In the formula t m1 m2 sdifference sdifference is

270

✪ The Distribution of Differences Between Means 271

✪ Hypothesis Testing with a t Test for Independent Means 278

✪ Assumptions of the t Test for Independent Means 286

✪ Effect Size and Power for the t Test for Independent Means 288

✪ Review and Comparison of the Three Kinds of t Tests 290

✪ Controversy: The Problem of Too Many t Tests 291

In the previous chapter, you learned how to use the t test for dependent means tocompare two sets of scores from a single group of people (such as the same menmeasured on communication quality before and after premarital counseling). In this chapter, you learn how to compare two sets of scores, one from each of

two entirely separate groups of people. This is a very common situation in psychol- ogy research. For example, a study may compare the scores from individuals in an experimental group and individuals in a control group (or from a group of men and a group of women). This is a t test situation because you don’t know the population variances (so they must be estimated). The scores of the two groups are indepen- dent of each other; so the test you learn in this chapter is called a t test for inde- pendent means.

✪ The t Test for Independent Means in Research Articles 292

✪ Advanced Topic: Power for the t Test for Independent Means When Sample Sizes Are Not Equal 293

✪ Summary 294

✪ Key Terms 295

✪ Example Worked-Out Problems 295

✪ Practice Problems 298

✪ Using SPSS 305

✪ Chapter Notes 309

The t Test for Independent Means

Chapter Outline

CHAPTER 8

t test for independent means hypothesis-testing procedure in which there are two separate groups of people tested and in which the popula- tion variance is not known.

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The t Test for Independent Means 271

Let’s consider an example. A team of researchers is interested in the effect on physical health of writing about thoughts and feelings associated with traumatic life events. This kind of writing is called expressive writing. Suppose the researchers recruit undergraduate students to take part in a study and randomly assign them to be in an expressive writing group or a control group. Students in the expressive writ- ing group are instructed to write four 20-minute essays over four consecutive days about their most traumatic life experiences. Students in the control group write four 20-minute essays over four consecutive days describing their plans for that day. One month later, the researchers ask the students to rate their overall level of physi- cal health (on a scale from to ). Since the expressive writing and the control group contain different students, a t test for inde- pendent means is the appropriate test of the effect of expressive writing on physical health. We will return to this example later in the chapter. But first, you will learn about the logic of the t test for independent means, which involves learning about a new kind of distribution (called the distribution of differences between means).

The Distribution of Differences Between Means In the previous chapter, you learned the logic and figuring for the t test for dependent means. In that chapter, the same group of people each had two scores; that is, you had a pair of scores for each person. This allowed you to figure a difference score for each person. You then carried out the hypothesis-testing procedure using these dif- ference scores. The comparison distribution you used for this hypothesis testing was a distribution of means of difference scores.

In the situation you face in this chapter, the scores in one group are for different people than the scores in the other group. So you don’t have any pairs of scores, as you did when the same group of people each had two scores. Thus, it wouldn’t make sense to create difference scores, and you can’t use difference scores for the hypothesis- testing procedure in this chapter. Instead, when the scores in one group are for differ- ent people than the scores in the other group, what you can compare is the mean of one group to the mean of the other group.

So the t test for independent means focuses on the difference between the means of the two groups. The hypothesis-testing procedure, however, for the most part works just like the hypothesis-testing procedures you have already learned. Since the focus is now on the difference between means, the comparison distribution is a distribution of differences between means.

A distribution of differences between means is, in a sense, two steps removed from the populations of individuals: First, there is a distribution of means from each population of individuals; second, there is a distribution of differences between pairs of means, one of each pair from each of these distributions of means.

Think of this distribution of differences between means as being built up as follows: (a) randomly select one mean from the distribution of means for the first group’s population, (b) randomly select one mean from the distribution of means for the second group’s population, and (c) subtract. (That is, take the mean from the first distri- bution of means and subtract the mean from the second distribution of means.) This gives a difference score between the two selected means. Then repeat the process. This creates a second difference score, a difference between the two newly selected means. Repeating this process a large number of times creates a distribution of differences be- tween means. You would never actually create a distribution of differences between means using this lengthy method. But it shows clearly what makes up the distribution.

100 = perfect health0 = very poor health

distribution of differences between means distribution of differences between means of pairs of samples such that, for each pair of means, one is from one population and the other is from a second population; the comparison distribution in a t test for independent means.

