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Statistics 1100 Sections 15 - 24 Test#3 Version 1

November 22, 2019

Instructions for Completing this test:

1. On this first page, put your name, your PeopleSoft Number and your Section Number.  If you do not know your section number please check with the Room monitor when you hand in your test.

2.  On the top left section of the bubble sheet, put your last name, a blank space and your first name.  Then, shade in the bubbles below the letters of your name.

3. At the bottom left section, enter your PeopleSoft Number as the Identification Number and shade in the bubbles below the numbers.

4. Next, in Column K enter your version number, which is ‘1’ for this test version and shade in the number ‘1’ in that column.

5. Last, in Columns L and M, enter your Discussion Section Number and shade in the bubbles below the numbers.

6. Now you are ready to start the test.  Your test is graded electronically using your bubble sheet so please double check all your answers and make sure they are correctly bubbled in.   When you have completed your test, hand in your completed test and your bubble sheet to one of the proctors.  Please show the proctor your UConn ID.  Your test will be returned to you in your discussion section. Please circle all your answers on the test itself.  Your bubble sheet will not be returned to you.

7. All the questions have been checked for errors so we believe that everything is correct.  If you think something is ambiguous please write your reasoning on the back of the test and you can make an appointment with Kate to address this.  There will be no questions answered during the test itself.

8. Your test will be graded out of ‘20’ questions.  So each question is worth 5 points.  If you get all ‘21’ questions correct, your score will be 105. Your grades will be posted in HuskyCT (lecture site) by Tuesday, November 26^th.  Answers to all versions will also be posted on the Homepage of HuskyCT (lecture site).

9. Have your Student ID ready to present to one of the proctors when you hand in your scantron.

1. Connecticut Lotto is a Connecticut lottery game in which the player picks 6 numbers ranging from 1 to 44.  The price of a ticket to play this game is $1.00.

            Shown here are the various prizes and the probability of winning each prize.

	Match all 6 digits	Match 5 out of 6 digits	Match 4 out of 6 digits	Match 3 out of 6 digits	Match 0, 1 or 2 digits out of 6
Prize (x)	$1,000,000	$2000	$50	$2	$0
Probability

Fill in the missing probability value in the table and then calculate the expected winnings for this game and decide whether it is worthwhile, in the long run, to play.

a.) The expected value is $0.33. This means that in the long run you lose $0.67 for

      every ticket you purchase so it is not worthwhile to play the game.

b.)  The expected value is $1.67. This means that in the long run you will make a profit of

     $0.67 for every ticket you purchase so it is worthwhile to play the game.

c.)  The expected value is $2.07. This means that in the long run you will make a profit

       of $1.07 for every ticket you purchase so it is worthwhile to play the game.

d.) The expected value is $0.55. This means that in the long run you will lose $0.45

     for every ticket you purchase so it is not worthwhile to play the game.

2. According to the Democratic Presidential candidates, 51% of young Americans support tuition-free college. Suppose a random sample of 30 UConn students is selected.  What is the probability that at least 10 of them support tuition-free college?

a.)  0.0164

b.)  (0.51)³⁰

c.)  0.9836

d.)  0.3849

e.)  0.6151

3. Elizabeth Warren, one of the current Democratic Presidential candidates, claims that 70% of all Americans support her “Medicare for All” proposal. Suppose a random sample of 20 members of a focus group in Iowa is selected.  What is the probability that fewer than 15 of them support her Medicare for all proposal?

a.) 0.0002

b.) 0.0171

c.) 0.0308

d.) 0.5836

e.) 0.1843

Information for Questions 4 and 5:  The question of building a wall between the U.S. and Mexico has been discussed and debated ever since President Trump made it a major campaign issue.  How do Americans feel about this wall? A summary of opinion polls conducted in 2013 showed that 65% of all Americans supported building a wall. Recently (March 2019), Quinnipiac University conducted a poll to see how Americans now feel about this controversial topic. A sample of 1120 randomly selected Americans was asked if they supported or opposed building the wall.  Let X = the number of respondents in the sample who support building the wall.

4. Let’s assume, for the moment, that American attitudes have not changed (that is,

     p  = 0.65). Find µ and s for the binomial random variable X.

a.)  µ = 728 and s = 27.66.

b.)  µ = 0.65 and s = 0.2275

  c.)  µ = 0.65 and s = 0.35

d.)  µ = 728 and s = 15.96

      e.)  µ = 728 and s = 254.8

5. If the sample in March 2019 resulted in 548 persons who said they supported building   a wall between the U.S. and Mexico, would you conclude that Americans have, in fact, become less willing to support the building of the wall?

a.)  Yes, because observing 548 respondents who support the wall in the sample of 1120 Americans is more than 3 standard deviations from the mean, µ, and this is highly unlikely.

b.) No, because the 548 respondents who say they support the wall is only 35% lower than the mean value in 2013 and that is within a (µ ± 2σ) interval.

  c.) No, because observing 548 respondents who support building the wall in the

                 sample of 1120 Americans is only 1 standard deviation from the mean, µ, and this

                 is not unlikely.

d.) Yes, because the average for the 2013 sample is 728 Americans who supported the wall so any value lower 728 in a sample of 1120 would indicate that American attitudes have changed since the survey of 2013.

        e.)  You cannot make any comments about this particular survey unless you know that the underlying population is approximately normal.

