get the disease; a son of a carrier has probability .5 of having DMD; a daughter has probability .5 of being a carrier. As many as one-third of DMD cases, however, are due to spontaneous mutations in sons of mothers who are not carriers. Toni has one son, who has DMD. In the absence of other information, the probability is 1/3 that the son is the victim of a spontaneous mutation and 2/3 that Toni is a carrier: There is a screening test called the CK test that is positive with a probability .7 if a woman is a carrier and with probability .1 if she is not. Toni's CK test is positive. What is the probability that she is a carrier?

A: This one is so tricky I hate it. We have to use Bayes's Rule, which I didn't remember; we really only need it to know how to do this one problem which is nice. P(Toni is a carrier given + test) = P(C, given +) *P(C)/ (P(+, given C)P(C)+P(+, given Healthy)P(Healthy)) = .467/.5 = .993 (2)

In the Excel data set email, the variable number describes whether no number (indicated by none), only small numbers (labeled small) or at least one big number (big) appeared in the corresponding email. Of the 3,921 emails, 549 had no numbers, 2,827 had only small numbers, and 545 had at least one big number. a) Are the outcomes none, small, and big disjoint? b) Determine the proportions of e-mails with value small and the proportion of e-mails with c) Use the Addition Rule for disjoint events to find the probability that an e-mail chosen at

A: Yes A: Proportion Small = 2,827/3,921 = .721; Proportion Big = 545/3,921 = .139 A: P(Small or Big) = P(Small) + P(Big) - P(Big or Small) In this case, small and big are disjoint, so the equation becomes: A: P(Small or Big) = P(Small) + P(Big) .139 + .721 = .860 (2)

Translate each of the following research questions into appropriate H0 and H1: a) U.S. Census Bureau data show that the mean household income in the area served by a shopping mall is $42,800 per year. A market research firm questions shoppers at the mall to find out whether the mean household income of mall shoppers is higher than that of the general population. b) Last year, your online registration technicians took an average of .4 hours to respond to trouble calls from students trying to register. Do this year's data show a different average response to this?

H0 = p = 42,800, H1 = p > 42,800 H0 = p = .4, H1 = p not equal .4 (4)

For each of the following, formulate quantitative H0 and H1 (a) We now buy sheet metal from A & 90% of the time it meets our specs. B claims more of their sheet metal meets our specs. (b) Is there strong evidence that on average girls score better than boys on achievement tests? (c) Can we reasonably dispute a claim that on average girls score better than boys on achievement tests? (d) Can we conclude that on average girls score differently from boys on achievement tests? (e) Is there strong evidence for disputing a claim that on average girls score differently from boys on achievement tests? (f) Test a claim that 70% of consumers prefer Brand A. (g) Test the validity of a claim that at least 70% of consumers prefer Brand A.

H0: p ≤ 0.9 H1: p > 0.9 H0 = p=.5, H1= p >=.5 H0>=.5, H1=.5 H0=p=.5, H1=p does not equal .5 H0 = p not equal .5, H1 = p = .5 H0 = p = .7, H1 = p doe not equal .7 H0 = p >= .7 H1 = p < .7 (4)

For X Uniformly Distributed over [0, 2], Find c, so that 0.8 = P[X <= c]

Height = .5 .8 * 2 = 1.6 (8)

Suppose only 60% of the US teenagers used a fee-based website to download music. A) if you interview 15 US teenagers at random, what is the mean of the count X who use a fee-based website to download music? What is the mean of the proportion p in your sample who used a fee-based website to download music? b) Repeat the calculations in part a for samples of size 150 and 1500. What happens to the mean count of successes as the sample size increases? What happens to the mean proportion of successes?

Mean of counts X: np = (.6*15) = 9 Mean of proportion = p = .6 Counts, np, (150*.6) = 90 Mean of proportion = p = .6 Counts, np, (1500*.6) = 900 Mean of proportion = p = .6 (7)

Mean score = 115, SD = 30. A teacher gives test to older 25 older students, mean score is 127.8. a) assuming that SD = 30 for population of older students, carry out a test of H0: miu = 115, Ha: miu >115. b) What were two important assumptions you had to make in part a and why important?

