1. Chapter Thirteen Part I
Hypothesis Testing:
Basic Concepts and Tests of Association,
Chi-Square Tests

2. Basic concepts - Example
GEICO feels that consumers are bored with the gecko ad campaign (mean liking = 2; (1 (strongly dislike) – 5 (strongly like) scale).
GEICO wants to verify this feeling so they take a sample and measure liking levels. The mean in the sample is 4
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. mean attitudes are 2 measured on a 1 – 5 scale
Purpose of Hypothesis Testing: To make a judgment about the difference between the sample statistic and the population parameter
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 – 5 scale)
Observed mean sample attitudes are 4 (on a 1 – 5 scale)
Is the difference between the two a chance event or are we really wrong about our hypothesis?
This is statistically evaluated.

5. Clearly state the null and alternative hypotheses.
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 sample statistics
A mirror-image of the alternative (research) hypothesis
Alternative hypothesis (Ha or H1) – the hypothesis of differences or relationships
Example
Ho: Mean population attitudes = 2
Ha: Mean population attitudes are not = 2

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 < 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 H1 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 H1 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 Data Analysis conclusion is:
Null hypothesis in population is
True
False
Type 1 error
Prob: alpha
(Significance level)
Correct decision (Power of the test)
Correct decision (Confidence level)
Type 2 error
Prob: beta (weakness of the test)
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. 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 increase the risk of committing 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. 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 t (if  is unknown)
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
df = 4 x2
F(x2)
 = .05

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 / Don’t eat
Y N
A 10 8
School B C
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 School Eat at SU Don’t Eat
Total Pr A
O1 = 10
E1 = 12
O2 = 8
E2 = 6 18
0.12 B
O3 = 20
E3 = 24
O4 = 16
E4 = 12 36
0.24 C
O5 = 45
E5 = 42
O6 = 18
E6 = 21 63
0.42 D
O7 = 16
E7 = 15
O8 = 6
E8 = 7 22
0.15 F
O9 = 9
E9 = 7
O10 = 2
E10 = 4 11
0.07
100 50
150
1.00 Pc
0.67
0.33

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 < critical chi-square (5.42 < 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 <= CVx2, accept H0 Thus, for 4 df and  = .05
If x2 is > CVx2, reject H0 If If x2 is <= 9.49, accept H0