Chapter Thirteen Part I

Hypothesis Testing: Hypothesis Testing:

Basic Concepts and Tests of Basic Concepts and Tests of Association,Association,

Chi-Square TestsChi-Square Tests

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?

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

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.

Problem Definition

Clearly state the null and alternative

hypotheses.Choose the relevant

test and the appropriate probability distributionChoose the critical

value

Compare test statistic and critical value

Reject null

Does the test statistic fall in the critical

region?

Determine the significance

level

Compute relevant test

statistic

Determine the degrees of freedom

Decide if one-or two-tailed

test

Do not reject null

No

Yes

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

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)

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

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

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

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

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

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

Type 1 and Type 2 errors

Type 1 errorProb: alpha(Significance level)

Correct decision (Power of the test)

Correct decision (Confidence level)

Type 2 errorProb: beta (weakness of the test)

Null hypothesis in population is

True False

Data Analysis conclusion is:

Reject Null hypothesis

Do not reject Nullhypothesis

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

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

Hypothesis Testing

Tests in this classStatistical Test

• Frequency Distributions 2

• Means (one) z (if is known)

t (if is unknown)

• Means (two) t • Means (more than two) ANOVA

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

Chi-square Step-by-Step

1) Formulate Hypotheses

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

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

The chi-square distribution

df = 4

x2

F(x2)

= .05

Chi-square Step-by-Step

1) Formulate Hypotheses2) Calculate row and column totals3) Calculate row and column proportions4) Calculate expected frequencies (Ei)

5) Calculate 2 statistic

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

i

iin

i E

EO 2

1

2 )(

Chi-square Step-by-Step1) Formulate Hypotheses2) Calculate row and column totals3) Calculate row and column proportions4) Calculate expected frequencies (Ei)

5) Calculate 2 statistic6) Calculate degrees of freedom

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

Chi-square Step-by-Step

1) Formulate Hypotheses2) Calculate row and column totals3) Calculate row and column proportions4) Calculate expected frequencies (Ei)

5) Calculate 2 statistic6) Calculate degrees of freedom7) Obtain Critical Value from table

The chi-square distribution

• Ex: Significance level = .05Degrees of freedom = 4CVx

2 = 9.49

df = 4

x2

F(x2) Critical value = 9.49

5% of area under curve

= .05

Chi-square Step-by-Step

1) Formulate Hypotheses2) Calculate row and column totals3) Calculate row and column proportions4) Calculate expected frequencies (Ei)

5) Calculate 2 statistic6) Calculate degrees of freedom7) Obtain Critical Value from table8) Make decision regarding the Null-

hypothesis

Example of Chi-square as a Test of Independence

Eat / Don’t eat

Y N

A 10 8

School B 20 16

C 45 18

D 16 6

E 9 2

This is a ‘Cell’

This is the

observed value

Chi-square example

School Eat at SU Don’t Eat Total Pr

A O1 = 10E1 = 12

O2 = 8E2 = 6

18 0.12

B O3 = 20E3 = 24

O4 = 16E4 = 12

36 0.24

C O5 = 45E5 = 42

O6 = 18E6 = 21

63 0.42

D O7 = 16E7 = 15

O8 = 6E8 = 7

22 0.15

F O9 = 9E9 = 7

O10 = 2E10 = 4

11 0.07

Total 100 50 150 1.00

Pc 0.67 0.33 1.00

36/1500.24 * 0.67 *

150

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.

The chi-square distribution

Ex: Significance level = .05Degrees of freedom = 4CVx

2 = 9.49

The decision rule when testing hypotheses by means of chi-square distribution is:

If x2 is <= CVx2, accept H0Thus, for 4 df and = .05

If x2 is > CVx2, reject H0 If If x2 is <= 9.49, accept

H0

df = 4

x2

F(x2) Critical value = 9.49

5% of area under curve

= .05