T I P F O R S U C C E S S The comparison distributions for the t test for dependent means and the t test for independent means have similar names: a distribution of means of difference scores, and a distribution of differences be- tween means, respectively. Thus, it can be easy to confuse these com- parison distributions. To remember which is which, think of the logic of each t test. The t test for depen- dent means involves difference scores. So, its comparison distrib- ution is a distribution of means of difference scores. The t test for independent means involves differences between means. Thus, its comparison distribution is a dis- tribution of differences between means.

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272 Chapter 8

Distributions of means

Populations

Samples

Distribution of differences between

means

Figure 8–1 Diagram of the logic of a distribution of differences between means.

The Logic Figure 8–1 shows the entire logical construction for a distribution of differences between means. At the top are the two population distributions. We do not know the characteristics of these population distributions, but we do know that if the null hy- pothesis is true, the two population means are the same. That is, the null hypothesis is that . We also can estimate the variance of these populations based on the sample information (these estimated variances will be and ).

Below each population distribution is the distribution of means for that popula- tion. Using the estimated population variance and knowing the size of each sample, you can figure the variance of each distribution of means in the usual way. (It is the estimated variance of its parent population divided by the size of the sample from that population that is being studied.)

Below these two distributions of means, and built from them, is the crucial distribution of differences between means. This distribution’s variance is ultimately based on estimated population variances. Thus, we can think of it as a t distribution. The goal of a t test for independent means is to decide whether the difference be- tween the means of your two actual samples is a more extreme difference than the cutoff difference on this distribution of differences between means. The two actual samples are shown (as histograms) at the bottom.

Remember, this whole procedure is really a kind of complicated castle in the air. It exists only in our minds to help us make decisions based on the results of an actual ex- periment. The only concrete reality in all of this is the actual scores in the two samples. You estimate the population variances from these sample scores. The variances of the two distributions of means are based entirely on these estimated population variances (and the sample sizes). And, as you will see shortly, the characteristics of the distribu- tion of differences between means are based on these two distributions of means.

Still, the procedure is a powerful one. It has the power of mathematics and logic behind it. It helps you develop general knowledge based on the specifics of a particu- lar study.

With this overview of the basic logic, we now turn to six key details: (1) the mean of the distribution of differences between means, (2) the estimated population variance, (3) the variance of the two distributions of means, (4) the variance and standard deviation of the distribution of differences between means, (5) the shape of the distribution of differences between means, and (6) the t score for the difference between the two means being compared.

S22S 2 1

�1 = �2

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The t Test for Independent Means 273

Mean of the Distribution of Differences Between Means In a t test for independent means, you are considering two populations: for example, one population from which an experimental group is taken and one population from which a control group is taken. In practice, you don’t know the mean of either popu- lation. You do know that if the null hypothesis is true, these two populations have equal means. Also, if these two populations have equal means, the two distributions of means have equal means. (This is because each distribution of means has the same mean as its parent population of individuals.) Finally, if you take random samples from two distributions with equal means, the differences between the means of these random samples, in the long run, balance out to 0. The result of all this is the follow- ing: whatever the specifics of the study, you know that, if the null hypothesis is true, the distribution of differences between means has a mean of 0.

Estimating the Population Variance In Chapter 7, you learned to estimate the population variance from the scores in your sample. It is the sum of squared deviation scores divided by the degrees of freedom (the number in the sample minus 1). To do a t test for independent means, it has to be reasonable to assume that the populations the two samples come from have the same variance (which, in statistical terms, is called homogeneity of variance). (If the null hypothesis is true, they also have the same mean. However, whether or not the null hypothesis is true, you must be able to assume that the two populations have the same variance.) Therefore, when you estimate the variance from the scores in either sample, you are getting two separate estimates of what should be the same number. In practice, the two estimates will almost never be exactly identical. Since they are both supposed to be estimating the same thing, the best solution is to average the two estimates to get the best single overall estimate. This is called the pooled estimate of the population variance ( ).

In making this average, however, you also have to take into account that the two samples may not be the same size. If one sample is larger than the other, the estimate it provides is likely to be more accurate (because it is based on more information). If both samples are exactly the same size, you could just take an ordinary average of the two estimates. On the other hand, when they are not the same size, you need to make some adjustment in the averaging to give more weight to the larger sample. That is, you need a weighted average, an average weighted by the amount of infor- mation each sample provides.

