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

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Hypothesis Testing:

Basic Concepts and Tests of Association,

Chi-Square Tests

- 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?

- 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 population’s attitude?

- The mechanism adopted to make this objective judgment is the core of hypothesis testing

- 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.

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

- 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

- the hypothesis of no difference
- 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

- 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)

- 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

- 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

- 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 Ha look in the tail of the direction of the Research Hypothesis

- 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

- 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

- 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

Data Analysis

conclusion is:

Null hypothesis in population is

True

False

Reject Null

hypothesis

Do not reject Null

hypothesis

- 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

- 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

Tests in this class

Statistical Test

- Frequency Distributions2
- Means(one)z (if is known)
t (if is unknown)

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

- 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

1) Formulate Hypotheses

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 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

F(x2)

df = 4

= .05

x2

1) Formulate Hypotheses

2) Calculate row and column totals

3) Calculate row and column proportions

4) Calculate expected frequencies (Ei)

5) Calculate 2 statistic

- 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 rowsc = 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

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

Degree of Freedom

v = (r - 1) * (c - 1)

r = number of rows in contingency table

c = number of columns

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

F(x2)

Critical value = 9.49

df = 4

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

CVx2 = 9.49

5% of area under curve

= .05

x2

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

Eat / Don’t eat

YN

A108

SchoolB2016

C4518

D166

E92

This is the observed value

This is a ‘Cell’

0.24 * 0.67 * 150

36/150

- 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.

F(x2)

Critical value = 9.49

df = 4

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 H0Thus, for 4 df and = .05

If x2 is > CVx2, reject H0If If x2 is <= 9.49, accept H0

5% of area under curve

= .05

x2