TRICKY TRICKY! a) First, find z-score, (127.8-115)/6 (account for new sample size of 25 and divide 30 by 5 to get new SD) = 2.13. Then, =1-NORM.DIST(2.13,0,1,TRUE) b) That this is an SRS and a Normal Distribution (9)

A study of 14 subjects reported a result that failed to achieve statistical significance at the 5% level. The p-value was .051. Write a short summary of how you would interpret these findings.

With a larger sample, we might have more significant results. (10)

The heights of women aged 20-29 are approximately Normal with mean 64 inches and s.d. 2.7 inches. Men the same age have a mean height 69.3 inches and s.d. 2.8 inches.(Round to 2 decimal places) a. What is the z score for a woman 6 feet tall? b. What is the z score for a man 6 feet tall?

Z = (X-mean)/SD --> (72-64)/2.7 = 2.96 Z = (72-69.3)/2.8 --> .96 (8)

For a certain population IQ-scores are approximately normally distributed with a mean of 102 points and a standard deviation of 14 points. (a) What is the z-score for a member of this population that has an IQ of 139? (Give your answer to two decimal places.) (b) What is the probability that a person chosen at random from this group has an IQ in excess of 139? (Give your answer to two decimal places.) (c) What IQ-score is at the 85th percentile? (Give your answer to two decimal places.) (d) Find the median, quartiles and interquartile range for these IQ scores. (Give your answers to two decimal places.)

a = (139-102)/14 = 2.64 b =1-NORM.DIST(139,102,14,TRUE) = 0.00 c =NORM.INV(.85,102,14) = 116.51 d =NORM.INV(.5,102,14) = 102 =NORM.INV(.25,102,14) = 92.56 =NORM.INV(.75,102,14) = 111.44 IQR = Q3 - Q1 = 18.88 (8)

A survey of Inuits reported that 3274 out of 5000 respondents said that at least half of the meat they eat comes from their own country.(Round 3 decimal places) a. Based on the information, what is the sample proportion? b. Give a 95% Best Guess Confidence Interval for the population proportion of Inuits who at least half their meat comes from their own country. Lower Bound of Interval: Upper Bound of Interval: c. Give a 95% Conservative Confidence Interval for the population proportion of Inuits who at least half their meat comes from their own country. Lower Bound of Interval: Upper Bound of Interval:

a) 3724/5000 = .655 b) =NORM.INV(0.975,0,SQRT(0.655*(1-0.655)/5000)) = .013 upper/lower = .655 +/- .013 = .642/.668 c) =NORM.INV(.975,0,1/(2*SQRT(5000)) = .014 upper/lower = .669/.641 (10)

In a sample of 159,949 college freshman, 42% report they plan to study abroad. a. Based on the information, how many students plan to study abroad? b. Give a 99% Best Guess Confidence Interval for the population proportion of college freshman who plan to study abroad. (Round 3 decimal places) Lower Bound of Interval: Upper Bound of Interval: c. Give a 99% Conservative Confidence Interval for the population proportion of college freshman who plan to study abroad. (Round 3 decimal places) Lower Bound of Interval: Upper Bound of Interval:

a) 67179 b) =NORM.INV(.995,0,SQRT(.42(1-.42)/159949))) = .003 UB = .423, LB = .417 b) =NORM.INV(.995,0,1/(2*SQRT(159949)) = same as b (10)

Suppose the speeds of cars along a stretch of I-40 is normally distributed with a mean of 71 mph and standard deviation of 6 mph. Use the 68-95-99.7 rule to answer the following questions. (a) Approximately what percent of cars are travelling between 65 and 77 mph? (b) If the speed limit on this stretch of highway is 65 mph, approximately what percent of cars are traveling faster than the speed limit? (c) What percent of cars are traveling at a speed greater than or equal to 83 mph? (Answer to one decimal place.) (d) What percent of cars are traveling at a speed greater than 83 mph? (Answer to one decimal place.) (e) 95% of cars are traveling between what two speeds? (Answer to two decimal places)

a) 68% (1 SD) b) 34+50 = 84% c) 5/2 = 2.5% d) 2.5% e) 59-83 (8)

