Heiman G. Basic Statistics for the Behavioral Sciences

Подождите немного. Документ загружается.

298 CHAPTER 13 / The One-Way Analysis of Variance

which produces a larger . Turning this around, the larger the , the more it

appears that is true. Putting this all together:

The larger the , the less likely it is that H

is true and the more likely it

is that H

is true.

If our is large enough to be beyond , we will conclude that is so unlikely to

be true that we will reject and accept .

REMEMBER If is true, should equal 1 or be close to 1. The larger the

, the less likely that is true and the more likely that is true.

Before moving on to the computations, we will briefly discuss the underlying com-

ponents that represents in the population.

The Theoretical Components of the F-ratio

To fully understand the -ratio, we need to understand what and represent

in the population. We saw that estimates the variance of individual scores in the

population Statisticians also call this variance the error variance, symbolized by

. Thus, the is an estimate of (The is also referred to as the “error

term” in the -ratio.)

When is true and we have only one population, the also estimates . We saw

that with one population, the variability of sample means depends on the variability of indi-

vidual scores. Thus, although is computed using sample means, it ultimately reflects

the variability among the scores, which is . Therefore, when is true, the reason that

should equal is because both reflect the error variance in that one population.

In symbols then, here is what the -ratio represents in the population when is true.

Both mean squares are merely estimates of the one value of , so they should be

equal, and so their ratio equals 1.

On the other hand, if is false and is true, then more than one population is

involved. By measuring the differences between the means of our conditions, will

measure this treatment effect. Statisticians refer to the differences between the popula-

tions produced by a factor as the treatment variance, which is symbolized as .

Thus, is an estimate of .

However, even if a factor does produce different populations, our samples will not

perfectly represent them because of sampling error. Therefore, to some extent, the dif-

ferences between our means, as measured by the , will still reflect the variability

in scores, which we call error variance. Thus, estimates both treatment variance

plus error variance. Altogether, here is what the -ratio represents in the population

when is false and is true.

Sample Estimates Population

obt





→

 F  1

In the denominator, the is still the same estimate of . In the numerator,

however, the larger the differences between the conditions, the larger is the σ

treat

error

1 σ

treat

error

treat

error

Sample Estimates Population

obt





→



 1

.MS

error

.MS

error

1σ

obt

crit

obt

Performing the ANOVA 299

component, and so the larger will be . A larger numerator produces an that is

greater than 1. Thus, when is true, regardless of whether we have a positive, nega-

tive, or curvilinear relationship, simply will be larger than , so that is

greater than 1. This is why we test only two-tailed hypotheses. (Technically, this for-

mula is always used to describe what represents, even when is true. The

always estimates the amount of that is present, but when is true, this amount is

zero. This produces the equivalent of the previous formula where .)

REMEMBER An equal to (or less than) 1 indicates that is true. An

greater than 1 may result from sampling error, or it may indicate a treat-

ment effect in the population. A significant indicates that the means from

the conditions are likely to represent two or more .s

obt

5 1

treat

obt

■

The measures the differences among level

means (which may reflect treatment variance in the

population).

■

The measures the differences among individ-

ual scores (which reflects error variance in the pop-

ulation).

■

MORE EXAMPLES

In a study, , , and . The

equals the only when all samples belong

to the same population. Therefore, we retain the

that all conditions represent the same population. Say,

instead, that , , and .

Because is so much larger than , at least

two conditions might represent different populations

(with reflecting ). If is beyond ,

these results are unlikely to be due to sampling error,

so we accept the that at least two conditions repre-

sent different populations.

crit

obt

treat

obt

5 4MS

5 6MS

5 24

obt

5 1MS

5 6MS

5 6

obt

For Practice

1. is the symbol for ____, and is the sym-

bol for _____.

2. Differences produced by the independent variable

are called the ____ effect, and in the population,

are measured as ____.

3. Differences among the scores in the data are meas-

ured by ____ in the sample to estimate the popula-

tion ____.