Also, to be precise, the amount of information each sample provides is not its number of scores, but its degrees of freedom (its number of scores minus 1). Thus, your weighted average needs to be based on the degrees of freedom each sample provides. To find the weighted average, you figure out what proportion of the total degrees of freedom each sample contributes and multiply that proportion by the pop- ulation variance estimate from that sample. Finally, you add up the two results, and that is your weighted, pooled estimate. In terms of a formula,

(8–1)

In this formula, is the pooled estimate of the population variance. is the degrees of freedom in the sample from Population 1, and is the degrees of freedom in the sample from Population 2. (Remember, each sample’s df is its number of scores minus 1.) is the total degrees of freedom . is theS2 1(dfTotal = df1 + df2)dfTotal

df2

df1S 2 Pooled

S2Pooled = df1

dfTotal (S21) +

df2 dfTotal

(S22)

S2Pooled

pooled estimate of the population variance ( ) in a t test for inde- pendent means, weighted average of the estimates of the population variance from two samples (each estimate weighted by the proportion consisting of its sample’s degrees of freedom divided by the total degrees of freedom for both samples).

S2Pooled

weighted average average in which the scores being averaged do not have equal influence on the total, as in figur- ing the pooled variance estimate in a t test for independent means.

The pooled estimate of the population variance is the de- grees of freedom in the first sample divided by the total degrees of freedom (from both samples), multiplied by the population estimate based on the first sample, plus the degrees of freedom in the second sample divided by the total degrees of freedom mul- tiplied by the population variance estimate based on the second sample.

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274 Chapter 8

estimate of the population variance based on the scores in Population 1’s sample; is the estimate based on the scores in Population 2’s sample.

Consider a study in which the population variance estimate based on an experi- mental group of 11 participants is 60, and the population variance estimate based on a control group of 31 participants is 80. The estimate from the experimental group is based on 10 degrees of freedom (11 participants minus 1), and the estimate from the control group is based on 30 degrees of freedom (31 minus 1). The total information on which the estimate is based is the total degrees of freedom—in this example, 40 (that is, ). Thus, the experimental group provides one-quarter of the infor- mation ( ), and the control group provides three-quarters of the informa- tion ( ).

You then multiply the estimate from the experimental group by , making 15 (that is, ), and you multiply the estimate from the control group by

, making 60 (that is, ). Adding the two gives an overall estimate of 15 plus 60, which is 75. Using the formula,

Notice that this procedure does not give the same result as ordinary averaging (without weighting).

Ordinary averaging would give an estimate of 70 (that is, ). Your weighted, pooled estimate of the population variance of 75 is closer to the esti- mate based on the control group alone than to the estimate based on the experimen- tal group alone. This is as it should be, because the control group estimate in this example was based on more information.

Figuring the Variance of Each of the Two Distributions of Means The pooled estimate of the population variance is the best estimate for both popula- tions. (Remember, to do a t test for independent means, you have to be able to as- sume that the two populations have the same variance.) However, even though the two populations have the same variance, if the samples are not the same size, the dis- tributions of means taken from them do not have the same variance. That is because the variance of a distribution of means is the population variance divided by the sam- ple size. In terms of formulas,

(8–2)S2M1 = S2Pooled

N1

360 + 804>2 = 70

= 1

4 (60) +

3

4 (80) = 15 + 60 = 75.

S2Pooled = df1

dfTotal (S21) +

df2 dfTotal

(S22) = 10

40 (60) +

30

40 (80)

80 * 3>4 = 603>4 60 * 1>4 = 15 1>430>40 = 3>4

10>40 = 1>410 + 30

T I P F O R S U C C E S S You know you have made a mis- take in figuring if it does not come out between the two esti- mates of the population variance. (You also know you have made a mistake if it does not come out closer to the estimate from the larger sample.)

S2Pooled

The variance of the distribu- tion of means for the first population (based on an estimated population vari- ance) is the pooled estimate of the population variance divided by the number of participants in the sample from the first population.