The variable Z has a standard Normal distribution. (Round to 2 decimal places) a. Find the number z that has cumulative proportion 0.78. b. Find the number z such that the event Z > z has proportion 0.22.

a) =NORM.INV(.78,0,1) = .77 b) =NORM.INV(.78,0,1) = .77 (yes they're the same not a typo) (8)

A study asks a total of 340 Millennials to indicate their average stress level (on a 10 point scale) during the past month. The mean score was 5.4. Assume the population standard deviation is 2.3. (Round to 3 decimal places) a. Give the margin of error and find the 95% confidence interval. lower/upper bound b. Repeat these calculations for 99% confidence interval. lower/upper bound

a) =NORM.INV(.975,0,(2.3/SQRT(340))) = .244 lower bound: 5.15 upper: 5.62 b) =NORM.INV(.995,0,(2.3/SQRT(340))) lower: 5.079 upper: 5.721 (10)

For T ~ t, with degrees of freedom: (a) 3 (b) 12 (c) 150 (d) N(0,1) Find: P{T> 1.7} P{T < 2.14} (0.939, 0.973, 0.983, 0.984) P{T < -0.74} (0.256, 0.237, 0.230, 0.230) P{T > -1.83} (0.918, 0.954, 0.965, 0.966) P{|T| > 1.18} (0.323, 0.261, 0.240, 0.238) P{|T| < 2.39} (0.903, 0.966, 0.982, 0.983) P{|T| < -2.74} (0, 0, 0, 0) C so that 0.05 = P{|T| > C} (3.18, 2.17, 1.98, 1.96) C so that 0.99 = P{|T| < C} (5.84, 3.05, 2.61, 2.58)

a) =T.DIST.RT(1.7,3) b) =1-T.DIST.RT(3,2.14) c)

How much education children get is strongly associated with the wealth and social status of their parents, termed SES. The SES, however, has little influence on whether childre who have graduated from college continue their education. One study looked at whether college graduates took the graduate admissions tests for business, law, and other graduate programs. The effects of the parent's SES on taking the LSAT test for law school were both "statistically insignificant" and small." a) What does "statistically insignificant" mean? b) Why is it important that the effects were small in size as well as statistically insignificant?

a) The differences could occur by chance, without SES being a factor. b) The statistical insignificance did not occur merely because of a small sample size.

Suppose the speeds of cars along a stretch of I-40 is normally distributed with a mean of 75 mph and standard deviation of 5 mph. (a) What is the z-score for a car travelling at 70 mph? (Give your answer to two decimal places.) (b) What is the z-score for a car travelling at 86 mph? (Give your answer to two decimal places.) (c) What would be the speed for a car whose z-score is z=−1.6? (Give your answer to two decimal places.) (d) If a car is chosen at random on this stretch of highway, what is the probability it will be travelling at a speed below 70 mph? (Give your answer to 2 decimal places.) (e) If a car is chosen at random on this stretch of highway, what is the probability it will be travelling at a speed greater than or equal to 86 mph? (Give your answer to 2 decimal places.) (f) If a car is chosen at random on this stretch of highway, what is the probability it will be travelling at a speed between 70 mph and 86 mph? (Give your answer to 2 decimal places.) (g) What is the probability that a car chosen at random on this highway will be traveling between 65 mph and 80 mph? (Give your answer to 2 decimal places.)

a) Z = (70-75)/5 = -1 b) Z = (86-75)/5 = 2.2 c) -1.6 = (X-75)/5 = 67 d) =NORM.DIST(70,75,5,TRUE) = .15 e) =1-NORM.DIST(86,75,5,TRUE) = .01 f) =NORM.DIST(86,75,5,TRUE)-NORM.DIST(70,75,5,TRUE) = .83 g) =NORM.DIST(80,75,5,TRUE)-NORM.DIST(65,75,5,TRUE) = .82 (8)

