Assignment 6 Mixed strategy

profileACETUTORS
 (Not rated)
 (Not rated)
Chat

Assignment #6

Mixed Strategy

 

  1. Observe at least 50 events where the strategy chosen and the outcome is observable.  The strategy choice must be observable; e.g. serve to left or right.  The outcome must be observable; e.g. point won or lost.  Both must be dichotomous.  You can observe a sporting event or get data from web sites like espn.com or mlb.com. Also, this must be a contest with opponents.  For example, a golf test involving chip and run vs pitching does not have an opponent.

 

Possible topics

            1.  pitcher and batter.  strategy:  first pitch is fast ball or some other pitch; outcome: hit or out.

            2.  football play selection:  3rd and short (< 3 yds).   strategy: run or pass; outcome: make the first down or not.

            3.  shootouts in hockey or soccer (one team in several games or matches): strategy: left or right; outcome goal or not.

            4.  you can also stage a contest.  For example, ask a friend to play one-on-one basketball.  Strategy:  drive to the basket or shoot a jumper; outcome: make or miss.

 

  1. Briefly describe your observations.  For example, “I watched a tennis match between John and Jane Doe.”

 

  1. Record you observations in sequence. 

L, L, R, L, R, R, R, R, L, R, R, L, L, . . . .  etc.

 

  1. Count the number of times each strategy was chosen and the number of runs. In the example it is

Left 37, right 37, runs 31

 

  1. Test for random choice.  Go to  http://www.quantitativeskills.com/sisa/statistics/ordinal.htm

 

  1. Enter the numbers and calculate the test statistic.

 

Expected Number of Runs: 38; sd: 4.2716

z-value= -1.6387; approx. probability: 0.05063

z-val= -1.521(continuity correct.); p: 0.06404

exact probability of 31 or fewer runs= 0.06355

 

      If the choice were random the expected number of runs would be 38.  The number of runs is a normal distribution with a standard deviation of 4.2716.  This outcome is -1.521 standard deviations below the mean.  The p value of .06355 means that a little over 6 percent of the time one would observe fewer runs than 31 if the choice were random.

 

  1. In a table record the strategies on the vertical and the outcomes on the horizontal.

 

Won    Lost

 

            Left     23        14

 

            Right   26        11

 

  1. Go to http://www.graphpad.com/quickcalcs/contingency1.cfm


 

  1. Enter the numbers and calculate the (two tailed) chi square test with correction.

 

 

P value                        .62

 

  1. Interpret the results.

 

The p value answers the question, “How likely is an association as strong as this one, if there really is no association?”  In this example, 62 percent of the time one would expect this much or a greater difference in success if the outcomes were drawn from the same distribution; that is, if the server were equally successful when serving to the left or right.

 

 


 

Name _____________

 

Assignment Sheet

Mixed Strategy

 

1.  Briefly describe your observations.

 

 

2.  How many runs?                                                                            _________________ 

 

3.  Expected number of runs                                                               _________________

 

4.  z-statistic                                                                                        _________________

 

5.  p-value                                                                                           _________________

 

6.  Interpretation

 

 

 

 

 

 

 

 

 

7.  Table for chi-square test

 

 

 

 

 

 

 

 

 

 

8.  p-value                                                                                           _________________

 

9.  Interpretation

 

 

    • 9 years ago
    Assignment 6 Mixed strategy
    NOT RATED

    Purchase the answer to view it

    blurred-text
    • attachment
      assignment_6.docx