Practice Questions 8.3

For the ANOVA method, select all choices below that are assumptions of this method.

A. Individuals are chosen randomly from each population.
B. Each data value is independent of all other data values.
C. Each population has the normal shape.
D. All populations have the same variance.

With a computer, which statistical method is used to find out if the shape of a population is normal?

A.  The Shapiro-Wilk test, using the p-value approach.
B. The Bartlett's test, by comparing two values.
C. The Levene's test, using the p-value approach.
D. The Tukey's test, by comparing two values

Using the computer output below, should this population be considered to have the normal shape?

Goodness-of-Fit Test
	W	Prob<W
Shapiro-Wilk	0.9186235	0.4188

A. Yes, the shape is normal because 0.9186 is greater than 0.4188.
B. No, the shape is not normal because the p-value (0.4188) is greater than alpha (0.05).
C. Really can't tell because this is not the appropriate statistical method to use.
D.  Yes, the shape is normal because the p-value (0.4188) is greater than alpha (0.05).

With a computer, which statistical method is used to find out if all the populations have the same variance?

A. The Shapiro-Wilk, test using the p-value approach.
B. The Tukey's test, by comparing two values.
C. The Bartlett's test, by comparing two values.
D.  The Levene's test, using the p-value approach.

Using the computer output below, is it appropriate to consider that all populations have the same variance?

Test	F Ratio	DFNum	DFDen	Prob > F
O'Brien[.5]	1.4419	2	21	0.2590
Brown-Forsythe	1.1006	2	D21	0.3511
Levene	1.3125	2	D21	0.2903
Bartlett	0.4683	2	.	0.6261

A.  Yes, the population variances should be considered equal because the p-value (0.2903) is greater than alpha (0.05).
B. Really can't tell because this is not the appropriate statistical method to use.
C. Yes, the population variances should be considered equal because the F-Ratio (1.3125) is greater than one (1).
D. No, the population variances should be considered not equal because the p-value (0.4188) is greater than alpha (0.05).

The two variances of the ANOVA fraction are shown below, what two variances are used in the ANOVA table?
(s²x) and (s²pooled / n)

A.  (s²x * n) and (s² pooled)
B. (s²x) and (s²pooled * n)
C.  (s²x) and (s²pooled)
D. (s²x + s²pooled) and (n)

For the ANOVA table, select all choices that is a column heading in the ANOVA table.

A. Source column.
B. Degrees of Freedom column.
C. Sum-of-Squares column
D. Mean Squares column.
E. F-Value column.
F. P-Value column.

In the ANOVA table below, are all the degrees of freedom correct if there are three (3) populations and ten (10) data values from each population?

Analysis of Variance

Source	DF	Sum of Square	Mean Square	F Ratio	Prob> F
Day	2	7129434	3564717	18.3965	<.0001*
21	21	4069197	193771
C. Total	23	11198631

A.  No, because the degrees of freedom should be, Error=27 and C. Total=29.
B. No, because the degrees of freedom should be, DAY=3.
C. Yes, because the degree of freedom check is met, 2+21= 23.
D. Yes, because the computer is not wrong.

In the ANOVA table below, what is the value of the F-Ratio.

Analysis of Variance

Source	DF	Sum of Squares	Mean Square	F Ratio	Prob> F
Day	2	7129434	3566717	jQuery1124037259809705227087_1712104222390?	<.0001*
Error	21	4069197	196771
C. Total	23	11198631

A. F-Ratio = 1.7520
B. F-Ratio = 190.3254
C.  F-Ratio = 18.1262
D. F-Ratio = 18.3965

In the ANOVA table below, match up the values of the two Mean Squares.

Analysis of Variance

Sourec	DF	Sum of Squares	Mean Square	F Ratio	Prob> F
DAY	2	7121434	jQuery1124018545071294071036_1712455621884?	18.3965	<.0001*
Error	21	4062197	jQuery112404324943061533939_1712455744412?
C. Total	23	11183631

A. MS(Errpr)
B. MS(DAY)

__ 3,560,717
__ 193,437

In the ANOVA table below, match up the values of the two Sum of Squares.

Analysis of Variance

Sourec	DF	Sum of Squares	Mean Square	F Ratio	Prob> F
DAY	2	jQuery112401157404587023958_1712521158377?	3564717	18.3965	<.0001*
Error	21	jQuery112406352651159208165_1712527155458?	193771
C. Total	23	11183631

A. SS(Error)
B. SS{DAY}

__ 7,129,434
__ 4,069,197

In the ANOVA table below, should the populations means be considered equal?