For X ~ Binomial(20,0.6) Calculate as: i.Exact Binomial ii.Simple Normal Approximation iii.Continuity Corrected Normal Approx. The probabilities: P[10 ≤ X ≤ 13] P[14 < X < 50] P[X = 11] P[|𝑝 - 0.6| < 0.15] P[|𝑝 - 0.6| > 0.03]

a) i. =BINOM.DIST(13,20,0.6,TRUE)-BINOM.DIST(9,20,0.6,TRUE) = .622 ii. (9)

Mean beetles trapped = .3, SD = .8 a_ Suppose that your state does not have the resources to check all the traps, and so it plans to check only an SRS of n = 100 traps. What are the mean and standard deviation of the average number of beetles x in 100 traps? b) Use the central limit theorem to find the probability that the average number of beetles in 100 traps is greater than .5. c) do you think it is appropriate in this situation to use the central limit theorem? Explain.

a) mean = .3, SD = .08 b) =1-NORM.DIST(0.5,0.3,0.08,TRUE) = .0062 c) Yes, n = 100 is a large enough sample (9)

a) You are interested in attitudes towards drinking among the 75 members of a fraternity (lol tru tho). You choose 30 members at random to interview. One question is "Have you had five or more drinks at one time during the last week?" Suppose that in fact 30% of the 75 members say yes. Explain why you cannot safely use the B(30,.3) distribution for the count X in your sample who say "yes." b) the national aids behavioral surveys found that .2% of adult heterosexuals had both received a blood transfusion and had a sexual partner from a group at high risk of aids. Suppose that this national proportion holds for your region. Explain why you cannot safely use the normal approximation for the sample proportion who fall in this group when you interview an SRS of 1000 adults.

a) ok when np >=10; np too small in this case, too few people b) 1000*.002 = 2; must be bigger than 10 (9)

Suppose that 500 randomly selected alumni of the university of Okoboji were asked to rate the university's academic advising services on a 1 to 10 scale. The sample mean xbar was found to be 8.6. Assume that the population standard deciation is known to be 2.2. a) Ima Bitlost computes the 95% confidence interval for the average satisfaction score as 8.6 +/- 1.96(2.2). What is her mistake? b) after correcting her mistake in part a, she states, I am 95% confident that the sample mean falls between 8.4 and 8.8" What is wrong with her statement? c) she quickly realizes her mistake in part b and instead states "The probability that the true mean falls between 8.4 and 8.8 is .95" What's wrong now d) Finally in her defense for using the normal distribution to determine the confidence interval she says, "beacuse the sample size is quite lareg, the population of alumni ratings will be approx. normal" Explain her misunderstanding and correct the statement.

a) she did not divide the standard deviation by root 500 b) confidence intervals concern the population mean c) .95 is a confidence level, not a probability d) The large sample size affects the distribution of the sample mean (by the central limit theorem), not the individual ratings. (10)

A bottling company uses a machine to fill cans with soda. The cans are suppose to contain 250 ml. However, the machine has variability, so the s.d. of the volume is 0.5 ml. A sample of 4 cans is inspected each hour and records are kept of the sample mean volume. a. If the process mean is exactly equal to the target value what will be the mean of the numbers recorded? b. What will be the standard deviation of the numbers recorded?

a) trick question, 250 :/ b) .5/SQRT(4) = .25 (9)

Fred keeps his savings in his mattress. He begins with $500 from his mother, and adds $100 each year. His total savings y, after x years are given by the equation: y = 500 + 100 x (a) Draw a graph of this equation. b) After 20 years, how much will Fred have? (c) If Fred adds $200 instead of $100 each year to his initial $500, what is the equation that describes his savings after x years?

b) 2500 c) (y = 500 + 200 x)

A new bone study is being planned that will measure the bio-marker TRAP. Assume the standard deviation is known to be 6.5. Find the sample size required to provide an estimate of the mean TRAP with a margin of error of 1.5 for 95% confidence.

n =((6.5/1.5)*(NORM.INV(0.975,0,1)))^2 = 72

An insurance company sells 1378 policies to cover bicycles against theft for 1 year. It costs $300 to replace a stolen bicycle and the probability of theft is estimated at 0.08. Suppose there is no chance of more than one theft per individual. a. Calculate the expected payout for each policy, to give a break even price for each policy. b. If two times the break even price is actually charged, what is the company's expected profit per policy, if the theft rate is actually 0.10?

