Basic Quantitative Methods in the Social Sciences (AKA Intro Stats)

1
Basic Quantitative Methods in the Social
Sciences(AKA Intro Stats)

• 02-250-01
• Lecture 10

2
Quantitative vs. Frequency Data

• Recall from our first lecture, that data could
  take the form of
• Quantitative data (AKA measurement data) whereby
  each observation represents a score on a
  continuum
• Most common statistics mean and SD
• Examples height, weight, IQ, rating of Chretien
  on a scale of 1-10.
• Categorical data (AKA frequency data) whereby
  frequencies of observations fall into one of two
  or more categories.
• Examples female vs. male, brown eyes vs. blue
  eyes, opposed to Chretien vs. support Chretien
• This type of data is not a measurement of
  anything per se, it is simply frequencies of
  occurrence in the nominal classes (e.g., of
  males vs. females, etc.)

3
What if we want to know whether the observed
frequencies were what we had expected?

• When it comes to frequency data, once we have
  counted our observations, we have
• Observed Frequencies frequencies that have
  actually occurred
• Expected Frequencies frequencies that we
  expected to occur if some assumption is true
• The Null Hypothesis (H0) is that the observed
  frequencies do not differ from what we had
  expected (the expected frequencies)

4
Chi - Square

• Examines the difference between the Observed and
  the Expected frequencies among groups
• Both variables are Nominal (therefore all we can
  measure is observed frequencies)
• E.g., we know how many males are in this class
  (the observed frequency), and how many we would
  expect to be in this class

5
Non-parametric tests

• The Chi-Square test is a non-parametric test.
• With non-parametric tests, we do not need to
  assume that the population data are normally
  distributed.
• Non parametric tests allow for nominal and
  ordinal scales of measurement.
• Although non-parametric tests are still
  inferential, they are less powerful than are
  parametric tests (e.g., t-tests, correlations).
  This means that parametric tests are not as
  likely to find significant results when the
  effect size is small.

6

• Cindy was hired at a fitness club in town. She
  was told by the boss that 68 of people who come
  in and inquire about becoming a member end up
  joining after they are shown around the club.
• Three months later, Cindy is fired because the
  boss feels that she is not good at selling the
  club to people inquiring about memberships. Over
  the three months, Cindy gave tours to 75 people.
  44 of them ended up joining.
• Cindy thinks that the boss just doesnt like her.

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Is the Boss Accusation True?

• The boss originally said that 68 of people
  typically join. Therefore, of the 75 people Cindy
  gave tours to, 51 of them should have joined if
  Cindy is on par with the other employees.

Join Dont Join
Observed 44 31
Expected 51 24
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The Chi-Square Goodness-of-Fit Test

• The Chi-Square Goodness-of-Fit Test is used when
  you have one classification variable (but it has
  2 or more categories).
• H0 The assumption that Cindy is on par with the
  other employees is true.
• Ha The assumption that Cindy is on par with the
  other employees is not true.
• The Chi-Square test allows a decision about
  whether observed and expected frequencies differ
  significantly.
• Rejection of H0 suggests that our assumption that
  led to the expected frequencies is wrong (or in
  this example, Cindy is not on par with the other
  club employees).

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? Greek letter Chi, O observed frequencies, E
expected frequencies
? 2
H0 O-E 0 (no difference between observed and
expected frequencies), therefore, ? 2 0.
10
Calculating Chi Square

• Create a table with columns for each category (so
  here, join and not join)
• Create a row for each of the following
• Observed frequencies (O)
• Expected frequencies (E)
• O E
• (O E)2
• (O E)2 / E
• The Chi Square statistic is then the sum of this
  final row of (O E)2 / E

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Chi Square Goodness of Fit Table
Join Not Join
O n join n not join
E
O E
(O E)2
(O E)2/E
Sum (O E)2/E Chi Square Statistic Value Chi Square Statistic Value
12
Calculating Chi-Squared ( )
Join Dont Join
Observed 44 31
Expected 51 24
(O-E) -7 7
(O-E)2 49 49
(O-E)2/E 0.9608 2.0417
Sum 0.9608 2.0417 3.0025
13
Testing the Significance of

• DF k-1 where k the number of outcome
  categories.
• Table E.1 in the text (p. 439) at the .050
  level of significance, df 1
• ?2crit 3.84
• Since ?2obt 3.00, we retain H0 (NOTE give obt
  value to 2 decimal places)
• Therefore, the result is not significant, Cindys
  recruiting performance at the fitness club is not
  significantly different than that of the other
  employees.

