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

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

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

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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.

- The two populations are independent
- The two populations each have a normal probability distribution.

Set Up Hypotheses

Equivalent hypotheses may be stated about standard deviations.

- 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 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.

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

- 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.

- 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

- 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.

- H0:1 = 2 = . . . = k
- H1:At least two of the means 1, 2, . . . , k are not equal.

- 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

- 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

Squares and Sums

Finding SS TOT

Find SS TOT

Finding SS BET

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 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.

Mean Square Between Groups =

Variance Estimate Within Groups

Mean Square Within Groups =

Mean Square Between Groups

Mean Square Within Groups

Finding the F Ratio

- 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

- 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.

- 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.