stat 200 exam 1 fall 2018 - university of...

Stat 200 Exam 1 Fall 2018

of 7 (10 problems)

Question 1 (6 pts.) A recent study was done to test the effectiveness of acupuncture in the treatment of tension-type headaches. The subjects were 270 adult volunteers who reported having had tension headaches for at least eight days a month in the previous three months. The subjects were randomly divided into 3 groups. One received 8 weeks of a traditional form of acupuncture, one received 8 weeks of a fake acupuncture (superficial needling at non-acupuncture points) and a third group were told they were on a waiting list and received no treatment for the 8 weeks. The subjects didn’t know if they were receiving the traditional or superficial acupuncture. All subjects kept headache diaries for 8 weeks. The evaluators recorded the difference in the number of headaches before and after treatment for each subject not knowing which subjects were in which group. The average number of headaches after treatment decreased by 7 in both the real and fake acupuncture groups, but only decreased by 1.5 in the waitlist group. a) (2 pts.) Which of the following best describes this study? Choose one:

i) It’s an observational study ii) It’s a randomized controlled experiment with a placebo and “blind” evaluators. iii) It’s a non-randomized controlled double-blind experiment.

b) (2 pts.) Which of the following statements is best? Choose one:

i) This study is very strong evidence that traditional acupuncture works better than a placebo in the treatment of tension-type headaches. ii) This study is strong evidence that acupuncture (both fake and real) has a strong placebo effect since the real

acupuncture worked no better than the fake acupuncture but both worked better than nothing. iii) This study only shows an association between traditional acupuncture and reduced headaches. It does not prove or

disprove that traditional acupuncture caused fewer headaches since there are likely to be other differences between those who received the real and those who received the fake acupuncture that could confound the results.

c) (2 pts.) Which of the following are likely to confound the results of this study? Choose one:

i) Pain Tolerance- People who choose acupuncture may tolerate pain better and thus report fewer headaches. ii) Alternative Medicine- People who choose acupuncture are more likely to be taking alternative therapies such as

herbal cures and massage which could help alleviate headaches. iii) Tension -- People who seek acupuncture are more likely to have tension headaches than those who don’t. iv) All of the above are likely confounders. v) None of the above are likely confounders.

Question 2 (8 pts.) A company has 455 job openings- 70 white collar jobs and 385 blue collar jobs. 600 men and 300 women apply for the new jobs. Here’s the data: Men Women # Applied # Hired Hiring Rate # Applied # Hired Hiring Rate White Collar 200 30 15% 200 40 20% Blue Collar 400 300 75% 100 85 85% Total 600 330 55% 300 125 41.67%

a) (2pts.) Overall 55% of the men but only 41.67% of the women who applied were hired, raising the question of sexual discrimination. Assuming that the men and women were equally qualified, which job category was discriminating against women?

Choose one: i) White Collar Only ii) Blue Collar Only iii) Neither iv) Both b) (2pts.) Based only on the data above, if you’re applying for a white collar job are hiring rates better for males or females?

Choose one: i) Male ii) Female c) (2pts.) Based only on the data above, if you’re applying for a blue collar job are hiring rates better for males or females?

Choose one: i) Male ii) Female iii) Not possible to compare rates since 400 men applied, but only 100 women. d) (2pts.) Based only on the data above, are hiring rates better for males or females? Choose one: i)Better for males. ii) Better for females. iii) Same. iv) Depends on whether it’s a white or blue collar job.


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Questions 3 (4 pts.)Suppose I wanted to determine whether or not offering optional homework problems to supplement the required homework would help students learn. I randomly split the class into 2 groups—the treatment group was offered optional problems to supplement the required homework while the control group was just given the required problems. At the end of the semester I compared the exam scores of the 2 groups and saw no significant difference. Both groups had an average exam score of 80% and a SD of 20%. But only half the students in the treatment group actually did the supplemental problems. Those who chose to do the optional homework did much better than those who chose not to do it. The table below gives the data:

Treatment group Exam Average Control Group Exam Average Chose to do extra homework 90% Not applicable

Chose not to do extra homework 70% Not applicable Total 80% 80%

a) (2 pts.) Which 2 percentages are most appropriate to compare to determine whether or not the extra problems helped?

Choose on i) 80% vs. 80% ii) 90% vs. 80% iii) 80% vs. 70% iv) 90% vs. 70%

b) (2pts.) Which conclusion is most appropriate to draw from the data? Choose one: i) Giving students the option of doing supplemental problems helps them learn. ii) Giving students the option of doing supplemental problems impairs learning. iii) Those who choose to do the supplemental problems are clearly helped by them and those who choose not to do the supplemental

problems are clearly hurt by them. iv) Offering optional supplemental problems makes no difference in exam scores, the difference we see is due to self-selection.

