madam siti aisyah zakaria eqt 271 sem 2 2014/2015 chapter 3 anova (extra note & exercise)

MADAM SITI AISYAH ZAKARIAEQT 271SEM 2 2014/2015

CHAPTER 3ANOVA

(EXTRA NOTE & EXERCISE)

REFRESH YOUR MIND

Agresti/Franklin Statistics, 3 of 82

The 1998 General Social Survey asked subjects how many friend they have. Is this associated with the respondent’s astrological sign (the 13 symbols of the Zodiac)? The ANOVA table for the data reports F=0.61

State the null hypothesis

a. The population means for all 12 Zodiac signs are the same.

b. At least two population means are different.



State the alternative hypothesis

a. The population means for all 12 Zodiac signs are the same.

b. At least two population means are different.



Based on what you know about the F distribution would you guess that the test value of 0.61 provides strong evidence against the null hypothesis?

a. Nob. Yes



The P-value associated with the F-statistic is 0.82. At a significance level of 0.05, what is the correct decision?

a. Reject Ho

b. Fail to Reject Ho

c. Reject Ha

d. Fail to Reject Ha

CONTINUE FROM PART 3 (FACTORIAL EXPERIMENT)

INTERACTION FACTOR


Exploring Interaction between Factors in Two-Way ANOVA

No interaction between two factors means that the effect of either factor on the response variable is the same at each category of the other factor.

[Factor A and Factor B] Do not Reject H0

Graphical

No interaction or have interaction



The lines in the graph is parallel or approximately parallel.

There is no significant interaction effect between the factor.

The difference in estimated means between the two fertilizer levels is the same for each manure level.

The effect of two factor is same



A graph showing interaction: Intersection occur

– disordinal interaction (SOME INTERACTION)

- Not more effect

No intersection – ordinal interaction (SIGNIFICANT INTERACTION)

- have more effect


Testing for Interaction

In conducting a two-way ANOVA, before testing the main effects, it is customary to test a third null hypothesis stating that their is no interaction between the factors in their effects on the response


Testing for Interaction

The test statistic providing the sample evidence of interaction is:

When H0 is false (reject H0), the F-statistic tends to be large

errorfor

ninteractiofor

MS

MSF


Example: Testing for Interaction with Corn Yield Data

ANOVA table for a model that allows interaction:


Example: Testing for Interaction with Corn Yield Data

The test statistic for H0: no interaction is• F = 1.10 with a corresponding P-value

• of 0.311

• There is not much evidence of interaction

• We would not reject H0 at the usual significance levels, such as 0.05


Check Interaction before Main Effects

In practice, in two-way ANOVA, you should first test the hypothesis of no interaction



If the evidence of interaction is not strong (that is, if the P-value is not small), then test the main effects (factor A and factor B) hypotheses and/or construct confidence intervals for those effects



If important evidence of interaction exists (have interaction in your hypothesis), plot and compare the cell means for a factor separately at each category of the other factor

REFRESH YOUR MIND


An experiment randomly assigns 100 subjects suffering from high cholesterol to one of four groups: low-dose Lipitor, high-dose Lipitor, low-dose Pravachol and high-dose Pravachol. After three months of treatment, the change in cholesterol level is measured.

What is the response variable?

a. Cholesterol levelb. Drug dosagec. Drug type


An experiment randomly assigns 100 subjects suffering from high cholesterol to one of four groups: low-dose Lipitor, high-dose Lipitor, low-dose Pravachol and high-dose Pravachol. After three months of treatment, the change in cholesterol level is measured.

What are the factors?

a. Cholesterol level and drug typeb. Drug dosage and cholesterol levelc. Drug type and drug dosage

ANOVAONE-WAY or TWO WAY

CONCLUSIONReject H0

There is difference(which one is difference???)

RELATIONSHIP BETWEEN HYPOTHESIS TESTING AND

CONFIDENT INTERVAL


For the single mean & proportion Confidence Interval vs Hypothesis Testing

0000 :@: ppHμμH

At the same level in confidence interval and hypothesis testing, when the null hypothesis is rejected, the confidence interval for the mean and proportion will not contain the hypothesized mean/proportion.

Likewise, when we do not reject null hypothesis the confidence interval will contain the hypothesized mean/ proportion.

•** Applies only for two-tailed test. •Allan Bluman, pg. 458

α


For the difference of means & proportions Confidence Interval vs Hypothesis Testing

[-8.5 , 8.5]

Contains zero

=If the CI contains zero, we do not reject H0

(Means that the there is NO DIFFERENCE in population means or proportions)

[5.45 , 12.45]No zero = If the CI does not contain zero, we reject H0

(Mean/proportion for population 1 is GREATER than the mean/proportion for population 2)

There is difference in population means

[-7.3 , -3.3]No zero =

If the CI does not contain zero, we reject H0

(Mean/proportion for population 1 is LESS than the mean/proportion for population 2)

There is difference in population means

** Applies only for two-tailed test. Eg. in Allan Bluman, pg. 477-478.

0:@0: 210210 ppHμμH

0:@0: 210210 ppHμμH

0:@0: 210210 ppHμμH

EXAMPLE

HYPOTHESIS TESTING&

CONFIDENT INTERVAL


Hog Weights

A researcher claims that adult hogs fed a special diet will have an average weight of 200 pounds. A sample of 10 hogs has an average weight of 198.2 pounds and a standard deviation of 3.3 pounds. At a 0.05, can the claim be rejected? Also, find the 95% confidence interval of the true mean.

27

Agresti/Franklin Statistics, 28 of 82 28


Sugar Production

Sugar is packed in 5-pound bags. An inspector suspects the bags may not contain 5 pounds. A sample of 50 bags produces a mean of 4.6 pounds and a standard deviation of 0.7 pound. Is there enough evidence to conclude that the bags do not contain 5 pounds as stated at a 0.05? Also, find the 95% confidence interval of the true mean.

Answer: Reject H0 and 95% confidence interval of μ does not contain the hypothesized value μ = 5.

Z = - 4.04

CI =[4.46, 4.79]

29

GOOD LUCK…

THE END….

madam siti aisyah zakaria eqt 271 sem 2 2014/2015 chapter 3 anova (extra note & exercise)

Documents