4. When is true, should equal ____.

5. When is false, will be ____.

Answers

1. mean square within groups; mean square between

groups

2. treatment; treatment variance or

3. ; error variance, or

4. 1 or close to 1

5. greater than 1

error

treat

obt

A QUICK REVIEW

PERFORMING THE ANOVA

Now we can discuss the computations involved in performing the ANOVA. Recall from

the beginning of this chapter that we changed the formula for the variance into comput-

ing a mean square by dividing the sum of squares by the degrees of freedom. In sym-

bols, this is

MS 5

300 CHAPTER 13 / The One-Way Analysis of Variance

Factor A: Perceived Difficulty

Level A

: Level A

Easy Medium Difficult

941

12 6 3

484

825

7102 Totals

⌺X ⫽ 40 ⌺X ⫽ 30 ⌺X ⫽ 15 ⌺X

tot

⫽ 85

⌺X

⫽ 354 ⌺X

⫽ 220 ⌺X

⫽ 55 ⌺X

tot

⫽ 629

⫽ 5 n

⫽ 5 N ⫽ 15

苶

⫽ 8 X

苶

⫽ 6 X

苶

⫽ 3 k ⫽ 3

TABLE 13.4

Data from Perceived

Difficulty Experiment

Adding subscripts, we will compute the mean square between groups by com-

puting the sum of squares between groups and dividing by the degrees of freedom

between groups . Likewise, we will compute the mean square within groups

by computing the sum of squares within groups and dividing by the degrees of

freedom within groups . Once we have and , we can compute .

If all this strikes you as the most confusing thing ever devised, you’ll find an ANOVA

summary table very helpful. Here is its general format:

Summary Table of One-Way ANOVA

Source Sum of Squares df Mean Square F

Between SS

obt

Within SS

Total SS

tot

The source column identifies each source of variation, either between, within, or total.

In the following sections, we’ll compute the components for the other columns.

Computing the

Say that we performed the perceived difficulty study discussed earlier, telling partici-

pants that some math problems are easy, of medium difficulty, or difficult and then

measured the number of problems they solved. The data are presented in Table 13.4.

As shown in the following sections, there are four parts to computing , finding

(1) the sum of squares, (2) the degrees of freedom, (3) the mean squares, and (4)

So that you don’t get lost, fill in the ANOVA summary table as you complete each step.

(There will be a test later.)

Computing the Sums of Squares The first task is to compute the sum of squares.

Do this in four steps.

Step 1 Compute the sums and means. As in Table 13.4, compute , , and for

each level (each column). Then add together the from all levels to get the total, which

is . Also, add together the from all levels to get the total, which is .©X

tot

©X

s©X

tot

©Xs

X©X

©X

obt

1df

1SS

1MS

21df

1SS

1MS

Performing the ANOVA 301

Step 2 Compute the total sum of squares .1SS

tot

The is the sum of all , and is the sum of all squared . is the total

in the study.

Using the data from Table 13.4, , , and , so

and

Step 3 Compute the sum of squares between groups .1SS

tot

5 629 2 481.67 5 147.33

tot

5 629 2

7225

tot

5 629 2

1852

N 5 15©X

tot

5 85©X

tot

5 629

NNXs©X

tot

Xs©X

tot

Back in Table 13.4, each column represents a level of the factor. Thus, find the

for a level, square the , and then divide by the in that level. After doing this

for all levels, add the results together and subtract the quantity Thus,

we have

and

5 545 2 481.67 5 63.33

5 1320 1 180 1 4522 481.67

5 a

1402

1302

1152

b2 a

1852

1©X

tot

>N.

n©X

©X

The formula for the total sum of squares is

tot

5 ©X

tot

2 a

1©X

tot

The formula for the sum of squares between groups is

5 © a

1Sum of scores in the column2

n of scores in the column

b2 a

1©X

tot

302 CHAPTER 13 / The One-Way Analysis of Variance

In the example, is 147.33 and is 63.33, so

Filling in the first column of the ANOVA summary table, we have

Source Sum of Squares df Mean Square F

Between 63.33 df

obt

Within 84.00 df

Total 147.33 df

tot

As a double check, make sure that the total equals the between plus the within. Here,

Now compute the degrees of freedom.

Computing the Degrees of Freedom We compute , , and , so there

are three steps.

1. The degrees of freedom between groups equals , where is the number

of levels in the factor. This is because when computing we essentially

have a sample containing the means from our levels that we determine the vari-

ability of. For we reduce the number in the sample by 1, so out of means,

. In the example, with three levels of perceived difficulty, .

Thus, .

2. The degrees of freedom within groups equals , where is the total in the

experiment and is the number of levels in the factor. This is because when com-

puting we essentially find the estimated variance in each condition and pool

them. In each condition we compute the variance using , so out of the entire

experiment , we subtract one per condition, subtracting a total of In the

example, is 15 and is 3, so .

3. The degrees of freedom total equals , where is the total in the

experiment. This is because when computing the total we treat the experiment

as one sample. Then we have scores, so the is . In the example, is

15, so .

