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Chapter 12

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Chapter 12

Analysis of Variance

- List the characteristics of the F distribution
- Conduct a test of hypothesis to determine whether the variances of two populations are equal
- Discuss the general idea of analysis of variance
- Organize data into a ANOVA table
- Conduct a test of hypothesis among three or more treatment means

- Used to test whether two samples are from populations having equal variances
- Applied when we want to compare several population means simultaneously to determine if they came from equal population
- ANOVA
- Analysis of variance

- ANOVA
- In both situations:
- Populations must be Normally distributed
- Data must be Interval-scale or higher

- Family of F Distributions
- Each is determined by:
- df in numerator
- comes from pop. 1 which has larger sample variation

- df in denominator
- comes from pop. 2 which has smaller sample variation

- df in numerator

- Each is determined by:
- F distribution is continuous
- F Value can assume an infinite number of values from 0 to ∞

- Value for F Distribution cannot be negative
- Smallest value = 0

- Positively skewed
- Long tail is always to right
- As # of df increases in both the numerator and the denominator, the distribution approaches normal

- Asymptotic
- As X increases the F curve approaches the X-axis

What if two machines are making the same part for an airplane?

Do we want the parts to be identical or nearly identical? Yes!

We would test to see if the means are the same: Chapter 10 & 11

We would test to see if the variation is the same for the two machines: Chapter 12

What if two stocks have similar mean returns?

Would we like to test and see if one stock has more variation than the other?

- Remember Chapter 11: Assumptions for small sample tests of means:
- Sample populations must follow the normal distribution
- Two samples must be from independent (unrelated) populations
- The variances & standard deviations of the two populations are equal

Always list the sample

with the

larger sample variance

as population 1

(allows us to

use fewer

tables)

- To conduct a test:
- Conduct two random samples
- List population 1 as the sample with the largest variance:
- n1 = # of observations
- s1^2 = sample variance
- n1 – 1 = df1 = degree of freedom (numerator for critical value lookup)

- List population 2 as the sample with the smaller variance:
- n2 = # of observations
- s2^2 = sample variance
- n2 – 1 = df2 = degree of freedom (denominator for critical value lookup)

- List the population with the suspected largest variance as population 1
- Because we want to limit the number of F tables we need to use to look up values, we always put the larger variance in the numerator and the smaller variance in the denominator
- This will force the F value to be at least 1
- We will only use the right tail of the F distribution

- Examples of Step 1:

- Appendix G only lists significance levels: .05 and .01

Significance level = .10

.10/2 = .05

Use .05 table in

Appendix G

Significance level = .05

Use .05 table in

Appendix G

If you have a df

that is not listed

in the border,

calculate your F

by estimating

a value

between

two values.

HW #5:

df = 11,

use value

Between

10 & 12

Book says:

(3.14+3.07)/2 =

3.105 3.10

- Look up Critical value in Appendix G and draw your picture

- Step 4: Formulate a decision rule:
- Example:
- If our calculated test statistic is greater than 3.87, reject Ho and accept H1, otherwise fail to reject Ho

Larger variancein numerator,

always!!

- Step 5: Take a random sample, compute the test statistic, compare it to critical value, and make decision to reject or not reject null and hypotheses
- Test Statistic F:
- Example Conclusion for a two tail test:
- Fail to reject null
- “The evidence suggests that there is not a difference in variation”

- Reject null and accept alternate
- “The evidence suggests that there is a difference in variation”

- Fail to reject null

Let’s Look at

Handout

Colin, a stockbroker at Critical Securities, reported that the mean rate of return on a sample of 10 software stocks was 12.6 percent with a standard deviation of 3.9 percent.

The mean rate of return on a sample of 8 utility stocks was 10.9 percent with a standard deviation of 3.5 percent. At the .05 significance level, can Colin conclude that there is more variation in the software stocks?

Step 1: The hypotheses are

Step 2: The significance level is .05.

Step 3: The test statistic is the F distribution.

Step 4:H0 is rejected if F>3.68 or if p < .05. The degrees of freedom are n1-1 or 9 in the numerator and n1-1 or 7 in the denominator.

Step 5: The value of F is computed as follows.

H0 is not rejected. There is insufficient evidence to show more variation in the software stocks.

- Technique in which we compare three or more population means to determine whether they could be equal
- Assumptions necessary:
- Populations follow the normal distribution
- Populations have equal standard deviations ()
- Populations are independent

- Assumptions necessary:
- Why ANOVA?
- Using t-distribution leads to build up of type 1 error

- “Treatment” = different populations being examined

- 22 students earned the following grades in Professor Rad’s class. The grades are listed under the classification the student gave to the instructor
- Is there a difference in the mean score of the students in each of the four categories?
- Use significance level α = .01

- Step 1: State H0 and H1
H0 : µ1 = µ2 = µ3 = µ4

H1 : The Mean scores are not all equal (at least one treatment mean is different)

- Step 2: Significance Level?
α = .01

- If our calculated test statistic is greater than we reject H0 and accept H1, otherwise we fail to reject H0

Now we move on to Step 5: Select the sample, perform calculations, and make a decision…

Are you ready for a lot of procedures?!!

- The idea is: If we estimate variation in two ways and use one estimate in the numerator and the other estimate in the denominator:
- If we divide and get 1 or close to 1, the sample means are assumed to be the same
- If we get a number far from 1, we say that the means are assumed to be different

- The F critical value will determined whether we are close to 1 or not

Let’s go calculate this!

Let’s go calculate this!

Let’s go calculate this!

Simple Subtraction!

- Because is less than 5.09, we fail to reject H0
- The evidence suggests that the mean score of the students in each of the four categories are equal (no difference)

- List the characteristics of the F distribution
- Conduct a test of hypothesis to determine whether the variances of two populations are equal
- Discuss the general idea of analysis of variance
- Organize data into a ANOVA table
- Conduct a test of hypothesis among three or more treatment means