Nonparametric Tests of Significance

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Nonparametric Tests of Significance
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Nonparametric Tests

• Nonparametric tests are tests that do not test
  hypotheses about population parameters.
• We generally use nonparametric tests of
  significance under one of two conditions
• 1) The variable we are measuring can not be
  tested with parametric tests.
• 2) The underlying assumptions of parametric
  tests are not fulfilled. For example, the shape
  of the distribution is not normal (particularly
  if N is small).

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• This is typically the case when we have
  categorical variables (nominal scale data) or
  ordinally scaled variables.
• Nominal scale data represent classes or
  categories. They have no quantitative properties
  (e.g., political party, religion, gender, numbers
  assigned to horses in a horse race).
• Ordinal scale data measurements that represent
  position or order in a series (ranking political
  candidates on popularity, movie ratings, places
  runners finished in a race).

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Categorical Variables

• Sometimes nominal data can only be broken down
  into two categories.
• E.g., right or wrong, male or female, pass or
  fail, heads or tails
• The data are then referred to as binomial and we
  have a dichotomous population.
• In such a case, we define p as the probability of
  obtaining one category and q as the probability
  of obtaining the other.

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The Binomial Test

• Note, there is a special relationship between p
  and q.
• p q 1
• So q 1 p
• If p q 0.5, we can solve binomial problems
  using Binomial Table M on page 352.

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An Example

• E.g., Tom claims he has a coin fixed to turn up
  heads. He flips it 20 times and gets 16 heads.
  Is his claim correct? Use normal decision rules.
• We will call p the probability of getting heads
  in one toss.
• p 0.5
• We will call q the probability of getting tails
  in one toss.
• q 0.5
• N 20 X 16 (score on the trials)

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• H0 p 0.5
• H1 p gt 0.5
• The left column of the Table M has values of N (5
  - 50).
• There are four other columns. Two for one tailed
  test (? at 0.05 and 0.01) and two for a
  two-tailed test (? at 0.05 and 0.01).
• The numbers in these columns represent X or N X
  (which ever is larger).

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• We will use X (16) since it is larger than N - X
  ( 4)
• Go down the left column to where N 20.
• Go across to the one-tailed column where ? is
  0.05. If your obtained X (or N X), is greater
  than or equal to the number in the Table, reject
  the H0.
• The number listed is 15.
• Since our Xobs 16 gt Xcrit 15, we will reject
  the H0. The coin does appear to be fixed (X obs
  16, p lt 0.05).

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Normal Approximation to the Binomial

• What if p q 0.5
• If N is sufficiently large, we can treat the
  binomial distribution as though its a normal
  distribution.
• If Np and Nq are both greater than 10, we can use
  normal approximation to the binomial
• If p q 0.5, we can still use normal
  approximation to the binomial as long as N is at
  least 25.
• In these cases, we can obtain a z-score.

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• Heres the formula

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• E.g., In psychology departments in Canada, ¾ of
  the students are female and only ¼ are male. A
  university administrator believes his university
  has a different percentage of males and females.
  A sample of 48 is randomly chosen and 14 are
  male. Is his claim correct? Use ? 0.05.
• p 0.25 q 0.75 N 48 X 14
• H0 p 0.25
• H1 p 0.25

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Np 48(0.25) 12 Nq 48(0.75) 36
We may use normal approximation to the binomial.
Remember, zcrit for a two tailed test for
alpha at 0.05, is 1.96.
Since Zobs 0.67 lt zcrit 1.96 we do not
reject the H0. The university does not have a
different percentage of males and females (Zobs
0.67, p gt 0.05).
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Another Example

• A police inquiry claims that 68 of all drivers
  speed on highways. A patroller uses his radar
  gun on 100 passing cars and finds that 86 of them
  were speeding. Is the inquiry accurate? Use ?
  0.01.
• p 0.68 q 0.32 N 100 X 86
• H0 p 0.68
• H1 p 0.68

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Np 100(0.68) 68 Nq 100(0.32) 32
We may use normal approximation to the binomial.
Zcrit 2.58
Since zobs 3.86 gt zcrit 2.58, reject
H0. The inquiry was not accurate (zobs 3.86, p
lt 0.01).
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?2 One Variable Case

• If p q 0.5, we can also use ?2.
• Also referred to as a chi square or goodness of
  fit test.
• This technique provides a test of whether a
  significant difference exists between observed
  number of cases and expected number of cases.

