Your textbook:

• Gravetter, F. J., Wallnau, L. B. (1996). Statistics for the Behavioral Sciences:
  A First Course for Students of Psychology and Education, 4th Edition. New York: West Publishing.
  

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  Chapter 9 - Introduction to the t-statistic

  This chapter will (or should) remind you a lot of the previous chapter. The major difference is that in the previous chapter we were dealing with situations in which we knew the population m and s.However, in most situations we don't know s. We can still do hypothesis testing, but what we need to do is use an estimate of s.

  What do we already know about that we can use as an estimate of s? think back to chapter 4 s, the sample standard deviation

  so instead of using the population standard deviation we will be using the sample standard deviation.

  sample standard deviation = s =

  So from this, our formula for standard error is almost the same, with some notational changes and our computed test statistic is also very similar.

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

Generally:

	z or t = sample mean - population mean
		    (estimated) standard error

Rule: When you know the value of s ², use a z-score. If s ² is unknown, use s² to to estimate s ² and use the t-statistic. The same is true for making estimations of the population mean (e.g., confidence intervals).

The t statistic is used to test hypotheses about m when the value for s² 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. Because we are using the sample standard deviation (s) to estimate the population standard deviation (s), then we need to take into account the fact that it is an estimate. If you think back to Chapter 4, we must 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.

This means that the higher the value of n, the more representative the sample will be of the population, which in turn means that s will be a better estimate of s. It also has implications for the test statistic. The shape of the t-distribution varies as a function of the size of n (really it varies with the degrees of freedom). The bigger the n (the bigger the df), the closer the t- distribution is to a normal distribution.

Notice that we're talking about a new distribution here (or family of distributions, the t-distributions). This also means that we won't be using the unit normal table. Instead we'll have to use a different table, the t distribution table (Appendix table b2, right next to the last page of the unit normal table, pg a-27).

Proportion in one tail

0.250 0.100 0.050 0.025 0.010 0.005

Proportion in two tails

df 0.500 0.200 0.100 0.050 0.020 0.010

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 : : : : : : : : : : : : : :

Reading this table is different than reading the unit normal table. So let's talk about why first, and then how. Why? Because the unit normal table is describing one distribution (the normal distribution). The t-distribution table is actually describing several different t distributions. This is because the there is a different t- distribution for every different degrees of freedom (although when df gets large, the differences become really small). So, each row corresponds to different t-distributions. As a result of this, there also isn't enough space to put all of the probabilities corresponding to each possible t-score. Instead what is listed are the t-scores at commonly used critical regions (that is, at popular alpha levels).

The table also splits one-tailed and two-tailed critcal t-values up for you.

Okay, so how do we use the table?

Think back to last chapter. 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 one-tailed test with a = 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 one-tailed tests with an a = 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.

So, what you need to do to use this table is: step 1: state your H₀ and H₁ & figure out your critera: a = ? step 2: figure out if your test is one-tailed or two-tailed step 3: figure out the df for your test step 4: find the critical t-score from the table step 5: compute your t-score for your sample step 6: compare your t-score with the critical t-score step 7: make your conclusions about the H₀

note: that sometimes you will see the following notations:

t_obs = observed t = t_crit = critical t from the table

Okay, let's look at a few examples

Example:

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 a = 0.05.

step 1:

H₀: m > 10
H₁: m < 10;
a = 0.05
(note: the null is what they'd get if they were guessing which would be 10 out of 20).

step 2: one-tailed test (worse than guessing)

step 3: what is our df? n = 9, so df = 9 - 1 = 8

step 4: find the critical t from the table: df = 8, one-tailed test, a = 0.05, so t_crit = -1.86
(keep in mind that this is worse than, so the critical t is negative)

step 5: compute your t_obs
(notice how we are bringing together most of what we've learned all into this one example (computing means, standard deviations, estimatied standard error, etc.))

= (SX)/n = 72/9 = 8.0

SS = (SX²) - (SX)²/ n = 588 - 722/9 = 12.0

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

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

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

step 6: t_obs = -4.88 < t_crit = -1.86

step 7: reject the H₀ - so it looks as if the students would have been better off guessing.

Example 2

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 a = 0.05.

step 1: H₀: m = 9.0 and H₁: m not equal to 9.0; a = 0.05

step 2: two-tailed test (are they different)

step 3: what is our df? n = 9, so df = 9 - 1 = 8

step 4: find the critical t from the table: df = 8, two-tailed test, a = 0.05, so tcrit = ±2.306

step 5: compute your t_obs

= (SX)/n = 72/9 = 8.0

SS = (SX²) - (SX)²/ n = 588 - 722/9 = 12.0

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

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

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

step 6: t_obs = -2.44 < t_crit = ±2.306

step 7: reject the H₀ - so it looks as if the current students are different from past students (they are doing worse).

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Go to Chapter 8: Introduction to Hypothesis testing
Go to Chapter 10: Hypothesis tests with two independent samples


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If you have any questions, please feel free to contact me at cutting@main.psy.ilstu.edu.