View of /trunk/NationalProblemLibrary/Rochester/setStatistics7MultinomialContingency/ur_stt_7_5.pg

    1 ## DESCRIPTION
    2 ##   Multinomial Contingency
    3 ## ENDDESCRIPTION
    4 
    5 ## KEYWORDS('Multinomial', 'Contingency', 'Hypothesis', 'Test Statistic', 'Critical Value', 'Chi Squared', 'Rejection', 'Region')
    6 ## Tagged by nhamblet
    7 
    8 ## DBsubject('Statistics')
    9 ## DBchapter('Hypothesis Testing')
   10 ## DBsection('Multinomial Contingency')
   11 ## Date('')
   12 ## Author('')
   13 ## Institution('Rochester')
   14 ## TitleText1('')
   15 ## EditionText1('')
   16 ## AuthorText1('')
   17 ## Section1('')
   18 ## Problem1('')
   19 
   20 DOCUMENT();        # This should be the first executable line in the problem.
   21 
   22 loadMacros(
   23 "PG.pl",
   24 "PGbasicmacros.pl",
   25 "PGchoicemacros.pl",
   26 "PGanswermacros.pl",
   27 "PGgraphmacros.pl",
   28 "PGnumericalmacros.pl",
   29 "PGstatisticsmacros.pl"
   30 );
   31 
   32 TEXT(beginproblem());
   33 $showPartialCorrectAnswers = 1;
   34 
   35 $alpha = random(.01,.05,.04);
   36 
   37 $totx = 0;
   38 $toty = 0;
   39 $totz = 0;
   40 
   41 for($i=0; $i<3; $i++) {
   42   $x[$i] = random(40,80,1);
   43   $y[$i] = random(40,80,1);
   44   $z[$i] = random(40,80,1);
   45   $xyz[$i] = $x[$i]+$y[$i]+$z[$i];
   46   $totx = $totx + $x[$i];
   47   $toty = $toty + $y[$i];
   48   $totz = $totz + $z[$i];
   49 }
   50 $n = $totx+$toty+$totz;
   51 
   52 for($i=0; $i<3; $i++) {
   53   $ex[$i] = ($xyz[$i]*$totx)/$n;
   54   $ey[$i] = ($xyz[$i]*$toty)/$n;
   55   $ez[$i] = ($xyz[$i]*$totz)/$n;
   56 }
   57 
   58 $chi = 0;
   59 for($i=0; $i<3; $i++) {
   60   $chi =
   61   $chi+(($x[$i]-$ex[$i])**2/$ex[$i])+(($y[$i]-$ey[$i])**2/$ey[$i])+(($z[$i]-$ez[$i])**2/$ez[$i]);
   62 }
   63 
   64 $rej = chisqrdistr(4,$alpha);
   65 
   66 $mc = new_multiple_choice();
   67 
   68 @ans = ("We can reject the null hypothesis that A and B are independent and accept that A and B are
   69   dependent. ",
   70   "There is not sufficient evidence to reject the null hypothesis that A and B are independent. ");
   71 
   72 if ($chi > $rej ) {$tag = 0;} else {$tag = 1;}
   73 
   74 $mc -> qa('The final conclustion is', $ans[$tag]);
   75 
   76 $mc -> extra($ans[1-$tag]);
   77 
   78 BEGIN_TEXT
   79 Test the null hypothesis of independence of the two classifications, A and B, of the \( 3 \times 3\)
   80 contingency table shown below.  Test using \(\alpha = $alpha\) $BR
   81 \{begintable(5)\}
   82 \{row("\( \ \)", "\(B_1\)", "\(B_2\)", "\(B_3\)", "Total")\}
   83 \{row("\(A_1\)", "\($x[0]\)", "\($y[0]\)", "\($z[0]\)", "\($xyz[0]\)")\}
   84 \{row("\(A_2\)", "\($x[1]\)", "\($y[1]\)", "\($z[1]\)", "\($xyz[1]\)")\}
   85 \{row("\(A_3\)", "\($x[2]\)", "\($y[2]\)", "\($z[2]\)", "\($xyz[2]\)")\}
   86 \{row("Total", "\($totx\)", "\($toty\)", "\($totz\)", "\($n\)")\}
   87 \{endtable()\}
   88 $PAR
   89 
   90 \(\chi^2 =\) \{ans_rule(20)\} $PAR
   91 rejection region is \(\chi^2 > \) \{ans_rule(20)\}
   92 
   93 \{$mc->print_q()\}
   94 \{$mc->print_a()\}
   95 
   96 END_TEXT
   97 ANS(num_cmp($chi));
   98 ANS(num_cmp($rej));
   99 ANS(radio_cmp($mc->correct_ans));
  100 
  101 
  102 ENDDOCUMENT();       # This should be the last executable line in the problem.
  103