The Thus, to check our work, , which

equals 14, the .df

tot

1 df

5 2 1 12df

tot

5 df

1 df

tot

5 15 2 1 5 14

NN 2 1dfN

NNN 2 1

5 15–3 5 12kN

k.1N2

n 2 1

NNN 2 k

5 2

k 5 3df 5 k 2 1

kdf

kk 2 1

tot

63.33 1 84.00 5 147.33

5 147.33–63.33 5 84.00

tot

Step 4 Compute the sum of squares within groups . We use a shortcut to com-

pute . Mathematically, equals plus . Therefore, the total minus the

between leaves the within.

tot

1SS

The formula for the sum of squares within groups is

5 SS

tot

2 SS

Performing the ANOVA 303

Adding the to the summary table, it looks like this:

Source Sum of Squares df Mean Square F

Between 63.33 2 MS

obt

Within 84.00 12 MS

Total 147.33 14

Now find each mean square.

Computing the Mean Squares You can work directly from the summary table to

compute the mean squares. Any mean square equals the appropriate sum of squares

divided by the corresponding . Thus,df

From the summary table, we see that

63.33

5 31.67

For the example,

Do not compute the mean square for because it has no use.

Now in the summary table we have

Source Sum of Squares df Mean Square F

Between 63.33 2 31.67 F

obt

Within 84.00 12 7.00

Total 147.33 14

tot

5 7.00

The formula for the mean square between groups is

The formula for the mean square within groups is

304 CHAPTER 13 / The One-Way Analysis of Variance

Computing the F Finally, compute F

obt

Notice that in the formula, is “over” , and this is how they are positioned

in the summary table. In our example, is 31.67 and is 7.00, so

Now the completed ANOVA summary table is

Source Sum of Squares df Mean Square F

Between 63.33 2 31.67 4.52

Within 84.00 12 7.00

Total 147.33 14

The is always placed in the row labeled “Between” because the is testing

for significant differences between our conditions. Also, in the source column you

may see (1) the name of the factor at “Between,” and (2) the word “Error” in place of

“Within.”

Interpreting

We interpret by comparing it to , and for that we examine the -distribution.

The F-distribution is the sampling distribution showing the various values of that

occur when is true and all conditions represent one population. To create it, it is as

if, using our and , we select the scores for all of our conditions from one raw score

population (like says we did in our experiment) and compute , and then

. We do this an infinite number of times, and plotting the , produce the sampling

distribution, as shown in Figure 13.1.

The -distribution is skewed because there is no limit to how large can be, but it

cannot be less than zero. The mean of the distribution is 1 because, most often when

is true, will equal , so will equal 1. The right-hand tail shows that

sometimes, by chance, is greater than 1. Because our can reflect a relationship in

the population only when it is greater than 1, the entire region of rejection is in this

upper tail of the -distribution. (That’s right, ANOVA involves two-tailed hypotheses,

but they are tested using only the upper tail of the sampling distribution.)

obt

FMS

obt

FsF

obt

kns

crit

obt

31.67

7.00

5 4.52

The formula for is

obt

Performing the ANOVA 305

crit

= 3.88

α = .05

1.0FFFF F FF F FFFF

obt

= 4.52

FIGURE 13.1

Sampling distribution of

F when H

is true for

2 and df

1255

There are actually many -distributions, each having a slightly different shape,

depending on our degrees of freedom. However, two values of determine the shape

of an -distribution: the used when computing the mean square between groups

and the used when computing the mean square within groups . There-

fore, to obtain , turn to Table 5 in Appendix C, entitled “Critical Values of .”

Across the top of the tables, the columns are labeled “Degrees of Freedom Between

Groups.” Along the left-hand side, the rows are labeled “Degrees of Freedom Within

Groups.” Locate the appropriate column and row, using the from your study. The

critical values in dark type are for , and those in light type are for . For

our example, and . For , the is 3.88. (If your are

not in the table, then for from 30 to 50, your is the average of the two critical

values shown for the bracketing that are given. For above 50, compare to the

two critical values for the in the table that bracket your , using the same strategy

we discussed for t-tests in Chapter 11.)

Thus, in our perceived difficulty study, is 4.52 and is 3.88 as above in Fig-

ure 13.1. Our says that is greater than 1 because of sampling error and that actu-

ally we are poorly representing no relationship in the population. However, our is

beyond , so we reject : Our is so unlikely to occur if our samples were rep-

resenting no difference in the population that we reject that this is what they represent.

Therefore, we conclude that the is significant and that the factor of perceived diffi-

culty produces a significant difference in mean performance scores.