Question 4 (9 pts.) A study found that older mothers are more likely to have autistic children than younger mothers. The study examined nearly 5 million birth records from the 1990’s in California and found 12,159 cases of autism. The median maternal age was 30 for the autistic group and 27 for all other births. Mothers over 40 were 77% more likely to have given birth to an autistic child than mothers under 25. a) (2 pts.)Based only on the information above, this study is an example of …. Choose one.:

i) Observational Study ii) Randomized Controlled Experiment without a placebo iii) Randomized Double-Blind Controlled Experiment iv) Non-Randomized Controlled Experiment

b) (2pts.)Which of the following statements is best? Choose one.

i) This study is strong evidence that maternal age causes autism. ii) This study only shows that maternal age is associated with autism: it doesn’t show whether or not maternal

age causes autism. iii) This study shows that maternal age is associated with, but definitely does not cause autism.

c) (2 pts.) The study reported that they controlled for the mothers’ education level. This means they thought education

might be a confounder so they eliminated its confounding effect. How did they do that? Choose one:

i) They divided the mothers into 2 groups by age and compared the autism rate of children born to mothers over 40 to the autism rate of children born to mothers under 25.

ii) They divided the mothers into groups by education level and compared the autism rate of children born to older mothers to the autism rate of children born to younger mothers within each education level.

iii) They eliminated the confounding effect of education by carefully sifting through the 5 million birth records and eliminating all mothers with advanced degrees.

d) ( 3 pts.) Below are either confounders or causal links or neither. Circle which is which. i) Heredity—Mothers who have autistic tendencies themselves are both more likely to have autistic kids and more

likely to be older when they give birth (since they lack social skills.) Choose one: a) Confounder b) Causal Link c)Neither

ii) Damaged DNA—As people age their DNA deteriorates which makes them more likely to have babies with birth defects including autism: Choose one: a) Confounder b) Causal Link c)Neither

iii) Poor Diet—Poor nutrition can damage neurological development. Choose one: a) Confounder b) Causal Link c)Neither.


of 7 (12 problems)

Question 5 pertains to the histogram below: (11 pts.) The histogram below represents the age at death of a large population. The height of each block is given in parentheses. (2.5)

a) (1 pt.)What percent of the population died in the 0-40 interval? Choose one: i) 10% ii) 15% iii) 20% iv) 25% v) 30%

b) (1 pt.)What is the median age of death? ______years Fill in the blank with a number.

c) (1 pt.)The average is ________the median. Fill in the blank with > (greater than), < (less than), or = . d) (1 pt.) What is the age at the 75th percentile? ______years Fill in the blank with a number

e) (1 pt.) What % of the population died at 75 years? (Assume an equal distribution throughout the interval.) ______%

f) (3 pt.) If 5 years were added to everyone’s life span how would it affect the average, median and SD of the histogram above? i) The average would … Choose one: a) increase b) decrease c) stay the same d) Not enough info given ii) The median would Choose one: a) increase b) decrease c) stay the same d) Not enough info given iii) The SD would … Choose one: a) increase b) decrease c) stay the same d) Not enough info given

g) (3 pt.) If 5 years were only added to those in the 0-20 interval, how would it affect the average, median and SD of the original histogram? i) The average would.. Choose one: a) increase b) decrease c) stay the same d) Not enough info given ii) The median would.. Choose one: a) increase b) decrease c) stay the same d) Not enough info given iii) The SD would … Choose one: a) increase b) decrease c) stay the same d) Not enough info given Question 6 (3 pts.) pertains to the 2 histograms below which depict the male and female answers from this class to the survey question: “How many pairs of shoes do you own?” a) (1 pt.) Which histogram represents the female answers? Choose one: i) Histogram A ii) Histogram B (Hint: Use common knowledge- who do you think would own more shoes?)

b) (2 pts.) The following 4 numbers (in no particular order) are the averages and medians of the 2 histograms: 7, 19, 10.152, 20.168 Fill in each of the 4 blanks below with the 4 given numbers. (Hint: 2 of the numbers are integers and 2 are decimals.) i) (1 pt.) The median of Histogram A is ________ and the median of histogram B is ________ ii) (1 pt.) The average of Histogram A is ________ and the average of Histogram B is ________

(1.5) (1.25) (1) (1) (0.5) (0.25)

0 10 20 30 40 50 60 70 80 90 100 Age at death

2.5 2.0 1.5 1.0 0.5

% per age

Histogram A Histogram B


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Question 7 ( 7 pts.) pertains to the following list of 4 numbers: 0, 2, -2, 8

a) ( 2 pts.) The average is _______ and the median is _________.

b) ( 2 pts.)The deviations from the average are : _____ ______ ______ _______

c) (1 pt.)The sum of the deviations from the average must always = _____

d) ( 2 pts.) Compute the SD. Show work using the deviations you got in part (b) above. Round answer to 2 decimal

places. Circle answer.

