AP Statistics 2017

using   this    function    (randBin(4,0.3,500) →   L1  )   and counting    the number  of  each    outcome.    (Can

you think   of  an  efficient   way to  count   each    outcome?)   She obtained

Do  these   data    provide evidence    that    the randBin function    on  the calculator  is  correctly   generating

values  from    this    distribution?

Calculator  Tip: It’s   a   bit of  a   digression, but if  you actually    wanted  to  do  the experiment  in  question    1,

you would   need    to  have    an  efficient   way of  counting    the number  of  each    outcome.    You certainly   don’t

want    to  simply  scroll  through all 500 entries and tally   each    one.    Even    sorting them    first   and then

counting    would   be  tedious (more   so  if  n were  bigger  than    4). One easy    way is  to  draw    a   histogram   of

the data    and then    TRACE to    get the totals. Once    you have    your    500 values  from    randBin in  L1  ,   go  to

STAT    PLOTS and   set up  a   histogram   for L1  .   Choose  a   WINDOW  something   like    [–0.5,4.5,1,–

1,300,1,1   ].  Be  sure    that    Xscl is set to  1.  You may need    to  adjust  the Ymax from    300     to get a   nice

picture on  your    screen. Then    simply  TRACE across    the bars    of  the histogram   and read    the value   of  n for

each    outcome off of  the screen. The reason  for having  x go    from    –0.5    to  4.5 is  so  that    the (integer)

outcomes    will    be  in  the middle  of  each    bar of  the histogram.

2. 
   

            A   chi-square  test    for the homogeneity of  proportions is  conducted   on  three   populations and one

   
   

   categorical variable that has four values. Computation of the chi-square statistic yields X 2 = 17.2. Is
   this result significant at the 0.01 level of significance?

   
   


3. Which of the following best describes the difference between a test for independence and a test for
   homogeneity of proportions? Discuss the correctness of each answer.
   a. There is no difference because they both produce the same value of the chi-square test statistic.
   b. A test for independence has one population and two categorical variables, whereas a test for
   homogeneity of proportions has more than one population and only one categorical variable.
   c. A test for homogeneity of proportions has one population and two categorical variables, whereas
   a test for independence has more than one population and only one categorical variable.
   d. A test for independence uses count data when calculating chi-square and a test for homogeneity
   uses percentages or proportions when calculating chi-square.