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The t Test for Independent Means 275

(8–3)

Consider again the study with 11 in the experimental group and 31 in the control group. We figured the pooled estimate of the population variance to be 75. For the experimental group, the variance of the distribution of means would be , which is 6.82. For the control group, the variance would be , which is 2.42. Using the formulas,

The Variance and Standard Deviation of the Distribution of Differences Between Means The variance of the distribution of differences between means is the variance of Population 1’s distribution of means plus the variance of Population 2’s distribution of means. (This is because, in a difference between two numbers, the variation in each contributes to the overall variation in their difference. It is like sub- tracting a moving number from a moving target.) Stated as a formula,

(8–4)

The standard deviation of the distribution of differences between means ( ) is the square root of the variance:

(8–5)

In the example we have been considering, the variance of the distribution of means for the experimental group was 6.82, and the variance of the distribution of means for the control group was 2.42; the variance of the distribution of the differ- ence between means is thus 6.82 plus 2.42, which is 9.24. This makes the standard deviation of this distribution the square root of 9.24, which is 3.04. In terms of the formulas,

Steps to Find the Standard Deviation of the Distribution of Differences Between Means

●A Figure the estimated population variances based on each sample. That is, figure one estimate for each population using the formula ).S2 = SS>(N - 1

SDifference = 2S2Difference = 29.24 = 3.04. S2Difference = S2M1 + S

2 M2 = 6.82 + 2.42 = 9.24

SDifference = 2S2Difference

SDifference

S2Difference = S2M1 + S 2 M2

(S2Difference)

S2M2 = S2Pooled

N2 =

75

31 = 2.42.

S2M1 = S2Pooled

N1 =

75

11 = 6.82

75>31 75>11

S2M2 = S2Pooled

N2

T I P F O R S U C C E S S Remember that when figuring esti- mated variances, you divide by the degrees of freedom. But when fig- uring the variance of a distribution of means, which does not involve any additional estimation, you di- vide by the actual number in the sample.

The variance of the distribution of differences between means is the variance of the distribution of means for the first population (based on an estimated population variance) plus the variance of the distribution of means for the second population (based on an estimated population variance).

variance of a distribution of differ- ences between means ( ) one of the numbers figured as part of a t test for independent means; it equals the sum of the variances of the distribu- tions of means associated with each of the two samples.

S2Difference

standard deviation of the distribu- tion of differences between means ( ) in a t test for independent means, square root of the variance of the distribution of differences between means.

SDifference

The standard deviation of the distribution of differences between means is the square root of the variance of the distribution of differences between means.

The variance of the distribu- tion of means for the second population (based on an estimated population vari- ance) is the pooled estimate of the population variance divided by the number of participants in the sample from the second population.

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276 Chapter 8

●B Figure the pooled estimate of the population variance:

( and ; )

●C Figure the variance of each distribution of means: and

●D Figure the variance of the distribution of differences between means:

●E Figure the standard deviation of the distribution of differences between means:

The Shape of the Distribution of Differences Between Means The distribution of differences between means is based on estimated population vari- ances. Thus, the distribution of differences between means (the comparison distri- bution) is a t distribution. The variance of this distribution is figured based on population variance estimates from two samples. Therefore, the degrees of freedom for this t distribution are the sum of the degrees of freedom of the two samples. In terms of a formula,

(8–6)

In the example we have been considering with an experimental group of 11 and a control group of 31, we saw earlier that the total degrees of freedom is 40 (that is,

; ; and ). To find the t score needed for sig- nificance, you look up the cutoff point in the t table in the row with 40 degrees of freedom. Suppose you are conducting a one-tailed test using the .05 significance level. The t table in the Appendix (Table A–2) shows a cutoff of 1.684 for 40 degrees of freedom. That is, for a result to be significant, the difference between the means has to be at least 1.684 standard deviations above the mean difference of 0 on the distribution of differences between means.

The t Score for the Difference Between the Two Actual Means Here is how you figure the t score for Step ❹ of the hypothesis testing: First, figure the difference between your two samples’ means. (That is, subtract one from the other). Then, figure out where this difference is on the distribution of differences be- tween means. You do this by dividing your difference by the standard deviation of this distribution. In terms of a formula,

(8–7)

For our example, suppose the mean of the first sample is 198 and the mean of the second sample is 190. The difference between these two means is 8 (that is,

). Earlier we figured the standard deviation of the distribution of dif- ferences between means in this example to be 3.04. That would make a t score of 2.63 (that is, ). In other words, in this example the difference between the two means is 2.63 standard deviations above the mean of the distribution of dif- ferences between means. In terms of the formula,

t = M1 - M2 SDifference

= 198 - 190

3.04 =

8

3.04 = 2.63

8>3.04 = 2.63 198 - 190 = 8

t = M1 - M2 SDifference

10 + 30 = 4031 - 1 = 3011 - 1 = 10

dfTotal = df1 + df2

SDifference = 2S2Difference. S2Difference = S2M1 + S

2 M2.

S2M2 = S 2 Pooled > N2.

