20140303_test1.pdf

Upload: 2009028818

Post on 23-Feb-2018

214 views

Category:

Documents


0 download

TRANSCRIPT

  • 7/24/2019 20140303_Test1.pdf

    1/5

    University of the Free StateDepartment of Mathematical Statistics and Actuarial Science

    STS618 / STS718

    First Semester Test: 3 March 2014

    Total marks: 50

    Time: 90 min

    1. Question 1 [10]

    A study was carried out in July 2010 to investigate the profile of soccer enthusiasm

    in Bloemfontein. A random sample of 1000 adults in Bloemfontein was drawn.

    Among other questions, participants were asked whether or not they had tickets

    for at least one of the world cup matches. Participants were cross-classified

    according to gender, and whether or not they had at least one world cup ticket.

    Of the 329 males in the sample, 159 had a word cup ticket. Of the females, 210had a world cup ticket.

    Summarize the data in a 2 2 table and then answer the following questions

    about soccer enthusiasm in Bloemfontein:

    (a) Estimate the probability that a male has a world cup ticket. Is this a

    conditional, marginal or joint probability? [2]

    (b) Estimate the probability that somebody who has no ticket is a female. Is

    this a conditional, marginal or joint probability? [2](c) Estimate the probability that an adult in Bloemfontein has a world cup

    ticket. Is this a conditional, marginal or joint probability? [2]

    (d) Estimate the probability of being female and having a world cup ticket. Is

    this a conditional, marginal or joint probability? [2]

    (e) Who was more likely to have a word cup ticket, males or females? Motivate

    your answer. [2]

    1

  • 7/24/2019 20140303_Test1.pdf

    2/5

    2. Question 2 [7]

    Consider the following Model 2 2 table.

    Table 2.1: Model 2 2 Table

    Characteristic B

    Characteristic A Present Absent Total

    Present n11 n12 n1+

    Absent n21 n22 n2+

    Total n+1 n+2 n

    (a) Under cross-sectional sampling, specify the distribution of the cell frequen-

    cies n11, n12, n21and n22. [1]

    (b) Which total counts are considered fixed under cross-sectional sampling? [1]

    (c) Under prospective sampling, specify the distribution of the cell frequency

    n21 [1]

    (d) Which total counts are considered fixed under prospective sampling? [1]

    (e) Under retrospective sampling, specify the distribution of the cell frequency

    n12 [1]

    (f) Which total counts are considered fixed under retrospective sampling? [1]

    (g) Specify the distribution of the cell frequency n21 when all marginal totals

    are fixed. [1]

    2

  • 7/24/2019 20140303_Test1.pdf

    3/5

    3. Question 3 [4]

    Before and during the 2010 World Cup the most convenient way to acquire tickets

    for a World Cup match was via the internet, and credit card payment. There-

    fore, a study was carried out on 27 June 2010 to investigate whether credit card

    ownership was associated with having a ticket for Germanys famous World Cup

    victory over England (4:1 !!) in Bloemfontein. That is, the research question

    was the following: Is an owner of a credit card more likely to have a ticket and

    attend the Germany-England match than somebody who did not own a credit

    card. In the stadium during the match, a sample of 100 adult spectators were

    asked whether or not they owned a credit card. At the some time, but outside

    the stadium in the Waterfront shopping centre, a control sample of 100 adults

    were asked the same question.

    Was this a prospective, a retrospective or a cross-sectional study. Motivate your

    answer. [4]

    4. Question 4 [13]

    Let be the probability that a student registered for STK114 attends class.

    A random sample of size n = 120 of students registered for STK114 is taken,

    and the random variable Xdenotes the number of students in the sample who

    are actually found to be in class . We assume that X follows the binomial

    distribution. The probability that exactly X= n1 members of the sample are

    observed to be in class is denoted by Bin(n1,n ,), and is given by the probability

    function

    Prob(X= n1) = Bin(n1,n ,) =

    n

    n1

    n1(1 )nn1

    = n!

    n1!(n n1)!n1(1 )nn1

    Of the n= 120 students in the sample, n1= 85 were found to be in class.

    (a) Derive the maximum likelihood estimate for . [5]

    (b) We want to test the two-sided null-hypothesis H0 : = 0.8 against the

    alternative HA: = 0.8. As test statistic we use the number n1of students

    3

  • 7/24/2019 20140303_Test1.pdf

    4/5

    in the sample who were in class. State the exact distribution ofn1 under

    the null-hypothesis [2].

    (c) In order to test the above null hypothesis, derive the exact P-value, that is,

    the probability, given the null-hypothesis, of observing X = n1, or a more

    extreme outcome than n1. (State the results in terms of the formula for the

    probability function of distribution ofn1under the null-hypothesis; there is

    no need to work out the actual P-value). [6]

    5. Question 5 [16]

    A cross-sectional study was carried out to investigate the attitude of students at

    the UFS to compulsory class attendance. A random sample of students was drawn

    from the total student population, and cross-classified according to whether or

    not they were postgraduates, and whether or not they approved of compulsory

    class attendance. The data for the sub-sample of students who study Actuarial

    Science is as follows:

    Table 5.1: Student seniority and approval of compulsory classes

    Approve compulsory class attendance

    Seniority Yes No Total

    Postgraduate student 5 2 7

    Undergraduate student 2 8 10

    Total 7 10 17

    We want to test the null hypothesis H0 No association between the variables

    Seniorityand Approve compulsory class attendance. As test statistic wechoose the frequency n11.

    (a) What is the exact distribution ofn11 under the null-hypothesis? [2]

    (b) Why is it a good idea to carry out Fishers exact test (instead of a chi-square

    test)? Motivate your answer. [3]

    (c) Specify the 5 easy steps to test a null-hypothesis for Fishers exact test

    for a 2 2 table. [5]

    4

  • 7/24/2019 20140303_Test1.pdf

    5/5

    (d) Carrying out Fishers exact test for Table 5.1 above, the table probabilities

    are as follows:

    Table 5.2: Table Probabilities

    Table Cell Probability

    n11 n12 n21 n22

    7 0 0 10 0.0001

    6 1 1 9 0.0036

    5 2 2 8 0.0486

    4 3 3 7 0.21603 4 4 6 0.3779

    2 5 5 5 0.2721

    1 6 6 4 0.0756

    0 7 7 3 0.0062

    Calculate the exact two-sided P-value and the exact one-sided P-value for

    testing the null-hypothesis. [3]

    (e) What can you conclude from the data in Table 5.1 and from the result ofthe hypothesis test? [3]

    5