Question 8 (13 pts.) According to our survey, the weights of Stat 100 females follow the normal curve with Average = 130 lbs and SD= 20 lbs. (You may “round” Z scores and percentages to fit the closest line on the Normal table and you may round percentages on the table to the nearest whole number.) a) (5 pts.) What percentage of the female students weigh over 165 lbs?

i) (2 pts.) First, convert 165 lbs to a Z-score. Z-score = __________ (Show work, no work no credit)

ii) (1pt.) Which histogram’s shaded region correctly depicts the percentage of women who weigh over 165 lbs?

iii) (1 pt.) What percentage of women weigh over 165 lbs.? ______% Show calculation.

iv) (1 pt.) Women who weigh 165 lbs are at the __________ percentile of the weight distribution.

b) (3 pts.) If a student is 0.5 SD’s below average in weight. How much does she weigh and what percentile is she in? i) (1 pt.) She weighs _________ lbs. (Show work)

ii) (1 pt.) She’s in the _______ percentile. iii) (1 pt.) Which histogram’s shaded region correctly depicts the percentile?

c) (3 pts.) If a student is in the 60th percentile how much does she weigh? i) (1 pt.) To find the Z score, you should look at the middle area = _________% Fill in the blank with the correct percent. ii) (1 pt.) Z=_________ iii) (1 pt.) Weight = _________lbs. (Show work.)

d) (2 pts.) If 2 people have the same Z scores in absolute value but opposite signs then their percentiles must sum to Choose one: i) 0 ii) 25 iii) 50 iv) 100 v) Not enough info.

Choose one: a) Histogram A b) Histogram B c) Histogram C I d) Histogram D

Choose one: a) Histogram A b) Histogram B c) Histogram C d) Histogram D


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Question 9 (4 pts.) pertains to a well-shuffled deck of 52 cards. A deck of cards has 4 suits: clubs, diamonds, hearts and spades. There are 13 cards in each suit: 2 through 10, jack, queen, king, ace. (So there are 4 Aces, 4 Jacks and 13 Hearts.) a) (1 pt.) Draw 2 cards with replacement. What is the chance that the first card is an Ace and the second is a Heart? i) 4/52 + 13/52 ii) 4/52 + 13/52 -1/52 iii) 4/52*3/51 iv) 4/52*13/51 v) 4/52*13/52 b) (1 pt.) Draw 2 cards without replacement. What is the chance that the both cards are Aces? i) 4/52 + 4/52 ii) 4/52 + 4/52 -1/52 iii) 4/52*3/51 iv) 4/52*4/51 v) 4/52*4/52 c) (1 pt.) Draw 2 cards without replacement. What is the chance that the first card is an Ace and the second is a Jack? i) 4/52 + 4/52 ii) 4/52 + 4/52 -1/52 iii) 4/52*3/51 iv) 4/52*4/51 v) 4/52*4/52 d) (1 pt.) Draw one card. What is the chance that it’s either a Jack or a Heart? i) 4/52 + 13/52 ii) 4/52 + 13/52 -1/52 iii) 4/52*3/51 iv) 4/52*13/51 v) 4/52*13/52 Question 10 (7 pts.) pertains to rolling fair dice.

a) (1 pt.) Two dice are rolled. What is the chance that the sum of the spots is 4? i) 2/36 ii) 3/36 iii) 4/36 iv) 5/36 v) 1/6*1/6 vi) 1/6 + 1/6

b) (1 pt.) One die is rolled 4 times. What is the chance of getting all 4’s? i) (5/6)4 ii) (1/6)4 iii) 1- (5/6)4 iv) 1- (1/6)4 v) 4/6

c) (1 pt.) One die is rolled 4 times. What is the chance of not getting all 4’s?

i) (5/6)4 ii) (1/6)4 iii) 1- (5/6)4 iv) 1- (1/6)4 v) 4/6

d) (1 pt.) One die is rolled 3 times. What is the chance of getting no 2’s? i) (5/6)3 ii) (1/6)3 iii) 1- (5/6)3 iv) 1- (1/6)3 v) 3/6

e) (1 pt.) One die is rolled 3 times. What is the chance of getting at least one 2?