S2M1 = S 2 Pooled > N1

dfTotal = df1 + df2df2 = N2 - 1df1 = N1 - 1

S2Pooled = df1

dfTotal (S21) +

df2 dfTotal

(S22)

The total degrees of free- dom for a t test for indepen- dent means is the degrees of freedom in the first sample plus the degrees of freedom in the second sample.

The t score is the difference between the two sample means divided by the standard deviation of the distribution of differences between means.

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The t Test for Independent Means 277

(e) is the pooled estimate of the population variance; and are the degrees of freedom in the samples from the first and second populations, respectively; is the total degrees of freedom (the sum of and ); and are the population variance estimates based on the samples from the first and second populations, respectively; is the variance of the distribu- tion of means for the first population based on an estimated variance of the population of individuals; is the number of participants in the sample from the first population; is the variance of the distribution of differences between means based on estimated variances of the populations of individu- als; tis the tscore for a ttest for independent means (the number of standard deviations from the mean on the distribution of differences between means);

and are the means of the samples from the first and second popula- tions, respectively; and is the standard deviation of the distribution of differences between means based on estimated variances of the popula- tions of individuals.

4.(a) You assume that both populations have the same variance; thus the estimates from the two samples should be estimates of the same number. (b) We weight (give more influence to) an estimate from a larger sample because, being based on more information, it is likely to be more accurate. (c) The actual weighting is done by multiplying each sample’s estimate by the degrees of freedom for that sample divided by the total degrees of freedom; you then sum these two products.

5.(a) Standard deviation of the distribution of differences between means:

(b) Mean:0;(c)Shape:tdistributionwith;(d)ShouldlooklikeFigure8–1 withnumberswrittenin(seeFigure8–2foranexample).

df=50

SDifference=212.78=3.57. S2

Difference=7.62+5.16=12.78; S2

M1=160 > 21=7.62; S2M2=160 > 31=5.16; S2

Pooled=(20 >50)(100)+(30>50)(200)=40+120=160.

SDifference

M2 M1

S2 Difference

N1

S2 M1

S2 2

S2 1 df2 df1 dfTotal

df2 df1 S2 Pooled

How are you doing?

1. (a) When would you carry out a t test for independent means? (b) How is this different from the situation in which you would carry out a t test for dependent means?

2. (a) What is the comparison distribution in a t test for independent means? (b) Explain the logic of going from scores in two samples to an estimate of the variance of this comparison distribution. (c) Illustrate your answer with sketches of the distributions involved. (d) Why is the mean of this distribution 0?

3. Write the formula for each of the following: (a) pooled estimate of the population variance, (b) variance of the distribution of means for the first population, (c) variance of the distribution of differences between means, and (d) t score in a t test for independent means. (e) Define all the symbols used in these formulas.

4. Explain (a) why a t test for independent means uses a single pooled estimate of the population variance, and (b) why and (c) how this estimate is “weighted.”

5. For a particular study comparing means of two samples, the first sample has 21 participants and an estimated population variance of 100; the second sample has 31 participants and an estimated population variance of 200. (a) What is the standard deviation of the distribution of differences between means? (b) What is its mean? (c) What will be its shape? (d) Illustrate your an- swer with sketches of the distributions involved.

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278 Chapter 8

Answers

1.(a) You carry out a ttest for independent means when you have done a study in which you have scores from two samples of different individuals and you do not know the population variance. (b) In a ttest for dependent means you have two scores from each of several individuals.

2.(a) The comparison distribution in a ttest for independent means is a distrib- ution of differences between means. (b) You estimate the population variance from each sample’s scores. Since you assume the populations have the same variance, you then pool the two estimates (giving proportionately more weight in this averaging to the sample that has more degrees of freedom in its estimate). Using this pooled estimate, you figure the variance of the distribution of means for each sample’s popula- tion by dividing this pooled estimate by the sample’s number of participants. Finally, since your interest is in a difference between means, you create a comparison distribution of differences between means. This comparison dis- tribution will have a variance equal to the sum of the variances of the two dis- tributions of means. (Because the distribution of differences between means is made up of pairs of means, one taken from each distribution of means, the variance of both of these distributions of means contributes to the variance of the comparison distribution.) (c) Your sketch should look like Figure 8–1. (d) The mean of this distribution will be zero because, if the null hypothesis is true, the two populations have the same mean. So differences between means would on the average come out to zero.