Information for Questions 6 - 8: Imagine that a friend of yours has a habit of being late when he is supposed to meet you.  Let the random variable X represent the time from when you are supposed to meet your friend until he shows up.  Further suppose that your friend is always between 10 minutes late (X=10) and 20 minutes late (X=20) and each value of X is equally likely.  I used the random number generator in Minitab to generate 200 values for the Uniform random variable, X, on the interval [10, 20]. The output is displayed in the histogram shown here.  To make the frequencies easier to read I have displayed the values directly above each bar. 

6.  Use the relative frequency definition of probability with the data displayed in the histogram to calculate the probability that a number between 15 and 18 was generated.

a.) .325

b.) .650

c.) .840

d.) .420

e.) .642

Information for Questions 7 and 8: Model this uniform random variable (described in Question #6) with a uniform graph as shown below:

                                   10                                  20

7.  The height of the uniform function from 10 to 20 is:

a.) 1

b.) .10

c.) 1/2

d.) .25

e.) 0.375

8.  Use the uniform distribution function to find the theoretical probability that the random variable X lies between 15 and 18.  That is, find P(15 ≤ X ≤ 18).

            a.)  0.375

            b.)  .20

            c.)  .30

            d.)  .40

            e.)  .50

Information for Questions 9 and 10:  I asked all 500 students in Statistics last semester to flip a biased coin that I have 10 times and to record the number of Heads.  I know that this biased coin has P(Heads) = 0.80 and P(Tails) = 0.20.  Define the random variable X as the number of heads for each individual student’s experiment.  So, X is a binomial random variable with n = 10 and p = 0.80.

Here is a table of the results of the 500 experiments:

X:     No. of Heads	0	1	2	3	4	5	6	7	8	9	10
Freq:	0	0	0	5	5	25	20	115	170	100	60

9. Use the output in the above table to calculate a relative frequency estimate of

      .

a.)   0.196

b.)   0.320

c.)   0.213

d.)   0.426

e.)   0.25

10.  Now use your calculator and the values for n and p given in the information above to

     calculate the theoretical probability:   .

a.)   0.2206

b.)   0.2044

c.)   0.4547

d.)   0.3158

e.)   0.25

11.  Heart disease is the major cause of death in the U.S. One of the key factors that contributes to heart disease is High Blood Pressure. The American Heart Association reports that a Blood Pressure score of 170 or higher represents high risk of heart disease.  A study of overweight men in the 30 - 40 year old age group who exercised less than 30 minutes per week reported a mean of 180 and a standard deviation of 20.  If the total cholesterol scores have a normal distribution, what proportion of all men in this group fall in the high-risk category?

a.) .3085

b.) .6915

c.) .997

d.) .6815

e.) .2266

Information for Questions 12 - 15: Cardiologists often ask their patients to work with nutritionists to improve their diets.  One item they are interested in controlling is sugar consumption. According to the Center for Disease Control, sugar consumption in the U.S. remains at record high levels.  The mean amount of “added sugar” (not including naturally occurring sugars found in fruits and other whole foods) consumed per day is 16.1 teaspoons with a standard deviation of 3.5 teaspoons. The data are normally distributed.

12.  The first thing one cardiologist decides to do is to calculate a “mid-range” of values for the sugar consumption of his patients.  The cardiologist decides to find the cut-off  values for the middle 80% of all his patients.  Find these 2 cut-off  values, such that 80% of all his patients fall within this daily sugar consumption range.  (Round your answers to the nearest tenth.)

a.)  12.4 to 30.4 teaspoons

b.)  11.6 to 20.6 teaspoons.

c.)  13.7 to 18.5 teaspoons.

d.)  12.5 to 19.7 teaspoons.

e.)  13.2 to 19.0 teaspoons.

13.  Because of the ever-increasing problem of obesity and heart disease here in the United States, this cardiologist decides to discuss healthy eating habits with his patients who fall in the upper 5% in terms of daily sugar consumption.  What is this cut-off value for the upper 5% in terms daily sugar consumption? (Round your answer to the nearest tenth.)

a.)  20.6 teaspoons.

b.)  19.7 teaspoons.

c.)  21.9 teaspoons.

d.)  19.0 teaspoons.

e.)  15.6 teaspoons.