((1378*.08)*300)/1378 = $24 ((1378*.1)*300)/1378 = $30 2*a(24) = 48 48-30 = $18 (6) 𝐸_𝑎𝑋+𝑏𝑌_=𝑎𝐸𝑋 +𝑏𝐸𝑌 ?

Select a young adult (age 24 to 35) randomly. The probability is 0.12 that the person chosen did not complete high school, 0.31 that the person owns a high school diploma without further education, and 0.29 that the person has at least a bachelor's degree. What is the probability that a randomly selected young adult has experienced at least a high school education? (give the answer to 2 decimal places)

.111

Let the random variable X be a random number with the uniform density curve in Figure 4.9 (height is 1, length is 1) a) P(X>=.30) b) P(X=30) c)P(.30<X<1.3) d) P(.2<=X<=.25 or .7<=X<=.9) e) X is not in the interval .4 to .7

.7 0 .7 .05+.2 = .25 .7 (8)

Select a young adult (age 24 to 35) randomly. The probability is 0.12 that the person chosen did not complete high school, 0.31 that the person owns a high school diploma without further education, and 0.29 that the person has at least a bachelor's degree. What is the probability that a randomly selected young adult has experienced at least a high school education? (give the answer to 2 decimal places)

.88

Which of the following are appropriate for the Normal approximation to the binomial for a significance test on the population proportion p? (Check all that apply) n=30 and H0: p=0.4 n=200 and H0: p=0.04 n=40 and H0: p=0.2 n=100 and H0: p=0.15

1,4 (10)

The National Center for Health Statistics says that the heights of adult men have mean 176.8 centimeters. There are 2.54 centimeters in an inch. What is the mean in inches? (Round to two decimal places)

176.8/2.54= 69.61 somehow this is supposed to have something to do with expected value but idk it's just common sense? (6)

Below are the weights in kilograms of a sample of 7 Tule Elk (Cervus canadensis nannodes) 163 195 177 187 248 224 178 Manually calculate the mean and standard deviation of these weights using the following step-by-step process (give all answers to 1 decimal place): (a) First, find the mean weight by hand. (b) Second, find each of the deviations (xi − x) What is the sum of the deviations? (Round your answer to the nearest integer.) c) Third, find the squares of the deviations. What is the sum of the squares of the deviations? d) Fourth, calculate the variance as the sum of squared deviations divided by (n−1) (e) Fifth, Calculate s the standard deviation.

196 0 5344 890.67 29.84 (6)

Which binomial distributions admit a "good" normal approximation? Bi(30, 0.3) Bi(40, 0.4) Bi(20,0.5) Bi(30,0.7)

2, 3 (9)

One of your employees has suggested that your company develop a new product. You decide to take a random sample of your customers and ask whether or not there is interest in the product. You record the responses as No and Yes. What sample size would you use if you wanted the 95% margin of error to be 0.2 or less?

=((NORM.INV(.975,0,1)/.2)^2)*.25 = 25 (round up)

In a study of bone turnover in young women a specific serum, called TRAP, is used to measure bone resorption. TRAP was measured in 31 subjects. The mean was 13.2 units per liter. Assume that the standard deviation is known to be 6.5 units per liter. Give the margin of error and find the 95% confidence interval for the mean TRAP amount in young women represented by this sample. (Round to 2 decimal places) Margin of error: Lower Bound of Interval: Upper Bound of Interval:

=CONFIDENCE.NORM(0.05,6.5,31) = 2.29 UB: 15.49 LB: 10.91 (10)

Random number generators allow a range to be specified for the numbers produced. Suppose that you specify the range to be between 0 and 2. Let Y be the random number generated. Then the density curve of the random variable Y has constant height between 0 and 2, and height 0 everywhere else. a. What is the height of the density curve between 0 and 2? b. Find P(Y<=1.6)? c. Find P(0.5< Y< 1.7) c. Find P(Y=>0.95)

Areas = probabilities; area always has to equal 1 a) .5 b) (1.6*.5) = .8 c) (1.2*.5) = .6 d) 1.05*.5) = .525 (8)

A college wants 950 entering freshman. Experience shows that about 75% of admitted students accept. The college admits 1200 students. Hence assuming the decisions are made independently, the number who accept has a B(1200,0.75) distribution. What is the mean of the number of students who accept?