Analysis of Variance

Sourec	DF	Sum of Squares	Mean Square	F Ratio	Prob> F
DAY	2	7129434	3564717	18.3965	<.0001*
Error	21	4069197	193771
C. Total	23	11183631

A. No , the population means are not equal, because MS(DAY) (3,564,717) is greater than MS(Error) (193,771).
B.  No, the population means are not equal, because p-value (<0.0001)  is less than alpha.
C. Yes, the population means are equal because p-value (<0.0001) is less than alpha (0.05).
D. Yes, the population means are equal, because SS(DAY) (7,129,434) and SS(Error) (4,069,197)add up to SS(C. Total) (11,198,631).

In the ANOVA table below, should the populations variances be considered equal?

A. No, because there are no sample variances In the ANOVA table.
B. No , because the MS(DAY) (3,564,717) does not equal MS(Error) (193,771).
C.  Unknown, because the ANOVA table tests for means, not for variances.
D. No, because the p-value (<0.0001) is less than alpha (0.05).

Please match each table below with its purpose in the ANOVA method.

A. The Tukey table
B. The ANOVA table

__ Gives an overall test to tell if all population means are equal.
__ Gives a series of tests to tell which population means are equal.

Does the Tukey table provide new information when the null hypothesis is not rejected in the ANOVA table?

A. Yes, the Tukey table gives new information about the population variances.
B.  Not really, because one population mean means one letter in the Tukey table.
C. Of course, because any additional statistical test gives more information about the situation.
D. Not really, because the Tukey table always gives the same information as the ANOVA table.

Does the Tukey table provide new information when the null hypothesis is rejected in the ANOVA table

A. Yes, the Tukey table indicates which population means are equal, and which are not equal.
B. No, because the Tukey table will indicate different population means just as the ANOVA table. 
C. No, because the Tukey table should not be used when the null in the ANOVA table is rejected.
D. Yes, because any additional statistical test gives more information about the situation.

In the Tukey table shown below from a study of the number of births per weekday, what relationship between the values of the population means is NOT appropriate?

Tukey Table
Connecting Letters Report

Level		Mean
FRIDAY	A	12002.875
WEDNESDAY	A	11586.875
MONDAY	B	10696.250

A. Friday's, Wednesday's, and Monday's population mean number of births are equal.
B. Friday's and Monday's population number of births are both greater than Monday's population mean.
C. Friday's, and Monday's, population number of births are equal.
D. Friday's population mean number of births = Wednesday's ≠ Monday's.

In the Tukey table shown below from a study of the number of births per weekday, what relationship between the values of the population means is appropriate?

Tukey Table
Connecting Letters Report

Level		Mean
TUESDAY	A	12237.125
THURSDAY	A	11897.875
WEDNESDAY	A	11586.875

Given the information shown below from a study of the number of births per weekday, please match each of the following questions with their appropriate answers.


A. Which population means are equal / not equal.
B. Is the assumption of equal population variances met?
C. Is the assumption of Normality met?
D. Are all population means equal?

__ Yes, because all SW p-values are greater than 0.05.
__ Yes, because Levene' p-value is greater than 0.05.
__ No, because the ANOVA p-value is less than 0.05.
__ Friday & Wednesday are equal / not Monday.

Given the information shown below from a study of the number of births per weekday, please match each of the following questions with their appropriate answers.


A. Which population means are equal / not equal.
B. Are all population means equal?
C. Is the assumption of equal population variances met?
D. Is the assumption of Normality met?

__ Yes, because all SW p-values are greater than 0.05
__ Yes, because Levene' p-value is greater than 0.05
__ Yes, because the ANOVA p-value is greater than 0.05.
__ Tuesday = Wednesday = Thursday, because all days have the same Tukey letter.

Given the information shown below from a study of the number of births by season, please match each of the following questions with their appropriate answers.


A. Is the assumption of Normality met?
B. Which population means are equal / not equal.
C. Are all population means equal?
D. Is the assumption of equal population variances met?

__ No, SW indicates that Winter and Summer do not have the normal shape.
__ Yes, because Levene' p-value is greater than 0.05.
__ Yes, because the ANOVA p-value is greater than 0.05.
__ Tukey's table does not add any new information.