GEICO feels that the caveman ad campaign has become stal

1

• Hypothesis Testing
• Basic Concepts and Tests of Association,
• Chi-Square Tests

2
Basic concepts - Example

• GEICO feels that consumers do not like the
  caveman ad campaign so it needs to be changed
• GEICO wants to verify this feeling so they survey
  a sample and find that the campaign is well
  liked.
• Should GEICO conclude that their feeling is wrong
  or that the sample mean is a function of chance?

3
Hypothesis Testing Basic Concepts

• Hypothesis An assumption made about a population
  parameter (not sample statistic)
• E.g. Consumers dislike the caveman ad campaign
• Purpose of Hypothesis Testing To make a judgment
  about the difference between the sample statistic
  and the population parameter
• The sample likes the caveman campaign. Is this an
  accurate representation of the populations
  attitude?
• The mechanism adopted to make this objective
  judgment is the core of hypothesis testing

4
Hypothesis testing Logic

• Is the sample statistic a function of chance or
  luck rather than an accurate representation of
  the population parameter?
• Example
• Hypothesized mean attitudes are 2 (on a 1(SD)
  5(SL) scale)
• Observed mean attitudes are 4 (on a 1(SD) 5(SL)
  scale)
• Is the difference between the two a chance event
  or are we really wrong about our hypothesis?
• This is statistically evaluated.

5
Problem Definition
Clearly state the null and alternative hypotheses.
Choose the relevant test and the appropriate
probability distribution
Determine the degrees of freedom
Determine the significance level
Choose the critical value
Compare test statistic and critical value
Compute relevant test statistic
Decide if one-or two-tailed test
Does the test statistic fall in the critical
region?
No
Do not reject null
Yes
Reject null
6
1. Formulate Null Alternative hypotheses

• Null hypothesis (Ho)
• the hypothesis of no difference
• between the population parameter and sample
  statistic
• OR no relationship
• Between two population parameters
• A mirror-image of the alternative (research)
  hypothesis
• Alternative hypothesis (Ha or H1) the
  hypothesis of differences or relationships in the
  population
• Example
• Ho Mean population attitudes 2
• Ha Mean population attitudes are not 2 OR
• Ho Use of social media is not related to
  likelihood of response to online ads
• Ha Use of social media is positively related to
  likelihood of response to online ads

7
2. Choose appropriate test and probability
distribution

• Depends on whether we are
• Comparing means (Z distribution if population
  standard deviation is known t distribution if
  population standard deviation is not known)
• Comparing frequencies (chi-square distribution)

8
3. Determine significance level

• The level at which we want to make a judgment
  about the population parameter (the null
  hypothesis)
• Generally 10, 5, 1 (corresponding to 90, 95
  and 99 confidence levels) in social sciences
• The level at which the critical test statistic is
  identified

9
4. Determine degrees of freedom

• Number of bits of unconstrained data available to
  calculate a sample statistic
• E.g. for X bar, d.f. is n for s, d.f. is n-1,
  since 1 d.f. is lost due to the restriction that
  we need to calculate the mean first to calculate
  the standard deviation

10
5. Decide if it is a one / two tailed test

• One Tailed test If the Research Hypothesis is
  expressed directionally
• E.g. Head-On wants to test if consumers dislike
  their ad campaign (mean liking lt 3 (1 (strongly
  dislike) 5 (strongly like) scale).
• Ho Population mean attitudes are greater than or
  equal to 3.0
• Ha Population mean attitudes are less than 3.0
• For confirmation of Ha look in the tail of the
  direction of the Research Hypothesis

11
5. Decide if it is a one / two tailed test

• Two Tailed test If the Research Hypothesis is
  expressed without direction
• E.g. Head-On wants to test if consumers feel
  differently about their ad campaign than they
  felt a year ago. (mean liking 4.5 (1 (strongly
  dislike) 5 (strongly like) scale).
• Ho Population mean attitudes 4.5
• Ha Population mean attitudes are not equal to
  4.5
• For confirmation of Ha look in the tails on both
  sides of the distribution

12
6. Find the critical test statistic

• Critical z value requires knowledge of level of
  significance
• Critical t value requires knowledge of level of
  significance and degrees of freedom
• Critical chi-square requires knowledge of level
  of significance and degrees of freedom

13
7. Criteria for rejecting / not rejecting H0

• Compute observed test statistic
• Compare critical test statistic with observed
  test statistic
• If the absolute value of observed test statistic
  is greater than the critical test statistic,
  reject Ho
• If the absolute value of observed test statistic
  is smaller than the critical test statistic then
  Ho cannot be rejected.
• Regions of rejection / acceptance

14
Type 1 and Type 2 errors
Null hypothesis in population is
Data Analysis conclusion is
True
False
Reject Null hypothesis
Do not reject Null hypothesis
15
Type 1 and Type 2 errors

• The lower the confidence level, the greater the
  risk of rejecting a true H0 Type 1 error
  (alpha) i.e. you increase the chances of
  accepting a false research hypothesis
• i.e. if you reduce the confidence level from 95
  to 90 the chances of you declaring that the
  effect observed in the sample actually prevails
  in the population, are higher.
• If the effect in reality does not exist in the
  population, then you commit a Type 1 error.
• Therefore in Type 1 error you declare an effect
  which does not exist