EX = n*p = 1200 *.75 = 900 (6)

If the ESP study finds a subject does better than guessing, more testing is done. The experimenter looks at cards with one of five shapes that the subject can't see. The answers of a subject who doesn't have ESP should be independent observations, each with probability 1/5 success. 900 attempts are recorded. What is the mean of the number of the success count?

EX = n*p = 900*.2 = 180 (6)

Below are the weights in kilograms of a sample of 18 Tule Elk (Cervus canadensis nannodes). 230 215 173 183 233 240 177 227 180 178 219 202 237 172 214 194 204 221 Find the 5 number summary for this data. According to the IQR criterion for outliers, what would be the smallest weight larger than the median that would qualify as an outlier? (Round your answer to the nearest whole kilogram.)

Min: 172 mg/dL Q1 180.75 mg/dL Median 209 mg/dL Q3 225.5 mg/dL Max 240 mg/dL To find outliers for greater than median, do Q3+(IQR*1.5) = 225.5+(44.75*1.5) = 293 (7)

Motor vehicles are sold to individuals are classified as either cars or light trucks (including SUVs) and as either domestic or imported. In a recent year, .69 of vehicles sold were light trucks .78 were domestic, and .55 were domestic light trucks. Let A be the event that a vehicle is a car and B the event that it is imported. A) Given that a vehicle is imported, what is the conditional probability that it is a light truck? A) Are the events "vehicle is a light truck" and "vehicle is imported" independent? Justify answer.

P(Light Truck given imported) = P(light and imported)/P(imported) = .14/.22 = .636 A: No, they are not independent. Independent means that being one has no effect on the other. The answer to question A would have been 0 if they were independent. (3)

The scores of the WAIS IQ test are approximately Normal with mean 100 and standard deviation 15. People with a score below 70 are considered developmentally disabled. What percent of adults are developmentally disabled?(Round the percent to 2 decimal places)

PERCENTAGES = AREAS UNDER CURVE (because area under curve = 1) =NORM.DIST(70,100,15,TRUE) = 2.28% (8)

Changing the unit of length from inches to centimeters multiplies each length by 2.54 because there are 2.54 cm in an inch. This change of units multiplies our usual measures of spread by 2.54. This is true of IQR and the standard deviation. What happens to the variance when we change units in this way?

The variance is the unit squared: 2.54^2 = 6.45 (7)

Estimate the standard error of: a. The estimate of the population proportion, 𝑝, when the sample proportion is 0.9, based on a sample of size 100. b. The estimate of the population mean, 𝜇, when the sample standard deviation is 𝑠=15, based on a sample of size 25

a) SE = SD(phat) = SQRT(p(p-1)/n) = SQRT(.9(.1)/100) = .03 b) SE = SD(mean) = SD/SQRT(n) = 15/5 = 3 (9)

Suppose that 40% of adults get enough sleep, 46% get enough exercise, and 24% do both. Find the probabilities of the following events: A) Enough sleep and not enough exercise: B) Not enough sleep and enough exercise: C) Not enough sleep and not enough exercise D) For each of parts A, B, and C, state the rule that you used to find your answer

a. A: P(sleep and NOT exercise) = P(sleep) * P(NOT exercise) = (look at venn diagram) = .16 b. A: P(NOT sleep and exercise) = P(NOT sleep) * P(exercise) = (look at Venn diagram) = .22 c. A: P(NOT exercise and NOT sleep) = P(NOT exercise) * P(NOT sleep) = (from Venn diagram) = .38 d. A: A and B = addition rule for disjoint events C = addition rule (complements) (2)

If the ESP study finds a subject does better than guessing, more testing is done. The experimenter looks at cards with one of five shapes that the subject can't see. The answers of a subject who doesn't have ESP should be independent observations, each with probability 1/5 success. 900 attempts are recorded. What is the standard deviation of the number of the success count?

sqrt(n*p*(1 - p)). 12 (7)