3.(a) Pooled estimate of the population variance:

.

(b)Variance of the distribution of means for the first population:

.

(c)Variance of the distribution of differences between means:

(d)tscore in a ttest for independent means: t= M1 - M2 SDifference

.

S2 Difference=S2M

1+S2M 2.

S2 M1=

S2 Pooled

N1

S2 Pooled=

df1 dfTotal

(S2 1)+

df2 dfTotal

(S2 2)

Hypothesis Testing with a t Test for Independent Means Considering the five steps of hypothesis testing, there are three new wrinkles for a t test for independent means: (1) the comparison distribution is now a distribution of differences between means (this affects Step ❷); (2) the degrees of freedom for find- ing the cutoff on the t table is based on two samples (this affects Step ❸); and (3) your sample’s score on the comparison distribution is based on the difference between your two means (this affects Step ❹).

Example of a t Test for Independent Means Let’s return to the expressive writing study that we introduced at the start of the chapter. Twenty students were recruited to take part in the study. The 10 students

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randomly assigned to the expressive writing group wrote about their thoughts and feelings associated with their most traumatic life events. The 10 students randomly assigned to the control group wrote about their plans for the day. One month later, all of the students rated their overall level of physical health on a scale from

to . The scores and figuring for the t test are shown in Table 8–1. Figure 8–2 shows

the distributions involved. Let’s go through the five steps of hypothesis testing.

❶ Restate the question as a research hypothesis and a null hypothesis about the populations. There are two populations:

Population 1: Students who engage in expressive writing. Population 2: Students who write about a neutral topic (their plans for the day).

The researchers were interested in identifying a positive or a negative health effect of expressive writing. Thus, the research hypothesis was that Population 1 students would rate their health differently from Population 2 students: . The null hypothesis was that Population 1 students would rate their health the same as Population 2 students: .

❷ Determine the characteristics of the comparison distribution. The compari- son distribution is a distribution of differences between means. (a) Its mean is

�1 = �2

�1 Z �2

100 = perfect health0 = very poor health

The t Test for Independent Means 279

T I P F O R S U C C E S S Note that in previous chapters, Population 2 represented the pop- ulation situation if the null hypothe- sis is true.

Table 8–1 t Test for Independent Means for a Fictional Study of the Effect of Expressive Writing on Physical Health

Expressive Writing Group Control Writing Group

Score

Deviation from Mean (Score – M )

Squared Deviation

from Mean Score

Deviation from Mean (Score – M )

Squared Deviation

from Mean

77 4 87 19 361

88 9 81 77 9 81

77 4 71 3 9

90 11 121 70 2 4

68 121 63 25

74 25 50 324

62 289 58 100

93 14 196 63 25

82 3 9 76 8 64

79 0 0 65 9

: 790 850 680 1002

Needed t with df = 18, 5% level, two-tailed

Decision: Reject the null hypothesis.

t = (M1 - M2)>SDifference = (79.00 - 68.00)>4.54 = 2.42 = ;2.101

S2Difference = 2S2Difference = 220.58 = 4.54 S2Difference = S2M1 + S

2 M2 = 10.29 + 10.29 = 20.58

S2M2 = S 2 Pooled>N2 = 102.89>10 = 10.29

S2M1 = S 2 Pooled>N1 = 102.89>10 = 10.29

S2Pooled = df1

dfTotal (S21) +

df2 dfTotal

(S22) = 9

18 (94.44) +

9 18

(111.33) = 47.22 + 55.67 = 102.89

dfTotal = df1 + df2 = 9 + 9 = 18 N1 = 10; df1 = N1 - 1 = 9; N2 = 10; df2 = N2 - 1 = 9 M1 = 79.00; S21 = 850>9 = 94.44; M2 = 68.00 S21 = 1002>9 = 111.33 g

-3

-5 -10-17 -18-5 -5-11

-2

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280 Chapter 8

0 (as it almost always is in a t test for independent means, because we are inter- ested in whether there is more than 0 difference between the two populations). (b) Regarding its standard deviation, ●A Figure the estimated population variances based on each sample. As shown

in Table 8–1, comes out to 94.44 and ●B Figure the pooled estimate of the population variance: As shown in Table

8–1, the figuring for gives a result of 102.89. ●C Figure the variance of each distribution of means: Dividing by the

N in each sample, as shown in Table 8–1, gives and ●D Figure the variance of the distribution of differences between means:

Adding up the variances of the two distributions of means, as shown in Table 8–1, comes out to

●E Figure the standard deviation of the distribution of differences between means:

(c) The shape of this comparison distribution will be a t distribution with a total of 18 degrees of freedom.