Of course, had been less than , then the corresponding differences between

our means would not be too unlikely to occur when is true, so we would not reject

Then, as usual, we’d draw no conclusion about our independent variable, one way

or the other. We would also consider if we had sufficient power to prevent a Type II

error (missing the relationship). We increase the power of an ANOVA using the same

strategies discussed in previous chapters: maximize the differences in scores between

conditions, minimize variability within conditions, and maximize . (These strategies

also increase the power of post hoc comparisons.)

crit

obt

crit

obt

crit

obt

dfdf

obt

dfdf

crit

 5 .05df

5 12df

5 2

 5 .01 5 .05

dfs

crit

1df

2df1df

dfF

306 CHAPTER 13 / The One-Way Analysis of Variance

■

To compute , compute , , and and

, , and Dividing by gives

; dividing by gives Dividing

by gives Compare to

MORE EXAMPLES

We test participants under conditions and

1. Compute the sums of squares:

5 SS

tot

2 SS

5 32 2 24.5 5 7.5

5 a

b2 a

b5 24.5

5 © a

1©X in column2

n in column

1©X

tot

5 ©X

tot

1©X

tot

5 320 2 a

b5 32

N 5 8n

5 4n

5 4

©X

tot

5 320©X

5 245©X

5 75

©X

tot

X 5 7.75X

5 4.25

crit

obt

.MS

.df

tot

obt

2. Compute the degrees of freedom:

3. Compute the mean squares:

4. Compute :

5. Compare to : With , ,

and , The is beyond

Therefore, the means of the conditions

differ significantly.

For Practice

1. What two components are needed to compute any

mean square?

2. For between groups, to compute ____ we divide

____ by ____. For within groups, to compute ____

we divide ____ by ____.

3. Finally, equals ____ divided by ____.

Answers

1. The sums of squares and the

2. , , ; , ,

3. , MS

obt

crit

obt

crit

5 5.99.df

5 6

5 1 5 .05F

crit

obt

5 MS

>MS

5 24.5>1.25 5 19.60

obt

5 SS

>df

5 7.5>6 5 1.25

5 SS

>df

5 24.5>1 5 24.5

tot

5 N 2 1 5 8 2 1 5 7

5 N 2 k 5 8 2 2 5 6

5 k 2 1 5 2 2 1 5 1

A QUICK REVIEW

PERFORMING POST HOC COMPARISONS

Because we rejected and accepted , we return to the means from the levels of our

factor:

Perceived Difficulty

Easy Medium Difficult

We are confident that these means represent a relationship in the population, in which

increasing perceived difficulty is associated with fewer problems solved. However,

we do not know whether every increase in difficulty produces a significant drop in

5 3X

5 6X

5 8

Performing Post Hoc Comparisons 307

performance. Therefore, we must determine which specific means differ significantly,

and to do that, we perform post hoc comparisons.

Statisticians have developed a variety of post hoc procedures that differ in how likely

they are to produce Type I or Type II errors. We’ll discuss two procedures that have

acceptably low error rates. Depending on whether or not your are equal, perform

either Fisher’s protected t-test or Tukey’s HSD test.

Fisher’s Protected t-Test

Perform Fisher’s protected t-test when the in all levels are not equal.ns

This is a variation of the independent-samples -test. We are testing ,

where and are the means for any two levels of the factor and and are the

corresponding ns in those levels. The is from our ANOVA.

It is not incorrect to perform the protected -test even when all are equal. For

example, we can compare the mean from our easy level (8) to the mean from our diffi-

cult level (3). Each is 5, and our is 7. Filling in the formula gives

Then

Next, we compare to , which we obtain from the -tables (Table 2 in

Appendix C). The here equals the from the ANOVA. For the example, with

and , is . Because the of is beyond this , the

means from the easy and difficult levels differ significantly.

To complete these comparisons, perform the protected -test on all possible pairs of

means in the factor. Thus, we would also test the means from easy and medium, and

the means from medium and difficult.

If a factor contains many levels, the protected -test becomes very tedious. If you

think there must be an easier way, you’re right.

Tukey’s HSD Multiple Comparisons Test

Perform the Tukey HSD multiple comparisons test when the in all levels are equal.

The HSD is a rearrangement of the -test that computes the minimum difference between

two means that is required for the means to differ significantly (HSD stands for the hon-

estly significant difference). There are four steps to performing the HSD test.

crit

12.99t

obt

;2.179t

crit

5 12␣ 5 .05

crit

obt

271.42

22.80

1.673

512.99

obt

8 2 3

7 a

nst

: ␮

2 ␮

5 0t

The formula for Fisher’s protected t-test is

obt

2 X