i) (5/6)3 ii) (1/6)3 iii) 1- (5/6)3 iv) 1- (1/6)3 v) 3/6 f) (1 pt.) One die is rolled twice. What is the chance that the first roll is a 2 and the second roll is a 3?

i) 1/36 ii) 2/36 iii) 6/36 iv) 11/36 v) 12/36

g) (1 pt.) One die is rolled twice. What is the chance that the first roll is a 2 or the second roll is a 3? i) 1/36 ii) 2/36 iii) 6/36 iv) 11/36 v) 12/36

Question 11 (4 pts.) Only about 0.1% of young women who participate in routine screening have breast cancer. Suppose 90% of women who have breast cancer will correctly get a positive result and that 20% of women without breast cancer will also get a positive result (false positives). Fill in blanks 1 and 2 below for a typical sample of 10,000 young women (8 pts) Tests Positive Tests Negative Total Has Breast Cancer

Cell 1 10

Does NOT have Breast Cancer

Cell 2 7992 9,990

Total

2007 10,000

a) (1 pt.) The correct number for Cell 1 is a) .1 b) 1 c) 2 d)9 e)9.9

b) (1 pt.) The correct number for Cell 2 is a) .1 b) 20 c) 99 d) 999 e) 1998

c) (1 pt.) If a woman gets a positive result, the chance she really has breast cancer is closest to..

a) 90% b) 20% c) 50% d) .1% e) 0.45%

d) (1 pt.) If a woman gets a negative result, the chance she really has breast cancer is closest to.. a) 1% b) 0.1% c) 0.0125% d) .1% e) 0.45%


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Question 12 (24 pts.) 25 draws are made at random with replacement from the box containing 6 tickets: 0 4 4 6 6 10

(SD of box = 3) a) (2 pts.) The smallest the sum of the 25 draws could possibly be is ______ and the largest is _______.

b) (2 pts.)What is the EV of the sum of the 25 draws? (Show work, circle answer.) No work, no credit.

c) (2 pts.)What is the SE of the sum of the 25 draws? (Show work, circle answer.) No work, no credit.

d) (9 pts) Look at the 3 probability histograms below. They refer to the box above. One shows the contents of the box, one shows the sum of 2 draws with replacement from the box and one shows the sum of 25 draws with replacement from the box. Which is which?

i) (3 pts.)Histogram ____ is for the sum of 2 draws, Histogram___ is for the sum of 25 draws and Histogram ____ is for the contents. (Fill in the 3 blanks with A, B, or C)

ii) (6 pts.) The numbers on the X axes in the histograms below are missing. Fill in the blanks.

e) (3 pts.) 100 draws are made at random from the box at the top of the page. The EV of the sum = 500 and the SE of the sum = 30. Use the normal approximation to find the chance that the sum of 100 draws will be above 524?

i) (2 pts.) First calculate the Z score. Show work. Circle answer.

ii) (1 pt.) Now use the normal table to find the chance that the sum of the 100 draws will be above 524. Round the middle area given in the table to the nearest whole number.

Chance =__________%

f) (1 pt.)What is the EV of the average of the 100 draws from the same box above? _______

g) (1 pt.) What is the SE of the average of the 100 draws? ________ Show work. (Remember SD of box = 3)

h) (4 pts.) Now suppose you draw at random with replacement from the same box above, but this time you’re only interested in the percent of 10’s. What is the EV and SE of the percent of 10’s in 100 draws?

| | | | ___ ___ ___ ___ (2 pts.) Write the correct number in each of the 4 blanks.

____ ___ ___ (3 pts.) Write the correct number in each of the 3 blanks

i) (2 pts.) Draw a new box. Label the two tickets with the correct numbers, and write how many of each above them. ii) (1 pt.) EV% of 10’s in 100 draws =______________% (Round to 2 decimal places.) iii) (1 pt) SE% of 10’s in 100 draws is closest to… Show work below. Circle one: a) 0.37% b) 0.47% c) 0.5% d) 3.7% e) 4.7% f) 5%

Histogram A

Histogram B

Histogram C

__ (1 pt) Write the correct number that goes in the center (same as the EV of the sum)


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Question 1 (6 pts.) A Fox News Poll asked a random sample of 937 adults nationwide the following question: “Do you believe in God?” At the same time CNN posted the same question on their website as a “Quick Vote” question where anyone who wants to can cast their vote. Here’s the results of both surveys:

Yes No Sample Size CNN Quick Vote 52% 48% 7362 Fox New Poll 92% 8% 937

a) (2 pts.) As you can see, the results of the 2 polls are quite different. Which survey gives a better estimate of the percentage of all

US adults who would say they believe in God? Circle one: i) The Fox survey because the sample was randomly selected ii) The CNN Quick Vote survey because the sample size is much larger. iii) Both surveys have about the same level of accuracy since difference in sample sizes is compensated by the differences in selection method.

b) (2 pt) The SE of the percentage of people in the sample Fox.com who answered "YES" is closest to… Circle one:

i) .92 * .08

937*100% ii) 937 * .92 * .08 *100%

iii) Impossible to compute a SE for this sample because it can’t be translated into a box model. c) (2 pt) The SE of the percentage of people in the CNN Quick Vote who answered "YES" is closest to …Circle one:

i) .52 * .48

7362*100% ii) 7362 * .52 * .48 *100%

iii) Impossible to compute a SE for this sample because it’s not a random sample so it can’t be translated into a box model.

Question 2 (5 pts.) A recent poll asked a random sample of 1,056 registered voters nation-wide: Would you say that Donald Trump is of sound mind, or not? 30% responded that they would say “Yes, he is.”

a) (3 pts) Circle whether each of the statements below is true or false : i) The expected value for the % of all registered female voters nation-wide who would answer “Yes” is 30%. a)True b)False ii)The expected value for the % of all registered voters nation-wide who would answer “Yes” is 30%.

a)True b)False iii)The expected value for the % of all registered Republican voters nation-wide who would answer “Yes” is 30%.

a)True b)False

b)(2 pts.) Is it possible to compute an approximate 95% confidence interval for the percent of all registered voters nation-wide who would say “Yes? Choose one:

i) Yes, an approximate 95% confidence interval is 30% +/- 2.8% ii) No, because we’re not given the SD of the sample. iii) No, because we cannot infer with 95% confidence the attitude of 180 million registered voters from data based on a

sample of only 1,056 randomly selected Americans.

Question 4 (6 pts.)

a) (2 pst.) In a pre-election presidential poll in a close race, a polling organization wanted the margin of errors for a 95% confidence interval to be 4%, how many people should they poll? (Assume SD=0.5)

i) 400 ii) 625 iii) 1111 iv) 2500 v) 10,000

b) (2 pts.) If you wanted to divide your margin of error by 2, you’d have to ____________your sample size by ____. Fill in the first blank with either “multiply” or “divide” and the second blank with a number.

c) (2 pts. )In a national election about 160 million people vote. In Ohio only about 6 million people vote. How would you adjust your answer in part (a) for a pre-election poll in Ohio?

i) keep it about the same. ii) make it significantly larger iii) make it significantly smaller


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Question 5 (17 pts.) To estimate online dating use among all 40,000 UI undergrads. 400 randomly selected UI undergrads were asked “How many “dates” have you had through an online dating service?” The average response was 10 dates with a SD of 15. Part 1 (10 pts)

a) ( 2 pt.) Which most closely resembles the relevant box model? Circle one: i) The box has 400 tickets marked with “1”s and “0”s ii) The box has 40,000 tickets marked with “1”s and “0”s iii) The box has 40,000 tickets, the exact average is unknown but estimated from the sample. iv) The box has 400 tickets with an average of 10 and a SD of 15.

b) (1 pt.) How many draws from the box? ________ c) (1 pt.) The draws are made… Choose one: i) with replacement ii) without replacement d) (1 pt.) The best estimate for the average number of dates all UI students had using an online dating service is _____ e) (2 pt.) What is the SE of the sample average? ________

f) (2 pts) The responses do not follow the normal curve. Is it still possible to construct a 95% confidence interval for the average of all 40,000 students? If so, fill in the upper and lower limits in the blanks provided.