SDifference = 2S2Difference = 220.58 = 4.54. S2Difference = 20.58.

S2M2 = 10.29.S 2 M1 = 10.29

S2Pooled

S2Pooled

S22 = 111.33.S2 1

Students who engage in expressive writing Students who write about a neutral topic

Distributions of meansSM = 3.21 (SM = 10.29)

SM = 3.21 (SM = 10.29)

SDifference = 4.54

S2 = 94.44 S2 = 111.33

79.00 68.00

Samples

Distribution of differences between means (comparison distribution)

t Score = 2.420

2 2

Populations (SPooled = 102.89)

2

Figure 8–2 Distributions for a t test for independent means for the expressive writing example.

T I P F O R S U C C E S S Notice that, in this example, the value for is the same as the

value for This is because there was the same number of stu- dents in the two groups (that is, was the same as ). When the number of individuals in the two groups is not the same, the values for and will be different.S2M2S

2 M1

N2 N1

S2M2.

S2M1

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The t Test for Independent Means 281

❸ Determine the cutoff sample score on the comparison distribution at which the null hypothesis should be rejected. This requires a two-tailed test because the researchers were interested in an effect in either direction. As shown in Table A–2 (in the Appendix), the cutoff t scores at the .05 level are 2.101 and .

❹ Determine your sample’s score on the comparison distribution. The t score is the difference between the two sample means ( , which is 11.00), divided by the standard deviation of the distribution of differences between means (which is 4.54). This comes out to 2.42.

➎ Decide whether to reject the null hypothesis. The t score of 2.42 for the differ- ence between the two actual means is larger than the cutoff t score of 2.101. You can reject the null hypothesis. The research hypothesis is supported: students who engage in expressive writing report a higher level of health than students who write about a neutral topic.

Although the actual numbers in this study were fictional, the results are consistent with those from many actual studies that have shown beneficial effects of expressive writing on self-reported health outcomes, as well as additional outcomes such as psychological well-being (e.g., Pennebaker & Beall, 1986; Warner et al., 2006; see also Frattaroli, 2006).

Summary of Steps for a t Test for Independent Means Table 8–2 summarizes the steps for a t test for independent means.1

79.00 - 68.00

-2.101

Table 8–2 Steps for a t Test for Independent Means

❶ Restate the question as a research hypothesis and a null hypothesis about the populations.

❷ Determine the characteristics of the comparison distribution.

a. Its mean will be 0.

b. Figure its standard deviation.

●A Figure the estimated population variances based on each sample. For each population,

.

●B Figure the pooled estimate of the population variance:

●C Figure the variance of each distribution of means:

●D Figure the variance of the distribution of differences between means:

●E Figure the standard deviation of the distribution of differences between means:

c. Determine its shape: It will be a t distribution with degrees of freedom.

❸ Determine the cutoff sample score on the comparison distribution at which the null hypothesis should be rejected.

a. Determine the degrees of freedom ( ), desired significance level, and tails in the test (one or two).

b. Look up the appropriate cutoff in a t table. If the exact df is not given, use the df below it.

❹ Determine your sample’s score on the comparison distribution:

❺ Decide whether to reject the null hypothesis: Compare the scores from Steps ❸ and ❹.

t = (M1 - M2)>SDifference

dfTotal

dfTotal

SDifference = 2S2Difference

S2Difference = S2M1 + S 2 M2

S2M1 = S 2 Pooled>N1 and S2M2 = S2Pooled>N2

(df1 = N1 - 1 and df2 = N2 - 1; dfTotal = df1 + df2)

S2Pooled = df1

dfTotal (S21) +

df2 dfTotal

(S22)

S2 = SS>(N - 1)

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282 Chapter 8

A Second Example of a t Test for Independent Means Valenzuela (1997) compared the mothering received by poor children who either were or were not undernourished. One of her measures was systematic ratings of how well the mother assisted her child in a standard puzzle-solving task (observed during home visits). The mothers of the 43 adequately nourished children had a mean quality of assistance of 33.10 and an estimated population variance of 201.64. The mothers of the 42 chronically undernourished children had a mean of 27.00 on this measure, with an estimated population variance of 134.56.