16
Type 1 and Type 2 errors

• The higher the confidence level the greater the
  risk of accepting a false H0 Type 2 error
  (beta), i.e. you reduce the chances of accepting
  a true research hypothesis
• i.e. if you increase the confidence level from
  95 to 99, the chances that you miss the effect
  which may actually be there in the population,
  are higher.
• the power of the test to spot the effect is
  reduced
• Therefore power 1 beta
• Therefore in Type 2 error you miss an effect
  which exists

17
Hypothesis Testing

• Tests in this class
• Statistical Test
• Frequency Distributions ?2
• Means (one) z (if ? is known)
• t (if ? is unknown)
• Means (two) t
• Means (more than two) ANOVA

18
Chi-Square as a test of independence

• Statistical Independence if knowledge of one
  does not influence the outcome of the other
• E.g. Affiliation to school (nominally scaled)
  does not influence decision to eat at the student
  union
• Expected Value The average value in a cell if
  the sampling procedure is repeated many times
• Observed Value The value in the cell in one
  sampling procedure
• Only nominal / categorical variables

19
Chi-square Step-by-Step

• 1) Formulate Hypotheses

20
Chi-Square As a Test of Independence

• Null Hypothesis Ho
• Two (nominally scaled) variables are
  statistically independent
• There is no relationship between school
  affiliation and decision to eat at the student
  union
• Alternative Hypothesis Ha
• The two variables are not independent
• School affiliation does influence the decision to
  eat at the student union

21
Chi-square As a Test of Independence (Contd.)

• Chi-square Distribution
• A probability distribution for categorical data
• Total area under the curve is 1.0
• A different chi-square distribution is associated
  with different degrees of freedom

22
The chi-square distribution
23
Chi-square Step-by-Step

• 1) Formulate Hypotheses
• 2) Calculate row and column totals
• 3) Calculate row and column proportions
• 4) Calculate expected frequencies (Ei)
• 5) Calculate ?2 statistic

24
Chi-square Statistic (?2)

• Measures of the difference between the actual
  numbers observed in cell i (Oi), and number
  expected (Ei) under independence if the null
  hypothesis were true
• With (r-1)(c-1) degrees of freedom
• r number of rows c number of columns
• Expected frequency in each cell Ei pc pr n
• Where pc and pr are proportions for independent
  variables and n is the total number of
  observations

25
Chi-square Step-by-Step

• 1) Formulate Hypotheses
• 2) Calculate row and column totals
• 3) Calculate row and column proportions
• 4) Calculate expected frequencies (Ei)
• 5) Calculate ?2 statistic
• 6) Calculate degrees of freedom

26
Chi-square As a Test of Independence (Contd.)

• Degree of Freedom
• v (r - 1) (c - 1)
• r number of rows in contingency table
• c number of columns

27
Chi-square Step-by-Step

• 1) Formulate Hypotheses
• 2) Calculate row and column totals
• 3) Calculate row and column proportions
• 4) Calculate expected frequencies (Ei)
• 5) Calculate ?2 statistic
• 6) Calculate degrees of freedom
• 7) Obtain Critical Value from table

28
The chi-square distribution
F(x2)
Critical value 9.49
df 4
5 of area under curve
? .05
x2

• Ex Significance level .05
• Degrees of freedom 4
• CVx2 9.49

29
Chi-square Step-by-Step

• 1) Formulate Hypotheses
• 2) Calculate row and column totals
• 3) Calculate row and column proportions
• 4) Calculate expected frequencies (Ei)
• 5) Calculate ?2 statistic
• 6) Calculate degrees of freedom
• 7) Obtain Critical Value from table
• 8) Make decision regarding the Null-hypothesis

30
Example of Chi-square as a Test of Independence

• Eat / Dont eat
• Y N
• A 10 8
• School B 20 16
• C 45 18
• D 16 6
• E 9 2

This is the observed value
This is a Cell
31
Chi-square example
0.24 0.67 150
36/150
32
Chi-square example

• Observed chi-square (10 12)2 / 12 (8
  6)2 / 6 (20 24)2 / 24 (2 4)2 / 4
  5.42
• d.f. (r-1)(c-1) (5-1)(2-1) 4
• Critical chi-square at 5 level of significance
  at 4 degrees of freedom 9.49
• Since observed chi-square lt critical chi-square
  (5.42 lt 9.49), H0 cannot be rejected
• Hence decision to eat / not eat at the student
  union is statistically independent of their
  school affiliation. In other words there is no
  relationship between the decision to eat at the
  SU and the school they are in.

33
The chi-square distribution
F(x2)
Critical value 9.49
df 4
5 of area under curve
? .05
x2

• Ex Significance level .05
• Degrees of freedom 4
• CVx2 9.49
• The decision rule when testing hypotheses by
  means of chi-square distribution is
• If x2 is lt CVx2, accept H0 Thus, for 4 df and ?
  .05
• If x2 is gt CVx2, reject H0 If If x2 is lt 9.49,
  accept H0