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• It is easy to see that in the numerator of the
  formula, observed frequencies are compared to
  expected frequencies to assess how well the
  sample data match the hypothesized data. Why must
  we divide the numerator by the expected frequency
  for each category?
• Suppose you were going to throw a party and you
  expected 1000 people to show up. However, at the
  party, you counted the number of guests and
  observed that 1040 actually showed up. Forty more
  guests than expected are no major problem when
  all along you were planning for 1000. There will
  probably still be enough beer and chips for
  everyone.

15

• On the other hand, suppose you had a party and
  you expected 10 people to attend but instead 50
  actually showed up. Forty more guests in this
  case spell big trouble. How significant the
  discrepancy is depends in part on what you were
  originally expecting.
• With very large expected frequencies, allowances
  are made for more errors between observed and
  expected frequencies. This is accomplished in the
  chi-square formula by dividing the squared
  discrepancy for each category by its expected
  frequency.

16
What About When There are More Than Two
Categories?

• In the preceding example, observed frequencies
  fell into one of two categories joined or did
  not join.
• What if there are more than two categories?

17
Example

• Suppose a study showed that of 90 people in
  trauma-induced comas who were treated with
  traditional medicine, 30 died, 30 woke up and
  fully recovered, and 30 remained comatose
  indefinitely. (Note These data were made up).
• Dr. X, a naturopathic doctor who works with
  patients with trauma-induced comas, claims that
  alternative approaches result in superior
  recovery rates. To test his claim, 90 comatose
  people were treated with his alternative approach
  and were then observed. 40 of them woke up and
  were fully recovered, 30 died, and 20 remained
  comatose indefinitely.

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Chi- Square
Stayed in Coma
Woke
Died
40
20
Total O 90
O
O
30
30
E
E
Whats H0?
19
Chi- Square
Stayed in Coma
Died
Woke
40
30
20
Total O 90
O
O
O
30
30
30
E
E
E
n 30 30 30 90
20
Chi- Square
? 2
Figure for Each Cell
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Chi- Square
Stayed In Coma
Woke
Died



22
Chi- Square
Stayed In Coma
Died
Woke
( 20 - 30 ) 2



fe
fe
23
Chi- Square
Stayed In Coma
Died
Woke
( -10 ) 2



fe
fe
24
Chi- Square
Stayed in Coma
Died
Woke
100



fe
fe
25
Chi- Square
Stayed In Coma
Woke
Died
100
100



30
30
30
26
Chi- Square
Stayed In Coma
Woke
Died

3.3333
0
3.3333


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Chi- Square
?2obt 6.6666 6.67 df k - 1 2 ?2crit
5.99 Therefore We reject H0, Dr. Xs
alternative approach does indeed generate
significantly more recoveries.

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Distribution of violent crimes in the United
States, 1995
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Sample results for 500 randomly selected
violent-crime reports from last year
30
Expected frequencies if last years violent-crime
distribution is the same as the1995 distribution
31
Calculating the goodness of fit (Chi-Square)
?2obt 4.219. With dfk-13, at .05, ?2crit
7.81 Therefore, we retain H0, last years crime
distribution is not significantly different from
that in 1995.
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The Chi-Square Test for Independence

• The ?2 statistic can also be used to test whether
  or not there is a relationship between two
  categorical (nominal) variables.
• Each individual in the sample is measured or
  classified on two separate variables.
• Also known as the Contingency Table Analysis

33
Do people with cell phones have more car
accidents than people without cell phones?

• The Department of Transportation wanted to see if
  cell phone users have more car accidents than
  non-cell phone users. The following data are a
  sample of 50 people who have had car accidents
  over the past month, and 50 randomly sampled
  drivers who have not had car accidents over the
  past month

34
Chi-Square
Cell Phone
No Cell Phone
Car Accident
No Car Accident
35
So what now?

• Notice that here, only observed frequencies are
  given to you. You have to calculate the expected
  frequencies.
• First, total your rows and columns.