Choose one: i) No, if the data does not follow the normal curve, it’s never possible to construct confidence intervals

ii) No, it’s not possible to construct confidence intervals for averages.

iii) Yes, even though the data doesn’t follow the normal curve, the probability histogram for the average of 400 draws will come pretty close to following the normal curve, so a 95% confidence interval is: (_______ to _______)

g) (1 pt.) Suppose our sample size was < 25 with the same average and SD, should we use Z or t curves to compute Confidence Intervals? Choose one: i) Z ii) t iii) neither

Part 2 (3 pts.) The study also asked the 400 students what percent of their dates were arranged from an online service. The average was 45 percent with a SD of 10 percent and the data roughly followed the normal curve. a) (1 pt.) The relevant box model for this question contains tickets with Choose one:

i) Numbers ranging from 0 to 100 ii) Only “1”s and “-1”’s iii) Only “1”s and “0”s

b) (1 pt.) An approximate 95% CI for the average percent for all UI students is (_________ to ________) c) (1 pt.) Suppose our sample size was < 25 with the same average and SD, should we use Z or t curves to compute Confidence Intervals? Choose one: i) Z ii) t iii) neither

Part 3 (4 pts) The study also asked the 400 students whether or not they ever used an online service. 45% of the students answered “Yes”. a) The relevant box model for this question contains tickets with Choose one:

i) Numbers ranging from 0 to 100 ii) Only “1”s and “-1”’s iii) Only “1”s and “0”s .

b) What is the SE of the sample percent? __________ c) An approximate 95% CI for the % of all 40,000 students who would answer “Yes” is (_______% to ______%) d) Suppose our sample size was < 25 and 45% answered “Yes”. Would it be appropriate to use the t curve to compute Confidence Intervals? Choose one: i) Yes ii) No


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Question 6 (15 pts.) A random sample of 10 male mice was taken from a population of 500 male mice. The average weight of the mice in the sample was 19.2 grams with a SD=1.8 grams. Part 1 (6 pts.) Compute a 95% CI for the average weight of all 500 mice using the t distribution. (Assume the weights of the male mice in the population follow the normal curve.) a) (2 pts.) SE+ = ___________ Show work.

b) The degrees of freedom = 9. What is t* (the critical value of t) for a 95% Confidence? t*= ______

c) (2 pts.) 95% CI = (___________ to ____________) Put the lower number first.

d) If you computed a 95% CI using the Normal curve it would be ___________ the CI in part (c) Choose one: i) wider than ii) narrower than iii) the same width as

For Parts 2-4, perform the stated significance tests using the same sample data as in Part 1. (Assume we formulated our hypotheses before looking at the sample data.) Part 2: (3 pts.) Perform a t-test to test Ho : µ= 18 grams. vs Ha : µ ≠ 18 grams. (µ is the average weight of all 500 mice in the population.) a) (2 pts.) t stat= _________ b) (1 pt.) Reject the null at α = 0.05? Choose one: i) Yes ii) No Part 3 (2 pts.) Repeat Part 2 for a 1-sided test: Ho : µ= 18 grams. vs Ha : µ > 18 grams.

a) Reject the null at α = 0.05 ? Choose one: i) Yes ii) No Part 4 (4 pts.) Suppose you knew the SD of the 500 mice = 1.8 grams, then you could perform a Z test. Repeat Part 3 by using the Z test. a) (1 pt.) SE = ________ b) (1 pt.) Z= _________ c) (1 pt.) p-value = _______ d) (1 pt.) The Z test will yield a __________ p-value than the t test.

Choose one: i) smaller ii) larger iii) exactly the same

Question 7 (5 pts.) A nation-wide poll asked a random sample of 400 men and 400 women if they thought sexual harassment would always be a problem in the US. 55% of the women answered “Yes” and 44% of the men answered “Yes”. The null hypothesis is that the 11% difference between female and male responses is just due to chance and doesn’t reflect a real difference between females and males in the population. a) Which of the following most accurately describes the null box(es)?

a) There are 2 null boxes, one with 800 tickets marked with "0"s and "1"s and one with 400 tickets with 0’s and 1’s. b) There are 2 null boxes, each with millions of tickets, and each with the same percentage of “1”’s. c) There are 2 null boxes, each with millions of tickets. One box has 55% "1"s, and the other has 44% "1"s.

b) Assuming the null to be true, the SE for the females’ sample percentage is about 2.5% and the SE for the males’ sample percentage is about 2.5%. The SE for the difference of the 2 sample percentages is closest to … a) 0.75% b) 1.75% c) 2.5% d) 3.5% e) 5% c) The Z statistic for testing the null hypothesis is closest to … a) .75 b) 1.5 c) 2.8 d) 3.14 e) 11 d) Suppose the P-value (for the 1-sided test) is about 0.085%, what do you conclude?

i) Cannot reject the null. It’s plausible that there is no female/male difference on this question among US adults ii) Reject the null and conclude that there is extremely strong evidence that our sample difference reflects

a real female/male difference on this question among US adults. e) If we did a chi-square test for independence to test the null hypothesis that people’s answers to this question are independent of whether they are male of female our p-value would be about … i) 0.085% ii) 0.17% iii) 0.00425% iv) not enough info


