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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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### 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 equal) sample variance

Set Up Hypotheses equal) sample variance

### Equivalent hypotheses may be stated about standard deviations.

Use the F Statistic deviations.

The F Distribution 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 deviations.

Degrees of Freedom for Test of Two Variances deviations.

• 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 deviations.

Given in Table 8 of Appendix II

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

d.f.N = 3

d.f.D = 5

Right tail area =  = 0.025

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

### Testing Two Variances deviations.

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

Hypotheses that the population variances are

The Sample Test Statistic that the population variances are

Degrees of Freedom for Test of Two Variances that the population variances are

• 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 that the population variances are

• 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 that the population variances are

Area = /2

F = 4.10

Our Test Statistic Does not fall in the Critical Region that the population variances are

Area = /2

F = 1.108

F = 4.10

### Conclusion that the population variances are

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

P Value Approach that the population variances are

• 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 that the population variances are

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

Therefore P is greater than 0.100.

### Analysis of Variance that the population variances are

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

### One Way Analysis of Variance that the population variances are

Groups are based on values of only one variable.

### ANOVA that the population variances are

Analysis of Variance

Assumptions for ANOVA that the population variances are

• 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 that the population variances are

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

### Null Hypothesis that the population variances are

All the group populations are the same.

All sample groups come from the same population.

### Alternate Hypothesis that the population variances are

Not all the group populations are equal.

Hypotheses that the population variances are

• H0: 1 = 2 = . . . = k

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

Steps in ANOVA that the population variances are

• 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 that the population variances are

• 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

Data for three groups: that the population variances are

Sample sizes that the population variances are

• 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 that the population variances are

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

Hypotheses and Level of Significance that the population variances are

• H0: 1 = 2 = 3

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

Use  = 0.05.

Find SS that the population variances are TOT= the sum of the squares of entire collection of data

Find SS that the population variances are TOT= the sum of the squares of entire collection of data

Find SS that the population variances are TOT= the sum of the squares of entire collection of data

Find SS that the population variances are TOT= the sum of the squares of entire collection of data

Squares and Sums that the population variances are

Finding SS that the population variances are TOT

Find SS that the population variances are TOT

Find SS that the population variances are BET

Finding SS that the population variances are BET

Variability Within that the population variances are ith Group

Variation Within First Group that the population variances are

Variation Within Second Group that the population variances are

Variation Within Third Group that the population variances are

### Variability Within All Groups that the population variances are

SS W = SS 1 + SS 2 + SS 3

Finding that the population variances are SS W

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

4.75 + 2.00 + 2.75 =

9.50

### Note: that the population variances are

SS TOT = SS BET + SS W

### We can check: that the population variances are

SS TOT = SS BET + SS W =

94 = 84.5 + 9.5

Degrees of Freedom that the population variances are

• 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 that the population variances are

Mean Square Between Groups =

Variance Estimate Within Groups that the population variances are

Mean Square Within Groups =

Mean Square Between Groups that the population variances are

Mean Square Within Groups that the population variances are

Find the F Ratio that the population variances are

Finding the F Ratio that the population variances are

• 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 distribution

• 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 distribution

• 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 distribution

• 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 distribution

• 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.