14.  This cardiologist gets a report on his patients’ sugar consumption from the nutritionist.  He becomes concerned that his patients seem to consume even more sugar that what is in a typical American diet.  He decides to take a random sample of 100 of his patients and calculate the mean sugar consumption, , of these 100 patients.  The mean sugar consumption for these 100 patients is 17.3 teaspoons.  Let’s assume, for the moment, that his patients are no different than the general population of Americans with µ (daily sugar consumption) = 16.1 teaspoons and       σ = 3.5 teaspoons and that his sample of 100 patients is one random sample from this population.  What is the probability of obtaining a mean sugar consumption of 17.3 teaspoons (or more) in his sample of 100 patients if his patients are similar to the general population?  That is, calculate P(≥ 17.3).

a.) 0.0025

b.) 0.0051

c.) 0.3006

d.) 0.3949

e.) 0.0003

15.  Based on the probability that you obtained in Example 14, do you think the cardiologist is justified in his concern that his patients consume more sugar than what is the typical amount in an American diet?

a.) Yes, he is justified in his concern because P(≥ 17.3) is so small.  It is very unlikely to obtain a mean sugar consumption of 17.3 teaspoons (or more) if, in fact, the mean sugar consumption for his patients is actually equal to 16.1 teaspoons.

b.)  Yes, he is justified in his concern because P(≥ 17.3) is fairly large.  It is not unusual to obtain a mean sugar consumption of 17.3 teaspoons (or more) if, in fact, the mean sugar consumption for his patients is actually equal to 16.1 teaspoons.

c.) No, he is not justified in his concern because P(≥ 17.3) is fairly large.  It is not unusual to obtain a mean sugar consumption of 17.3 teaspoons (or more) if, in fact, the mean sugar consumption for his patients is actually equal to 16.1 teaspoons.

d.) No, he is not justified in his concern because P(≥ 17.3) is so small.  It is very unlikely to obtain a mean sugar consumption of 17.3 teaspoons (or more) if, in fact, the mean sugar consumption for his patients is actually equal to 16.1 teaspoons.

16. Suppose you take random samples from a population and calculate  for each of your random samples.  What effect does your sample size have on the mean () and the standard deviation () of the sampling distribution of .  

a.) As n increases, the mean and standard deviation will increase.

b.)  As n increases, the mean increases and the standard deviation decreases.

c.)  As n increases, the mean will be unchanged, but the standard deviation will increase.

d.) As n increases, the mean and standard deviation will decrease.

e.)  As n increases, the mean will be unchanged, but the standard deviation will decrease.

Information for Questions 17 - 19:  One of the most controversial issues Americans face today is the issue of gun control.  Consider the following two questions that were asked in a recent Gallup poll.  Each poll used a sample size of n = 1000.

Question#1: Do you support a ban on the possession of assault rifles?

        Result:  48% (480 out of 1000) of those surveyed said ‘Yes.”

Question#2:  Do you support a ban on the possession of handguns?

      Result: 28%  (280 out of 1000) of those surveyed said “Yes.”

17.  Construct a 90% confidence interval for the percentage of all Americans who support a ban on the possession of assault rifles.

a.) (41.8% to 56.2%)

b.) (44.9% to 51.1%)

c.) (45.4% to 50.6%)

d.) (41.5% to 56.5%)

e.) (40.4% to 57.6%)

18.  Construct a 90% confidence interval for the percentage of all Americans who support a ban on the possession of handguns.

a.) (20.3% to 33.7%)

b.) (25.7% to 30.3%)

c.) (19.0% to 35.0%)

d.) (25.2% to 30.8%)

e.) (19.4% to 34.6%)

19.  Do the results of these two questions indicate that Americans feel differently about the right to possess assault rifles and the right to possess handguns?

a.)  Yes, because the confidence intervals constructed for the two survey questions do not overlap.

b.) No, because the confidence intervals constructed for the two survey questions overlap.

c.) Yes, because the two sample percentages are different.

d.) No, because the two sample percentages fall within a ( ± 2 standard deviation interval) which indicates an insignificant difference in the two results.

e.) Based on the Central Limit Theorem, no conclusions can be drawn unless the sample sizes are increased to n = 1500.

Information for Question 20:  When preparing a national standardized test to be given to sixth graders, The Standard Test Company gave a version of the test to a random sample of 50 sixth graders and timed how long it took them to finish the test.  The average time () required to finish the test for the sample was 135 minutes with a standard deviation (s) of 20 minutes.

20.  Estimate the true average time, µ, for all students to finish the test with a 95% confidence interval.

a.) (126.2, 143.8)

b.) (127.42, 142.58)

c.) (129.3, 140.7)

d.) (134.1, 135.9)

e.) (130.3, 139.7)

Information for Question 21: Many professional polls use a 95% Confidence Interval to estimate a population proportion and report a margin of error of 2.5%.  How large a sample is necessary?

a.)  1068

b.)  664

c.)  1537

d.)  385

e.) 166

  You have just completed Test 3, Version1.   Bubble in ‘1’ in Column K and your Section # in Columns L and M.  Bubble in your Peoplesoft Number as the Identification Number starting in Column A.  Please take a moment to check that you have bubbled in everything correctly on your answer sheet.  Also, make sure that you have circled all your answers on the test itself.