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Question 8 (5 pts): Suppose the instructor of a large class with an enrollment of over a thousand students claims to always grade on a curve so that 20% of the students receive A’s, 40% B’s, 30% C’s and 10% D’s or F’s. To test the claim I take a random sample of 50 students from the course. Here are the results: Grade Percents Claimed by

Instructor Observed # Expected # Obs -Exp (Obs-Exp)2

A 20% 5 -5 25/10 B 40% 20 0 C 30% 20 D or F 10% 5 0 Total 100% 50 50 a) To test the null hypothesis that our observed data fits the letter grade percentages claimed by the instructor we’d do …

i) the one-sample z test ii) the two-sample z test iii) the chi-square test for "goodness -of-fit"

b) The table above is missing all 4 expected numbers, which of the following is the missing column? i) 10 ii) 20 iii) 25 iv) 12.5 20 40 25 12.5 15 30 25 12.5 5 10 25 12.5 c) The value for C is missing in the Obs –Exp column, fill in the missing blank. i) 0 ii) 5 iii) 25 iv) not enough information to determine d) To compute the proper test statistic you’d have to sum the 4 values in the last column. The test statistic is closest to

i) 0 ii) 2.5 iii) 4.17 iv) 5 v) not enough information to determine e) The number of degrees of freedom is 3. What is the p-value?

i) < 1% ii) between 1% and 5% iii) between 5% and 10% iv) between 10% and 30% v) 30% and 50%

Question 9 (2 pts.) Suppose a well-designed randomized controlled double-blind experiment is done to test a new drug. The null hypothesis is that the drug works no better than a placebo. A significance test is done and P is computed to be 5%. Which statement best describes what is meant by a P value of 5%?

a) It means that even if the null hypothesis was true and the drug didn’t work, we would still see evidence this strong or stronger 5% of the time just by chance.

b) It means there is a 5% chance that the null is true. c) It means that there is a 5% chance that the null is false. d) It means that we have proof the drug certainly works.

Question 10 (2 pts.) An experiment on ESP is repeated 500 times. Suppose there is no ESP, and the experiment is done correctly with no cheating. About how many of the experiments would you expect to find statistically significant evidence for ESP, that is how many of the results would get p-values < 5%? (Note, answer how many, not what percent.)

a) 0 experiments b) 0.05 experiments c) 5 experiments d) 25 experiments e) 50 experiments Question 11 (2 pts.) The convention is to reject the null when p < 5% and call the result “statistically significant”. Is there any particular mathematical justification for this?

a) Yes, the shape of the normal curve, the t-curves and the chi-square curves all have sharp dropping off points that make 5% a natural dividing line.

b) Yes, 5% is the most likely percent to avoid the mistake of rejecting the null when the null is really true. All other percents would yield a higher likelihood of making that mistake.

c) No, there’s no particular mathematical justification for choosing 5%. d) Yes, 5% is the most likely percent to avoid the mistake of not rejecting the null when the null is really false. For

example, if we chose a percent that was higher or lower than 5% we’d be more likely to not think a drug worked no better than a placebo when it really did.


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Question 12 pertains to a screening test for breast cancer (10 pts.) If a patient has breast cancer there’s a 90% chance that the screening test will correctly give a positive result. But if the patient does not have breast cancer, there’s a 20% chance that the test will incorrectly give a positive result.

Let’s put this situation into the context of a Hypothesis test. a) (2 pts) The conventional null hypothesis (H0) is that the patient …

i) does not have breast cancer ii) has breast cancer iii) may have breast cancer at significance level 5% iv) has a 10% chance of having breast cancer

b) (2 pts) A Type II error can only occur ….

i) when the patient has breast cancer. ii) when the patient does not have breast cancer. iii) Either when the patient has breast cancer and the significance level is set too high or when the patient does

not have cancer and the significance level is set too low. c) (2 pts) If a person has breast cancer, what’s the probability she’ll test negative? __________%

What type of error would this be? Circle one: i) Type I error ii) Type II

d) (2 pts) If a person does not have breast cancer, what’s the probability she’ll test positive? __________% What type of error would this be? Circle one: i) Type I error ii) Type II

e) (2 pts) If you change the cut-off of the screening test to decrease the probability of false negatives what would happen to the probability of false positives? i) Increases ii) Decreases iii) Stays the Same

