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# Chapter Eleven Part 3 (Sections 11. 4 & 11.5) Chi-Square and F Distributions - PowerPoint PPT Presentation

Understandable Statistics S eventh Edition By Brase and Brase Prepared by: Lynn Smith Gloucester County College. Chapter Eleven Part 3 (Sections 11. 4 & 11.5) Chi-Square and F Distributions. Testing Two Variances.

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Chapter Eleven Part 3 (Sections 11. 4 & 11.5) Chi-Square and F Distributions

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## Understandable StatisticsSeventh EditionBy Brase and BrasePrepared by: Lynn SmithGloucester County College

Chapter Eleven Part 3

(Sections 11. 4 & 11.5)

Chi-Square and F Distributions

## Testing Two Variances

Use independent samples from two populations to test the claim that the variances are equal.

### Assumptions for Testing Two Variances

• The two populations are independent

• The two populations each have a normal probability distribution.

### Set Up Hypotheses

Set Up Hypotheses

## Equivalent hypotheses may be stated about standard deviations.

### The F Distribution

• Not symmetrical

• Skewed right

• Values are always greater than or equal to zero.

• A specific F distribution is determined from two degrees of freedom.

An F Distribution

### Degrees of Freedom for Test of Two Variances

• Degrees of freedom for the numerator =

d.f. N = n1 - 1

• Degrees of freedom for the denominator =

d.f. D = n2 - 1

## Values of the F Distribution

Given in Table 8 of Appendix II

## Find critical value of F from Table 8 Appendix II

d.f.N = 3

d.f.D = 5

Right tail area =  = 0.025

F Distributiond.f.N = 3, d.f.D = 5,  = 0.025

## Testing Two Variances

Assume we have the following data and wish to test the claim that the population variances are not equal.

### Hypotheses

The Sample Test Statistic

Degrees of Freedom for Test of Two Variances

• Degrees of freedom for the numerator =

d.f. N = n1 - 1 = 9 - 1 = 8

• Degrees of freedom for the denominator =

d.f. D = n2 - 1 = 10 - 1 = 9

### Critical Values of F Distribution

• Use  = 0.05

• For a two-tailed test , the area in the right tail of the distribution should be  /2 = 0.025.

• With d.f. N = 8 and d.f. D = 9 the critical value of F is 4.10.

Critical Value of F: Two-Tailed Test

Area = /2

F = 4.10

Our Test Statistic Does not fall in the Critical Region

Area = /2

F = 1.108

F = 4.10

## Conclusion

At 5% level of significance, we cannot reject the claim that the variances are the same.

### P Value Approach

• Our sample test statistic was F = 1.108

• Looking in the block of entries in table 8 where d.f. N = 8 and d.f. D = 9, we find entries ranging from 3.23 to 5.47 for  ranging from 0.050 to 0.010.

• F = 1.108 is less than even the smallest of these results.

## P Value Conclusion

For a two-tailed test, double the area in the right tail.

Therefore P is greater than 0.100.

## Analysis of Variance

A technique used to determine if there are differences among means for several groups.

## One Way Analysis of Variance

Groups are based on values of only one variable.

## ANOVA

Analysis of Variance

### Assumptions for ANOVA

• Each of k groups of measurements is from a normal population.

• Each group is randomly selected and is independent of all other groups.

• Variables from each group come from distributions with approximately the same standard deviation.

## Purpose of ANOVA

To determine the existence (or nonexistence) of a statistically significant difference among the group means

## Null Hypothesis

All the group populations are the same.

All sample groups come from the same population.

## Alternate Hypothesis

Not all the group populations are equal.

### Hypotheses

• H0:1 = 2 = . . . = k

• H1:At least two of the means 1, 2, . . . , k are not equal.

### Steps in ANOVA

• Determine null and alternate hypotheses

• Find SS TOT= the sum of the squares of entire collection of data

• Find SS BET which measures variability between groups

• Find SS W which measures variability within groups

Steps in ANOVA

• Find the variance estimates within groups: MS W

• Find the variance estimates between groups: MS BET

• Find the F ratio and complete the ANOVA test

### Sample sizes

• The sample sizes for the groups may be the same or different from one another.

• In our example, each sample has four items.

## Population Means

Let 1, 2, and 3 represent the population means of groups 1, 2, and 3.

Hypotheses and Level of Significance

• H0:1 = 2 = 3

• H1:At least two of the means 1, 2, and 3 are not equal.

Use  = 0.05.

### Find SS TOT= the sum of the squares of entire collection of data

Find SS TOT= the sum of the squares of entire collection of data

Find SS TOT= the sum of the squares of entire collection of data

Find SS TOT= the sum of the squares of entire collection of data

Squares and Sums

Finding SS TOT

Find SS TOT

Finding SS BET

### Variation Within First Group

Variation Within Second Group

Variation Within Third Group

## Variability Within All Groups

SS W = SS 1 + SS 2 + SS 3

Finding SS W

SS W = SS 1 + SS 2 + SS 3=

4.75 + 2.00 + 2.75 =

9.50

## Note:

SS TOT = SS BET + SS W

## We can check:

SS TOT = SS BET + SS W =

94 = 84.5 + 9.5

### Degrees of Freedom

• Degrees of freedom between groups =

d.f. BET = k - 1, where k is the number of groups.

• Degrees of freedom within groups =

d.f. W = N - k, where N is the total sample size.

• d.f. BET + d.f. W = N - 1.

### Variance Estimate Between Groups

Mean Square Between Groups =

Variance Estimate Within Groups

Mean Square Within Groups =

Mean Square Between Groups

Mean Square Within Groups

### Find the F Ratio

Finding the F Ratio

### Null Hypothesis: All the groups are samples from the same distribution

• If H0 is true MS BET and MS W would estimate the same quantity.

• The F ratio should be approximately equal to one.

Using Table 8, Appendix II

• d.f. BET = number of groups - 1 = d.f. for numerator.

• d.f. W = total sample size - number of groups = d.f. for denominator.

• Rejection region is the right tail of the distribution.

Using Table 8, Appendix II

• d.f. BET = number of groups - 1 = d.f. for numerator = 2.

• d.f. W = total sample size - number of groups = d.f. for denominator = 9.

•  = 0.05.

• Critical F value = 4.26

### Conclusion

• Since our observed value of F (40.009) is greater than the critical F value (4.26) we reject the null hypothesis.

• We conclude that not all of the means are equal.

### P Value Approach

• For d.f. BET = d.f. for numerator = 2 and d.f. W = d.f. for denominator = 9, our observed F value (42.009) exceeds even the largest critical F value.

• Thus P < 0.01.

• We conclude that we would reject the null hypothesis for any   0.01.