EDRS 6208
EDRS 6208
Analysis and Interpretation of DataNon Parametric
TestsChi-square Distribution
OutlineWhat are nonparametric tests?The Chi Square Test
definitionThe Chi-Square DistributionA Goodness-of-fit
TestContingency Table
Parametric and Nonparametric TestsParametric Data Measured
dataUsed with interval or ratio level dataUsually assumed to be
normally or almost normally distributedScores are considered
parametric data[z, t-tests, ANOVA, Pearsons]
Non parametric DataCounted nominal or ranked dataDoes not assume
a normal distribution of population [ or Spearmans]Nonparametric
testsNon parametric tests also known as distribution-free tests.
They are appropriate when:The nature of the population distribution
from which the sample is drawn is not known to be normal.The
variables are expressed in nominal form (they are classified into
categories and represented by frequency counts).The variables are
expressed on ordinal form (They ranked in order, expressed as
first, second, third) (Best and Kahn, 2006, p.433).
THE CHI-SQUARE DISTRIBUTIONDefinition The chi-square
distribution has only one parameter called the degrees of freedom.
The shape of a chi-squared distribution curve is skewed to the
right for small df and becomes symmetric for large df. The entire
chi-square distribution curve lies to the right of the vertical
axis. The chi-square distribution assumes nonnegative values only,
and these are denoted by the symbol 2 (read as chi-square). Prem
Mann, Introductory Statistics, 7/E Copyright 2010 John Wiley &
Sons. All right reservedFigure 11.1 Three chi-square distribution
curves. Prem Mann, Introductory Statistics, 7/E Copyright 2010 John
Wiley & Sons. All right reserved
Example
Find the value of for 7 degrees of freedom and an area of .10 in
the right tail of the chi-square distribution curve.
Prem Mann, Introductory Statistics, 7/E Copyright 2010 John
Wiley & Sons. All right reservedChi-Square 2 for df = 7 and .10
Area in the Right Tail Prem Mann, Introductory Statistics, 7/E
Copyright 2010 John Wiley & Sons. All right reserved
Figure: Critical Value X2 Prem Mann, Introductory Statistics,
7/E Copyright 2010 John Wiley & Sons. All right reserved
Example 2Find the value of for 12 degrees of freedom and an area
of .05 in the left tail of the chi-square distribution curve.
Solution
Area in the right tail = 1 Area in the left tail = 1 .05 =
.95
Prem Mann, Introductory Statistics, 7/E Copyright 2010 John
Wiley & Sons. All right reservedTable 11.2 2 for df = 12 and
.95 Area in the Right Tail Prem Mann, Introductory Statistics, 7/E
Copyright 2010 John Wiley & Sons. All right reserved
Figure Prem Mann, Introductory Statistics, 7/E Copyright 2010
John Wiley & Sons. All right reserved
A GOODNESS-OF-FIT TESTDefinition An experiment with the
following characteristics is called a multinomial experiment.It
consists of n identical trials (repetitions).Each trial results in
one of k possible outcomes (or categories), where k > 2.The
trials are independent.The probabilities of the various outcomes
remain constant for each trial. Prem Mann, Introductory Statistics,
7/E Copyright 2010 John Wiley & Sons. All right reservedA
GOODNESS-OF-FIT TESTDefinitionThe frequencies obtained from the
performance of an experiment are called the observed frequencies
and are denoted by O. The expected frequencies, denoted by E, are
the frequencies that we expect to obtain if the null hypothesis is
true. The expected frequency for a category is obtained as E =
npwhere n is the sample size and p is the probability that an
element belongs to that category if the null hypothesis is true.
Prem Mann, Introductory Statistics, 7/E Copyright 2010 John Wiley
& Sons. All right reservedA GOODNESS-OF-FIT TESTDegrees of
Freedom for a Goodness-of-Fit TestIn a goodness-of-fit test, the
degrees of freedom are df = k 1 where k denotes the number of
possible outcomes (or categories) for the experiment. Prem Mann,
Introductory Statistics, 7/E Copyright 2010 John Wiley & Sons.
All right reservedTest Statistic for a Goodness-of-Fit TestThe test
statistic for a goodness-of-fit test is 2 and its value is
calculated as
where O = observed frequency for a categoryE = expected
frequency for a category = npRemember that a chi-square
goodness-of-fit test is always right-tailed.