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Example

• E.g., A recent survey by the Center for Applied
  Psychological Testing suggests that 55 of all
  people are introverts, whereas 45 are
  extraverts. A skeptical psychologist randomly
  selects a sample of 93 subjects and gives them a
  personality test. Based on the test, he finds
  that 38 are extraverts and 55 are introverts. Is
  he right to be skeptical? Use normal decision
  rules.

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• Well make a table that compares expected
  frequencies (fe) to observed frequencies (fo).
• H0 fo fe
• H1 fo fe

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• Now we calculate ?2.

? 2 ? (fo - fe)2 fe
?2 (38 - 42)2 42
(55 - 51)2 51

? 2 0.38 0.31 0.69
We now compare this to a ?2crit in Table B on
page 328.
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• The ?2crit changes as the degrees of freedom
  change.
• df k - 1 where k represents the number of
  categories.

df k - 1 2 - 1 1
? 2crit 3.841
Since ? 2 0.69 lt ? 2crit 3.841, do not
reject H0. The survey is accurate (?12 0.69,
p gt 0.05).
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?2 One Variable Case

• Importantly, ?2 can be used to analyze
  categorical data that is not binomial.
• That is, it can be used to analyze data which has
  more than two outcomes.
• E.g., A researcher wants to determine whether
  there is a difference among beer drinkers living
  in St. Johns in their preference for brands of
  light beer.

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An Example

• 150 beer drinkers taste three brands of light
  beer. Their preferences are provided below. Is
  there a difference in the preference of the
  brands? Use ? 0.01.
• H0 fo fe
• H1 fo fe

Brand A Brand B Brand C Total fo 45
40 65 150 fe
50 50 50
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?2 ? (fo - fe)2 fe
?2 (45 - 50)2 50
(40 - 50)2 50
(65 - 50)2 50


0.5 2.00 4.5 7.00
df k - 1
3 - 1 2
Since ?2obs 7.00 lt ?2crit 9.210, do not
reject H0. There is no difference in preference
of the three brands (? 22 7.00, p gt 0.01).
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?2 Test of Independence

• So far, weve looked at cases in which we were
  investigating one categorical variable (e.g.,
  brand of beer, introvert/extravert).
• However, we can also use ?2 to deal with cases
  with more than one categorical variable in the ?2
  test of independence.
• When carrying out this test, we assume that the
  two variables are independent, i.e., one variable
  does not influence the other.

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An Example

• A bill in the United States has been proposed to
  lower the legal age for drinking to 18. A
  political scientist is interested in determining
  whether there is a relationship between political
  affiliation and opinions on this bill. He
  samples 200 registered Republicans and 200
  Democrats. Their opinions on the bill are
  presented below. Is there a difference in
  opinions between the Democrats and Republicans?
  Use ? 0.05.

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• H0 Opinions of the bill are independent of
  political affiliation.
• H1 Opinions on the bill depends on political
  affiliation.

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Opinion For Undecided
Against Republican 68 22
110 Democrat 92 18 90
Row Totals 200 200 400
Column 100 40 200 Totals
Our expected frequencies will be based on
our obtained frequencies.
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fe (c) (row total)(column)/N
(200)(200)/400 100
fe (d) (row total)(column)/N
(200)(160)/400 80
fe (e) (row total)(column)/N
(200)(40)/400 20
fe (f) (row total)(column)/N
(200)(200)/400 100
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Now we follow the same formula as before to
calculate ?2.
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(68 - 80)2 (22 - 20)2 (110 - 100)2 80
20 100
(92 - 80)2 (18 - 20)2 (90 - 100)2 80
20 100
1.8 0.2 1.0 1.8 0.2 1.00 6.00
df (r - 1)(c - 1)
r rows c columns
(2 - 1)(3 - 1) 2
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?2 crit 5.991
Since ?2 obt 6.00 gt ?2 crit 5.991, we
reject the H0. Opinion of the bill depends
on political affiliation (?22 6.00, p lt 0.05).