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Chi-Square
Cell Phone
No Cell Phone
50
Car Accident
50
No Car Accident
55
45
100
37
Eij RiCj / N where Eij the expected
frequency at row i, column j. Ri Row is
total Cj Column js total N Grand total (all
cells included)
38
Chi-Square
Cell Phone
No Cell Phone
50
29
Car Accident
50 X 45 100
O
22.5
22.5
E
50
No Car Accident
55
45
100
39
Chi-Square
Cell Phone
No Cell Phone
50
29
Car Accident
50 X 55 100
O
27.5
22.5
27.5
E
50
No Car Accident
55
45
100
40
Chi-Square
Cell Phone
No Cell Phone
50
29
Car Accident
O
22.5
27.5
E
50
No Car Accident
22.5
27.5
55
45
100
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And then.
You now have four cells with expected and
observed frequencies. Now use the Chi Square
formula!
(29-22.5)2/22.5 1.8778 (21-27.5)2/27.5
1.5364 (16-22.5)2/22.5 1.8778 (34-27.5)2/27
.5 1.5364 ?2obt 6.8284 6.83
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To test this statistic

• Lets use the .05 level of significance.
• Variable 1 Cell phone?
• Variable 2 Car accident?
• Because we are dealing with frequency data of two
  categorical variables, we will perform a
  chi-square test of independence.
• Because it is a chi-square, it is a two-tailed
  test.
• H0 Cell phones and car accidents are independent
  
• Ha Cell phones and car accidents are not
  independent

43
Chi-Square

• DF (for test of independence) df
  (R-1) (C-1)
• Where R number of rows
• Where C number of columns
• df (2-1)(2-1)1, ?2crit 3.84 (from table E.1,
  page. 439)
• ?2obt 6.83, so reject the H0.

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SO.

• Results are significant. The frequency of being
  in a car accident depends on whether or not one
  uses a cell phone.

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Note

• When the expected frequencies are too small,
  chi-square may not be a valid test, therefore all
  expected frequencies should be at least 5
  (dependent on sample size).
• The chi-square test is also only valid when the
  observations are independent from each other,
  therefore N should be equal to the number of
  subjects (every subject should only be measured
  once).

46
What about when you have more than two
categories?

• A fast-food marketing consultant wanted to know
  whether men and women had different preferences
  for fast-food restaurants. She randomly sampled
  150 men and 100 women and asked each to declare
  his or her preference for four fast foods
  restaurants. Here are her data

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Burger King
Total 100 150 ----- 250
Subway
Harveys
McDonalds
Women
Men
Total 55 55 85
55
48
Now remember.

• To calculate expected frequencies for each cell,
  Eij RiCj / N
• So (Row sum) (Column sum) / N
• Do this for each cell to get expected
  frequencies.

49
Burger King
Total 100 150 ----- 250
Subway
Harveys
McDonalds
Women
22
22
22
34
Men Total
33
33
33
51
55 55 85
55
50
You now have what you need to calculate ?2
(35-22)2/22 7.6818 (25-22)2/22
0.4091 (15-34)2/34 10.6176 (25-22)2/22
0.4091 (20-33)2/33 5.1212 (30-33)2/33
0.2727 (70-51)2/51 7.0784 (30-33)2/33
0.2727
Add these up! ?2obt 31.8626
31.86 df (R-1)(C-1) (2-1)(4-1) 3 ?2crit
7.82 (at .05)
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So.

• H0 Gender and fast food preference are
  independent.
• Ha Gender and fast food preference are
  dependent.
• We reject the null hypothesis.
• SO Gender and preference for fast food are
  dependent.
• Stated differently, men and women do not prefer
  the same fast food joints.

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Time for Some Review
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• Final Exam Info
• Wednesday, June 25 from 700 pm to 1000 pm
• REMINDERS Bring pencils, calculator, TEXTBOOK,
  student ID