Question 13 (7 pts.) Look at the histograms below. Label the 3 areas (indicated by 2 arrows and 1 blank) that represent Type I and Type II errors and Power by writing “Type I”, “Type II”, or “Power” above each arrow and blank. Question 14 (8 pts) An Instructor claims that his take home exams only take his 500 students an average of 60 minutes to complete with a SD=20 minutes (and roughly follow the normal curve), but I think the average is higher, at least 70 minutes with the same SD=20 minutes. I plan to take a random sample of students in the class and ask them how long it took them to complete the exam. a) (4 pts.) How big a sample (n) would I need to detect an effect size of at least 10 minutes with 80% Power at significance level α= 0.05. i) Zα = ____ ii) Zβ = ____ iii) n = _______ Show work.

b) (4 pts.) Suppose I could only manage to take a sample of n=16. What’s the Power of the test? i) SE = _____ ii) Zβ = ______ iii) Power = ________ Show work.

(4 points) Which of the following will increase the Power of the test? Circle either “yes” or “no”. a) Increasing the probability of a Type I error i) Yes ii) No b) Increasing D, the effect size (HA-H0) i) Yes ii) No c) Increasing the sample size i) Yes ii) No d) Increasing the SD i) Yes ii) No

(1pt.)

(1pt.) (1pt.)


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z Area z Area z Area 0.00 0.05 0.10 0.15 0.20

0.25 0.30 0.35 0.40 0.45

0.50 0.55 0.60 0.65 0.70

0.75 0.80 0.85 0.90 0.95

1.00 1.05 1.10 1.15 1.20

1.25 1.30 1.35 1.40 1.45

0.00 3.99 7.97

11.92 15.85

19.74 23.58 27.37 31.08 34.73

38.29 41.77 45.15 48.43 51.61

54.67 57.63 60.47 63.19 65.79

68.27 70.63 72.87 74.99 76.99

78.87 80.64 82.30 83.85 85.29

1.50 1.55 1.60 1.65 1.70

1.75 1.80 1.85 1.90 1.95

2.00 2.05 2.10 2.15 2.20

2.25 2.30 2.35 2.40 2.45

2.50 2.55 2.60 2.65 2.70

2.75 2.80 2.85 2.90 2.95

86.64 87.89 89.04 90.11 91.09

91.99 92.81 93.57 94.26 94.88

95.45 95.96 96.43 96.84 97.22

97.56 97.86 98.12 98.36 98.57

98.76 98.92 99.07 99.20 99.31

99.40 99.49 99.56 99.63 99.68

3.00 3.05 3.10 3.15 3.20

3.25 3.30 3.35 3.40 3.45

3.50 3.55 3.60 3.65 3.70

3.75 3.80 3.85 3.90 3.95

4.00 4.05 4.10 4.15 4.20

4.25 4.30 4.35 4.40 4.45

99.730 99.771 99.806 99.837 99.863 99.885 99.903 99.919 99.933 99.944 99.953 99.961 99.968 99.974 99.978 99.982 99.986 99.988 99.990 99.992 99.9937 99.9949 99.9959 99.9967 99.9973 99.9979 99.9983 99.9986 99.9989 99.9991


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Student’s curve, with degrees of freedom shown at the left of the table

The shaded area is shown along the top of the table is shown in the body of the table

Student’s t-TABLE


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Chi-SquareTable

Degreesoffreedom↓ 30% 10% 5% 1% 0.1% ← p-value

1 1.07 2.71 3.84 6.63 10.832 2.41 4.61 5.99 9.21 13.823 3.66 6.25 7.81 11.34 16.274 4.88 7.78 9.49 13.28 18.475 6.06 9.24 11.07 15.09 20.526 7.23 10.64 12.59 16.81 22.467 8.38 12.02 14.07 18.48 24.328 9.52 13.36 15.51 20.09 26.129 10.66 14.68 16.92 21.67 27.8810 11.78 15.99 18.31 23.21 29.5911 12.90 17.28 19.68 24.72 31.2612 14.01 18.55 21.03 26.22 32.9113 15.12 19.81 22.36 27.69 34.5314 16.22 21.06 23.68 29.14 36.1215 17.32 22.31 25.00 30.58 37.7016 18.42 23.54 26.30 32.00 39.2517 19.51 24.77 27.59 33.41 40.7918 20.60 25.99 28.87 34.81 42.3119 21.69 27.20 30.14 36.19 43.8220 22.77 28.41 31.41 37.57 45.3121 23.86 29.62 32.67 38.93 46.8022 24.94 30.81 33.92 40.29 48.2723 26.02 32.01 35.17 41.64 49.7324 27.10 33.20 36.42 42.98 51.18

←Chi-square

stat 200 exam 1 fall 2018 - university of...

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