For example, the manager of a baseball team needs to assign nine players to particular spots in the batting order but only has to make eight decisions (N − 1). Why? Because only one option remains after making the first eight decisions. So before the manager makes any decisions, there are N − 1, or 9 − 1 = 8, degrees of freedom. After the second decision, there are N − 1, or 8 − 1 = 7, degrees of freedom, and so on.

As in the baseball example, there is always one score that cannot vary once all of the others have been determined. For example, if we know that the mean of four scores is 6 and we know that three of the scores are 2, 4, and 8, then the last score must be 10. So the degrees of freedom is the number of scores in the sample minus 1. Degrees of freedom is written in symbolic notation as df, which is always italicized. The formula for degrees of freedom for a single-sample t test, therefore, is:

df = N − 1

9-4: The formula for degrees of freedom for a single-sample t test is: df = N − 1. To calculate degrees of freedom, we subtract 1 from the sample size.

9-3: As sample size increases, the t distributions more and more closely approximate the z distribution. You can think of the z statistic as a single-blade Swiss Army knife and the t statistic as a multi-blade Swiss Army knife that includes the single blade that is the z statistic.

The study: A researcher knows the mean number of calories lab rats will consume in half an hour if unlimited food is available. She wonders whether a new food will lead rats to consume a different number of calories—either more or fewer. She studies 38 rats and uses a p level of 0.05.

The cutoff(s): This is a two-tailed test because the research hypothesis allows for change in either direction. There are 38 rats, so the degrees of freedom is:

df = N − 1 = 38 − 1 = 37

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We want to look in the t table under two-tailed tests, in the column for 0.05 and in the row for a df of 37; however, there is no df of 37. In this case, we err on the side of being more conservative and choose the more extreme (i.e., larger) of the two possible critical t values, which is always the smaller df. Here, we look next to 35, where we see a value of 2.030. Because this is a two-tailed test, we will have critical values of −2.030 and 2.030.

Chapter 4 presented data that included the mean number of sessions attended by clients at a university counseling center. We noted that one study reported a mean of 4.6 sessions (Hatchett, 2003). Let’s imagine that the counseling center hoped to increase participation rates by having students sign a contract to attend at least 10 sessions. Five students sign the contract and attend 6, 6, 12, 7, and 8 sessions, respectively. The researchers are interested only in their university, so treat the mean of 4.6 sessions as a population mean.

STEP 1: Identify the populations, distribution, and assumptions.

Population 1: All clients at this counseling center who sign a contract to attend at least 10 sessions. Population 2: All clients at this counseling center who do not sign a contract to attend at least 10 sessions.

The comparison distribution will be a distribution of means. The hypothesis test will be a single-sample t test because we have only one sample and we know the population mean but not the population standard deviation.

This study meets one of the three assumptions and may meet the other two: (1) The dependent variable is scale. (2) We do not know whether the data were randomly selected, however, so we must be cautious with respect to generalizing to other clients at this university who might sign the contract. (3) We do not know whether the population is normally distributed, and there are not at least 30 participants. However, the data from the sample do not suggest a skewed distribution.

STEP 2: State the null and research hypotheses.

Null hypothesis: Clients at this university who sign a contract to attend at least 10 sessions attend the same number of sessions, on average, as clients who do not sign such a contract—H₀: µ₁ = µ₂.

Research hypothesis: Clients at this university who sign a contract to attend at least 10 sessions attend a different number of sessions, on average, than do clients who do not sign such a contract—H₁: µ₁ ≠ µ₂.

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STEP 3: Determine the characteristics of the comparison distribution.

µ_M = 4.6; s_M = 1.114

Calculations:

µ_M = µ = 4.6

The numerator of the standard deviation formula is the sum of squares:

STEP 4: Determine the critical values, or cutoffs.

df = N − 1 = 5 − 1 = 4

For a two-tailed test with a p level of 0.05 and df of 4, the critical values are −2.776 and 2.776 (as seen in the curve in Figure 9-2).

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STEP 5: Calculate the test statistic.

STEP 6: Make a decision.

Reject the null hypothesis. It appears that counseling center clients who sign a contract to attend at least 10 sessions do attend more sessions, on average, than do clients who do not sign such a contract (Figure 9-3).

