Outline

• The t-distribution
• A 1-sample t-test
• Degrees of freedom
• Using SPSS to do a 1-sample t-test

The t-distribution

We're now going to move back up to the top portion of our diagram to at cases where we are making comparisons and looking for differences. We've already looked at one test that will allow us to do that: one-sample z-test. Today, we'll look at another test that allows us to compare a treatment population (represented with a single sample) with a known population mean. Let's a quick look at our diagram:

Which test?

We're going to consider in this lab the case where we know the original population mean, but we don't know the population standard deviation (σ) so we have to estimate it from our sample. We do that in the 1-sample t-test. Find the string of decisions that lead to a one-sample t-test.

Here are the relevant formulas.

If we know σ:	If we don't know σ:
standard error of =	estimated standard error of =
test statistic: z-score	test statistic: t-score

Rule: When you know the value of σ, use a z-score. If σ is unknown, use s to to estimate σ and use the t-statistic.


The t statistic is used to test hypotheses about μ when the value for σ² is not known. The formula for the t statistic is similar in structure to that for the z-score, except that the t statistic uses estimated standard error.

Consider the following scenarios. For each determine which formula (z or t) is the appropriate one to use to answer the question asked. (You don't need to do any computations.)

(1) Pat, a personal trainer would like to examine the effects of humidity on exercise behavior. It is known that the average person in the United States exercises an average of μ = 21 minutes each day. The personal trainer selects a random sample of n = 100 people and places them in a controlled atmosphere environment where the relative humidity is maintained at 90%. The daily amount of time spent exercising for the sample averages = 18.7 minutes with s = 5.0. So Pat wants to know if humidity affects exercise behavior.

(2) In an attempt to regulate the profession, the US Department of Fitness has developed a fitness test for personal trainers. The test requires that the trainers must perform a series of exercises within a certain period of time. Normative data, collected in a nationwide test, reveal a normal distribution with an average completion time of of μ = 92 minutes and of σ = 11. Pat, and four other Hollywood personal trainers (so n = 5) take the test. For these trainers, the average time to complete the task is averages = 115 minutes. Pat is worried that the Holywood personal trainers (in this sample) differ significantly from the norm.

Using the t table

Because we are using the sample standard deviation to estimate the population standard deviation (σ), we need to take the degrees of freedom into account. Degrees of freedom describe the number of scores in a sample that are free to vary. Because the sample mean places a restriction on the value of one score in the sample, there are n - 1 degrees of freedom for the sample.

Notice that we're talking about a new distribution here (or family of distributions, the t-distributions), found in the t distribution table. Part of it is shown here, the same as the one appended to your packet.

One tail probability p .. 0.25 0.10 0.05 0.025 0.01 .005 Two tail probability p df 0.50 0.20 0.10 0.05 0.02 0.01 1 1.00 3.078 6.314 12.706 31.821 63.657 2 0.816 1.886 2.920 4.303 6.965 9.925 3 0.765 1.638 2.353 3.182 4.541 5.841 4 0.741 1.533 2.132 2.776 3.747 4.604 5 0.727 1.476 2.015 2.571 3.365 4.032 6 0.718 1.440 1.943 2.447 3.143 3.707 : : : : : : : : : : : : : : z* 0.674 1.282 1.645 1.96 2.326 2.576 CI% 50% 80% 90% 95% 98% 99%

t-distribution table

Think back to the 1-sample z-test. One of the ways that we would make our decision about whether or not to reject the H₀ was to figure out what z-score corresponded to the critical region (e.g., 1.65 = critcal z for 1-tailed test with α = to 0.05), then look at our the z that we computed and see if it was greater than (or equal to) that critical z. If if was, then we rejected the H₀, if it wasn't then we failed to reject H₀.

But keep in mind that for z-scores we use the unit normal table which only describes one distribution. So, for all 1-tailed tests with α = 0.05, the critical value of z will be 1.65. The logic is the same here with the t-table. But now, the critical values are going to change as a function of which t-distribution that we are looking at, which is in turn dependent on df.

The t-table clearly shows the relationship between a 1-tailed and 2-tailed p-value. Any given t-value has half the p-value 1-tailed compared to 2-tailed. That is why the p =.025, 1-tailed, and p = .05, 2-tailed, are in the same column. If t (df=6) = 2.0 and a = .05, it is in the critical region 1-tailed (critical t = 1.943) but not 2-tailed (critical t = 2.447).

1-sample t-test

We'll go through our 5-step procedure for this test.

step 1 : Hypotheses: H₀ and H_A (1-tailed or 2-tailed?)

step 2 : Criterion for decision: alpha= ?

step 3 : Sample statistics

step 4 : Test statistic (compute your observed t-test)

step 5 : Make a decision about the null hypothesis (Compare observed t to critical value for t)

NOTE: We will use the following notations:

tobs =

tcrit = critical t from the table

Suppose that your physics professor, Dr. M. C. Squared, gives a 20 point true-false quiz to 9 students and wants to know if they did worse than guessing. Their scores were: 6, 7, 7, 8, 8, 8, 9, 9, 10. We'll assume a significance level of α = 0.05.

step 1:

one-tailed test (worse than guessing)
H₀: μ > 10
H₁: μ < 10;
α = 0.05
(note: the null is what they'd get if they were guessing which would be 10 out of 20).

step 2: compute sample stats

  = 72/9 = 8.0

SS  = 12.0

s = sqroot(SS/n-1) = sqroot(12/8) = 1.225

step 3: compute t_obs

est standard error = s/sqroot(n) = 1.225/sqroot(9) = 0.41

t_obs = = (10 - 8) / 0.41 = -4.88

and df: n = 9, so df = 9 - 1 = 8

step 4: find the critical t from the table to compare with t_obs and make decision:

df = 8, one-tailed test, α = 0.05, so t_crit = -1.86
(keep in mind that this is worse than, so the critical t is negative)

t_obs = -4.88 < t_crit = -1.86

reject the H₀ - so it looks as if the students would have been better off guessing.