The figuring for the t test comparing the quality of assistance scores for the two conditions is shown in Table 8–3. The distributions involved are shown in Figure 8–3. Next, we go through the five steps of hypothesis testing.

❶ Restate the question as a research hypothesis and a null hypothesis about the populations. There are two populations:

Population 1: Mothers of adequately nourished poor children. Population 2: Mothers of chronically undernourished poor children.

The research hypothesis was that Population 1 mothers would score differently from Population 2 mothers on the quality of assistance to their chil- dren. Valenzuela predicted that Population 1 would score higher than Popula- tion 2. However, following conventional practice in studies like this, she used a nondirectional (two-tailed) significance test. (This had the advantage of al- lowing the possibility of finding significant results in the direction opposite to her prediction.) Thus, the research hypothesis actually tested was that Popula- tion 1 mothers would score differently from Population 2 mothers: . The null hypothesis was that the Population 1 mothers would score the same as Population 2 mothers: .

❷ Determine the characteristics of the comparison distribution. (a) Its mean will be 0. (b) Figure its standard deviation (see Table 8–3 for the figuring for each step below), ●A Figure the estimated population variances based on each sample. These

are already figured for us: and S22 = 134.56.S2 1 = 201.64

�1 = �2

�1 Z �2

Table 8–3 t Test for Independent Means in Study of Quality of Assistance of Mothers of Adequately Nourished Versus Chronically Undernourished Poor Chilean Children

Adequately Nourished Children:

Chronically Undernourished Children:

Needed t with (using in the table), 5% level,

Decision: Reject the null hypothesis; the research hypothesis is supported.

Source: Data from Valenzuela (1997).

t = (M1 - M2)>SDifference = (33.10 - 27.00)>2.82 = 6.10>2.82 = 2.16 two-tailed = ;1.990df = 80df = 83

SDifference = 2S2Difference = 27.94 = 2.82 S2Difference = S2M1 + S

2 M2 = 3.92 + 4.02 = 7.94

S2M2 = S 2 Pooled>N2 = 168.77>42 = 4.02

S2M1 = S 2 Pooled>N1 = 168.77>43 = 3.92

= .51(201.64) + .49(134.56) = 102.84 + 65.93 = 168.77

S2Pooled = df1

dfTotal (S21) +

df2 dfTotal

(S22) = 42 83

(201.64) + 41 83

(134.56)

dfTotal = df1 + df2 = 42 + 41 = 83 N2 = 42; df2 = N2 - 1 = 41; M2 = 27.00; S 22 = 134.56

N1 = 43; df1 = N1 - 1 = 42; M1 = 33.10; S 21 = 201.64

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The t Test for Independent Means 283

Populations (SPooled = 168.77)

Adequately nourished poor children Chronically undernourished poor children

Distributions of meansSM = 1.98 (SM = 3.92)

SM = 2.00 (SM = 4.02)

SDifference = 2.82

S2 = 201.64 S2 = 134.56

33.10 27.00

Samples

Distribution of differences between means (comparison distribution)

t Score = 2.160

2 2

2

Figure 8–3 Distributions for a t test for independent means for the mothers of adequately nourished versus chronically undernourished poor children.

Source: Data from Valenzuela, 1997.

●B Figure the pooled estimate of the population variance: The figuring for gives a result of 168.77.

●C BFigure the variance of each distribution of means: Dividing by the N in each sample gives and

●D Figure the variance of the distribution of differences between means: Adding up the variances of the two distributions of means comes out to

●E Figure the standard deviation of the distribution of differences between means:

(c) The shape of this comparison distribution will be a t distribution with of 83.

❸ Determine the cutoff sample score on the comparison distribution at which the null hypothesis should be rejected. The cutoff you need is for a two-tailed test (because the research hypothesis is nondirectional) at the usual .05 level, with 83 degrees of freedom. The t table in the Appendix (Table A–2) does not have a listing for 83 degrees of freedom. Thus, you use the next lowest df avail- able, which is 80. This gives cutoff t scores of and .-1.990+1.990

dfTotal