Prem Mann, Introductory Statistics, 7/E Copyright 2010 John
Wiley & Sons. All right reservedExample A bank has an ATM
installed inside the bank, and it is available to its customers
only from 7 AM to 6 PM Monday through Friday. The manager of the
bank wanted to investigate if the percentage of transactions made
on this ATM is the same for each of the 5 days (Monday through
Friday) of the week. She randomly selected one week and counted the
number of transactions made on this ATM on each of the 5 days
during this week. The information she obtained is given in the
following table, where the number of users represents the number of
transactions on this ATM on these days. For convenience, we will
refer to these transactions as people or users. Prem Mann,
Introductory Statistics, 7/E Copyright 2010 John Wiley & Sons.
All right reserved At the 1% level of significance, can we reject
the null hypothesis that the number of people who use this ATM each
of the 5 days of the week is the same? Assume that this week is
typical of all weeks in regard to the use of this ATM. Prem Mann,
Introductory Statistics, 7/E Copyright 2010 John Wiley & Sons.
All right reserved
SolutionStep 1:H0 : p1 = p2 = p3 = p4 = p5 = .20H1 : At least
two of the five proportions are not equal to .20Step 2: There are 5
categories 5 days on which the ATM is usedMultinomial experimentWe
use the chi-square distribution to make this test.
Prem Mann, Introductory Statistics, 7/E Copyright 2010 John
Wiley & Sons. All right reserved19Step 3:Area in the right tail
= = .01k = number of categories = 5df = k 1 = 5 1 = 4The critical
value of 2 = 13.277 Prem Mann, Introductory Statistics, 7/E
Copyright 2010 John Wiley & Sons. All right reserved
Calculating the Value of the Test Statistic Prem Mann,
Introductory Statistics, 7/E Copyright 2010 John Wiley & Sons.
All right reserved
Step 4:All the required calculations to find the value of the
test statistic 2 are shown in Table 11.3.
Prem Mann, Introductory Statistics, 7/E Copyright 2010 John
Wiley & Sons. All right reservedStep 5:The value of the test
statistic 2 = 23.184 is larger than the critical value of 2 =
13.277It falls in the rejection regionHence, we reject the null
hypothesisWe state that the number of persons who use this ATM is
not the same for the 5 days of the week. Prem Mann, Introductory
Statistics, 7/E Copyright 2010 John Wiley & Sons. All right
reservedExample 2In a July 23, 2009, Harris Interactive Poll, 1015
advertisers were asked about their opinions of Twitter. The
percentage distribution of their responses is shown in the
following table. Prem Mann, Introductory Statistics, 7/E Copyright
2010 John Wiley & Sons. All right reserved
Assume that these percentage hold true for the 2009 population
of advertisers. Recently 800 randomly selected advertisers were
asked the same question. The following table lists the number of
advertisers in this sample who gave each response. Prem Mann,
Introductory Statistics, 7/E Copyright 2010 John Wiley & Sons.
All right reserved
Test at the 2.5% level of significance whether the current
distribution of opinions is different from that for 2009.Step 1:H0
: The current percentage distribution of opinions is the same as
for 2009.H1 : The current percentage distribution of opinions is
different from that for 2009.Step 2: There are 4 categories 5 days
on opinionMultinomial experimentWe use the chi-square distribution
to make this test.
Prem Mann, Introductory Statistics, 7/E Copyright 2010 John
Wiley & Sons. All right reservedStep 3:Area in the right tail =
= .025k = number of categories = 4df = k 1 = 4 1 = 3The critical
value of 2 = 9.348 Prem Mann, Introductory Statistics, 7/E
Copyright 2010 John Wiley & Sons. All right reserved
Calculating the Value of the Test Statistic Prem Mann,
Introductory Statistics, 7/E Copyright 2010 John Wiley & Sons.
All right reserved
Step 4:All the required calculations to find the value of the
test statistic 2 are shown in Table 11.4.
Prem Mann, Introductory Statistics, 7/E Copyright 2010 John
Wiley & Sons. All right reservedSolutionStep 5:The value of the
test statistic 2 = 5.420 is smaller than the critical value of 2 =
9.348It falls in the nonrejection regionHence, we fail to reject
the null hypothesisWe state that the current percentage
distribution of opinions is the same as for 2009. Prem Mann,
Introductory Statistics, 7/E Copyright 2010 John Wiley & Sons.
All right reserved
