UsingWW: Tolerance setting for answers is unreliable

Greetings,

I am having a problem with the tolerance for students answers being much wider than set in the problem.

I have set the tolerance for correct answers at 0.1 with the command:

Context()->flags->set(tolerance => 0.1);

And yet the third and fourth answer in this problem set is accepting students answers that have a tolerance of plus or minus 1.4 for the students entry.

What I have I done to create this glitch, can you please help?

Tim

# DESCRIPTION

# Stat 109 Data summarization

# WeBWorK problem written by TimPayer <tsp1@humboldt.edu>

# ENDDESCRIPTION

## DBsubject(Statistics)

## DBchapter(Exploratory data analysis/descriptive statistics)

## DBsection(Summary statistics)

## Institution(Humboldt State University)

## Author(Tim Payer)

## KEYWORDS(summations, mean, sd)

DOCUMENT();

loadMacros(

"PGstandard.pl",

"PGbasicmacros.pl",

"MathObjects.pl",

"PGML.pl",

"niceTables.pl",

"parserPopUp.pl",

"PGcourse.pl",

"parserFormulaUpToConstant.pl"

);

Context("Numeric");

Context()->flags->set(tolerance => 0.1);

install_problem_grader(~~&std_problem_grader);

$showPartialCorrectAnswers = 1;

## The beginning of 3.1 Stat 109 problem:

## Generate 8 random values and sort. These values

## will form the "box" of the boxplot.

$parsum = 0;

@countpre = ();

foreach my $i (0..7) {

$countpre [$i] = random(14.3, 16.5,0.1);

$sum1 = Compute("$countpre[$i] + $parsum");

$parsum = $sum1;

}

sub num_sort {

PGsort sub {$_[0] <=> $_[1]}, @_;

}

@out = num_sort(@countpre);

## Create Quartiles and Outlier Threshold values

## based upon the 8 boxplot data points.

$Q1 = Compute(" ($out[0] + $out[1])/2");

$Q3 = Compute(" ($out[6] + $out[7])/2");

$step = Compute(" 1.5*($Q3 -$Q1)");

$LOT = Compute("$Q1 -$step");

$UOT = Compute("$Q3 + $step");

$LOTminus1 = $LOT -0.6;

$LOTplus1 = $LOT + 0.6;

$LOTplus2 = $LOT + 1.3;

$Q1minus1 = $out[0]-0.1;

$Q3plus1 = $out[7]+0.1;

$UOTminus1 = $UOT -0.2;

$UOTplus1 = $UOT + 0.2;

$LOTminus2 = $LOT -0.8;

$UOTplus2 = $UOT + 0.8;

$UOTminus2 = $UOT -0.8;

@count =();

$cnt1 = random( $LOTminus2, $LOTminus1, 0.1);# low outlier

$cnt10 = floor($cnt1*10);

$count[0] = Compute("$cnt10/10 ");

$cnt2 = random($LOTplus1, $Q1minus1, 0.1); # left whisker terminal.

$cnt20 = floor($cnt2*10);

$count[1] = Compute("$cnt20/10 ");

$cnt11 = random($Q3plus1, $UOTminus1, 0.1); # right whisker terminal.

$cnt110 = ceil($cnt11*10);

$count[10] = Compute("$cnt110/10 ");

$cnt12 = random( $UOTplus1, $UOTplus2, 0.1);# high outlier

$cnt120 = ceil($cnt12*10);

$count[11] = Compute("$cnt120/10 ");

$sum = Compute("$sum1 + $count[0] + $count[1] + $count[10] + $count[11] ");

##### Combine the separate box, outlier, whisker data into one array.

@boxa = ();

foreach my $i (0..7) {

$boxa [$i] = $out [$i];

}

$boxa [8] = $count [0]; # Lower outlier

$boxa [9] = $count [1]; # Lower whisker terminal

$boxa [10] = $count [10]; # Upper whisker terminal

$boxa [11] = $count [11]; # Upper outlier

######################################################

## For a problem that gives each student a different random sample,

## un-comment(#) the code above. For a fixed problem that permits each student to share

## a common answer, comment out the loop above and un-comment the 12

## lines of code below.

#$count [0] = 14.1;

#$count [1] = 14.4;

#$out [0] = 14.7;

#$out [1] = 14.8;

#$out [5] = 15.0;

#$out [6] = 15.2;

#$out [7] = 15.5;

#$out [8] = 15.7;

#$out [9] = 15.9;

#$out [10] = 16.1;

#$count [10] = 17.5;

#$count [11] = 17.9;

#$parsum = 0;

#foreach my $i (0..7) {

# $sum1 = Compute("$out[$i] + $parsum");

# $parsum = $sum1;

# $sum = Compute("$sum1 + $count[0] + $count[1] + $count[10] + $count[11] ");

#$Q1 = Compute(" ($out[0] + $out[1])/2");

#$Q3 = Compute(" ($out[6] + $out[7])/2");

#$step = Compute(" 1.5*($Q3 -$Q1)");

#$LOT = Compute("$Q1 -$step");

#$UOT = Compute("$Q3 + $step");

##############################################################

## Answers generated for both forms (randomized or fixed) problem sets.

$ans1 = Compute("$sum/12");

$ans2 = Compute("($out[3] + $out[4])/2");

$ans3 = $Q1;

$ans4 = $Q3;

$ans5 = $LOT;

$ans6 = $UOT;

$ans7 = $count[1];

$ans8 = $count[10];

$outliers = List($count[0],$count[11]);

$dif1 = $out[4]-$out[3];

$ans9 =sprintf("%0.1f",$dif1);

$ans10 = $out[0]+$out[1];

$dif2 = $ans4-$ans3;

BEGIN_PGML

*Drawn from lecture notes Week 2, Day 2.*

3.1) Beall et al. 2002, made an extensive survey of hemoglobin levels (g/dL) among male populations living at different elevations. Given that the concentration of oxygen at high elevations ( above 18,000 feet) is less than half that found at sea level, Beall's suspected that men living at higher elevations should have higher hemoglobin levels to compensate for the thinner air. As a base line for comparison, hemoglobin levels from over 1700 men living at sea level in the USA was recorded. A random sample of twelve of these hemoglobin levels are sorted and tallied in the table below.

Determine the mean and following eight boxplot values to summarize the twelve data points.

Source: Data set drawn from Beall et al. 2002, from the text: "The Analysis of Biological Data", Whitlock and Shluter, 2nd ed. 2015.

[@DataTable(

[

[ ["".PGML::Format('[`\text{Subject}`]')."",

halign => 'r|',

rowcss => 'border-bottom: 2px solid;',

cellcss => 'border-top: 2px solid;

border-left: 2px solid;

border-right: 2px solid; ',],

["".PGML::Format('[`\text{Hemoglobin: g/dL}`]')."",

rowcss => 'border-bottom: 2px solid;',

cellcss => 'border-top: 2px solid;

border-right: 2px solid; ',],

[ ["1",

midrule => '1',