After completing the hypothesis test, we want to present the primary statistical information in a report. There is a standard American Psychological Association (APA) format for the presentation of statistics across the behavioral sciences so that the results are easily understood by the reader.

In the counseling center example, the statistics would read:

t(4) = 2.87, p < 0.05

The statistics typically follow a statement about the finding. For example, “It appears that counseling center clients who sign a contract to attend at least 10 sessions do attend more sessions, on average, than do clients who do not sign such a contract, t(4) = 2.87, p < 0.05.” The report would also include the sample mean and standard deviation (not standard error) to two decimal points. Here, the descriptive statistics would read: (M = 7.80, SD = 2.49). By convention, we use SD instead of s to symbolize the standard deviation.

We can calculate a confidence interval with the single-sample t test data. The population mean was 4.6. We used the sample to estimate the population standard deviation to be 2.490 and the population standard error to be 1.114. The five students in the sample attended a mean of 7.8 sessions.

When we conducted hypothesis testing, we centered the curve around the mean according to the null hypothesis—the population mean of 4.6. Now we can use the same information to calculate the 95% confidence interval around the sample mean of 7.8.

STEP 1: Draw a picture of a t distribution that includes the confidence interval.

We draw a normal curve (Figure 9-4) that has the sample mean, 7.8, at its center (instead of the population mean, 4.6).

STEP 2: Indicate the bounds of the confidence interval on the drawing.

We draw a vertical line from the mean to the top of the curve. For a 95% confidence interval, we also draw two much smaller vertical lines that indicate the middle 95% of the t distribution (2.5% in each tail, for a total of 5%). We then write the appropriate percentages under the segments of the curve.

STEP 3: Look up the t statistics that fall at each line marking the middle 95%.

For a two-tailed test with a p level of 0.05 and a df of 4, the critical values are −2.776 and 2.776. We can now add these t statistics to the curve, as seen in Figure 9-5.

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STEP 4: Convert the t statistics back into raw means.

As we did with the z test, we can use formulas for this conversion, but first we identify the appropriate mean and standard deviation. There are two important points to remember. First, we center the interval around the sample mean, so we use the sample mean of 7.8 in the calculations. Second, because we have a sample mean (rather than an individual score), we use a distribution of means. So we use the standard error of 1.114 as the measure of spread.

Using this mean and standard error, we calculate the raw mean at each end of the confidence interval, and add them to the curve, as in Figure 9-6. The formulas are exactly the same as for the z test except that z is replaced by t, and σ_M is replaced by s_M.

M_lower = −t(s_M) + M_sample = −2.776(1.114) +7.8 = 4.71

M_upper = t(s_M) + M_sample = 2.776(1.114) +7.8 = 10.89

The 95% confidence interval, reported in brackets as is typical, is [4.71, 10.89].

STEP 5: Verify that the confidence interval makes sense.

The sample mean should fall exactly in the middle of the two ends of the interval.

4.71 − 7.8 = −3.09; and 10.89 − 7.8 = 3.09

We have a match. The confidence interval ranges from 3.09 below the sample mean to 3.09 above the sample mean. If we were to sample five students from the same population over and over, the 95% confidence interval would include the population mean 95% of the time. Note that the population mean, 4.6, does not fall within this interval. This means it is not plausible that this sample of students who signed contracts came from the population according to the null hypothesis—students seeking treatment at the counseling center who did not sign a contract. We conclude that the sample comes from a different population, one in which students attend more mean sessions than does the general population. As with the z test, the conclusions from the hypothesis test and the confidence interval are the same, but the confidence interval gives us more information—an interval estimate, not just a point estimate.

Let’s calculate the effect size for the counseling center study. Similar to what we did with the z test, we simply use the formula for the t statistic, substituting s for s_M (and µ for µ_M, even though these means are always the same). This means we use 2.490 instead of 1.114 in the denominator. Cohen’s d is based on the spread of the distribution of individual scores, rather than the distribution of means.

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The effect size, d = 1.29, tells us that the sample mean and the population mean are 1.29 standard deviations apart. According to the conventions we learned in Chapter 8 (that 0.2 is a small effect, 0.5 is a medium effect, and 0.8 is a large effect), this is a large effect. We can add the effect size when we report the statistics as follows: t(4) = 2.87, p < 0.05, d = 1.29.

Solutions to these Check Your Learning questions can be found in Appendix D.