Suppose that your psychology professor, Dr. I. D. Ego, gives a 20 point true-false quiz to 9 students and wants to know if they were different from groups in the past who have tended to have an average of 9.0. Their scores from the current group were: 6, 7, 7, 8, 8, 8, 9, 9, 10. Did the current group perform differently from those in the past. We'll assume a significance level of α = 0.05.

step 1:
two-tailed test (are they different)

H₀: μ = 9.0 and H₁: μ not equal to 9.0; α = 0.05

step 2: compute sample stats

=  8.0

SS  = 12.0

s = sqroot(SS/n-1) = sqroot(12/8) = 1.225

step 3: compute t_obs

est standard error = s/sqroot(n) = 1.225/sqroot(9) = 0.41

t_obs = = (9 - 8) / 0.41 = -2.44

what is our df? n = 9, so df = 9 - 1 = 8

step 4: find the critical t from the table to compare with t_obs and make decision:

df = 8, two-tailed test, α = 0.05, so tcrit = ±2.306

t_obs = -2.44 < t_crit = ±2.306

reject the H₀ - so it looks as if the current students are different from past students (they are doing worse).

(3) Ok, now let's try the examples from above. First, decide whether to use the one-sample z-test or t-test. Then do the hypothesis for that test, showing all steps.

(a) Pat, a personal trainer would like to examine the effects of humidity on exercise behavior. It is known that the average person in the United States exercises an average of μ = 21 minutes each day. The personal trainer selects a random sample of n = 100 people and places them in a controlled atmosphere environment where the relative humidity is maintained at 90%. The daily amount of time spent exercising for the sample averages = 18.7 minutes with a s = 5.0. So Pat want to know if humidity affects exercise behavior at the alpha = 0.05 level.

(b) In an attempt to regulate the profession, the US Department of Fitness has developed a fitness test for personal trainers. The test requires that the trainers must perform a series of exercises within a certain period of time. Normative data, collected in a nationwide test, reveal a normal distribution with an average completion time of of μ = 92 minutes and of σ = 11. Pat, and four other Hollywood personal trainers (so n = 5) take the test. For these trainers, the average time to complete the task is averages = 115 minutes. Pat is worried that the Hollywood personal trainers (in this sample) differ significantly from the norm.

Using SPSS to compute a one-sample t-test

Excel does not have a formula for this test. (It is rarely used except for teaching purposes. If, as in standard tests, the population mean is known, so is the population standard deviation.) We'll now go through the procedure in SPSS to compute one-sample t-tests.

Go to the Analyze menu and select the submenu Compare Means. In this submenu you'll see several tests. The one that we're interested in today is One-sample t-test.
After selecting One-sample t-test, you'll get a window that looks like this. Here you should select the Test Variable from your sample that you want to analyze. Then you enter the Test Value, which is the mean for H0 (10 for the Dr.MC Squared example; population mean for the other cases). This is very important: You must always enter the population mean on this screen in a one-sample t-test!
Here is what the output will look like.
The output includes the sample mean, the sample standard deviation, the standard error, the tobs (in the t column), the degrees of freedom, the mean difference (the sample mean - the test value), and a p-value for a 2-tailed test of significance. All of these results should match what we got above going through the problem by hand. check to see that they do. Note that SPSS does not tell you to reject or fail to reject the H0, nor does it give you the tcrit. To make your decision about the H0, you must compare the 2-tailed p-value with your α-level. If you are conducting a 1-tailed test, the p-value is half of what is reported. So if p (2-tailed) = .001, then p (1-tailed) = .0005. If the p-value is equal to or smaller than the your α-level (it's in the critical region), then you should reject the H0; otherwise you should fail to reject. So if a = .05 and p = .0005, p < a; therefore, reject the null hypothesis.

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(4) Use SPSS for the rest of the questions. Enter the data into SPSS and then conduct the one-sample t-test in order the answer the questions that follow. An example of how to address this question is provided in the instructions above.

Suppose that your psychology professor, Dr. I. D. Ego, gives a 20 point true-false quiz to 9 students and wants to know if they were different from groups in the past who have tended to have an average of 9.0. Their scores from the current group were: 6, 7, 7, 8, 8, 8, 9, 9, 10.

(a) Did the current group perform differently from those in the past? We'll assume a significance level of α = 0.05.

(5) Now try another problem with SPSS and write out each step of hypothesis testing, using the values on the SPSS output for t_obs, df, and the significance of t_obs.

The personnel department for a major corporation in the Northeast reported that the average number of absences during the months of January and February last year was μ = 7.4. In an attempt to reduce absences, the company offered free flu shots to all employees this year. For a sample of n = 10 people who took the flu shots, the number of absences this year were 6, 8, 10, 3, 4, 6, 5, 4, 5, 6. Do these data indicate a significant reduction in the number of absences? Use α = .05.