Applied Univariate, Bivariate, and Multivariate Statistics. Daniel J. Denis
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Recall that the form of the one‐sample z test for the mean is given by
where the numerator
In most research contexts, from simple to complex, we usually do not have direct knowledge of σ2. When we do not have knowledge of it, we use the next best thing, an estimate of it. We can obtain an unbiased estimate of σ2 by computing s2 on our sample. When we do so, however, and use s2 in place of σ2, we can no longer pretend to “know” the standard error of the mean. Rather, we must concede that all we are able to do is estimate it. Our estimate of the standard error of the mean is thus given by:
When we use s2 (where
was found to be distributed as a t statistic on n − 1 degrees of freedom. Again, the t distribution is most useful for when sample sizes are rather small. For larger samples, as mentioned, the t distribution converges to that of the z distribution. If you are using rather large samples, say approximately 100 or more, whether you evaluate your null hypothesis using a z or t distribution will not matter much, because the critical values for z and t for such degrees of freedom (99 for the one‐sample case) will be relatively alike, that practically at least, the two test statistics can be considered more or less equal. For even larger samples, the convergence is that much more fine‐tuned.
The concept of convergence between z and t can be easily illustrated by inspecting the variance of the t distribution. Unlike the z distribution where the variance is set at 1.0 as a constant, the variance of the t distribution is defined as:
where v are the degrees of freedom. For small degrees of freedom, such as v = 5, the variance of the t distribution is equal to:
Note what happens as v increases, the ratio
which is already quite close to the variance of a standardized normal variable z (i.e., 1.0). Hence, we can say more formally
That is, as v increases without bound, the variance of the t distribution equals that of the z distribution, which is equal to 1.0.
We demonstrate the use of the one‐sample t‐test using SPSS. Consider the following small, hypothetical data on IQ scores on five individuals:
IQ 105 98 110 105 95
Suppose that the hypothesized mean IQ in the population is equal to 100. The question we want to ask is—Is it reasonable to assume that our sampled data could have arisen from a population with mean IQ equal to 100? We assume we have no knowledge of the population standard deviation, and hence must estimate it from our sample data. To perform the one‐sample t‐test in SPSS, we compute:
T-TEST /TESTVAL=100 /MISSING=ANALYSIS /VARIABLES=IQ /CRITERIA=CI(.95).
The line /TESTVAL = 100
inputs the test value for our hypothesis test, which for our null hypothesis is equal to 100. We have also requested a 95% confidence interval for the mean difference.
One‐Sample Statistics | ||||
N | Mean | SD | SE Mean | |
IQ | 5 | 102.6000 | 6.02495 | 2.69444 |
We confirm from the above that the size of our sample is equal to 5, and the mean IQ for our sample is equal to 102.60 with standard deviation 6.02. The standard error of the mean reported by SPSS of 2.69 is actually not the true standard error of the mean. It is the estimated standard error of the mean, since recall that we did not have knowledge of the population variance (otherwise we would have been performing a z‐test instead of a t‐test).
One‐Sample Test | ||||||
---|---|---|---|---|---|---|
Test Value = 100 | ||||||
95% Confidence Interval of the Difference | ||||||
t | Df | Sig. (2‐tailed) | Mean Difference | Lower | Upper | |
IQ | 0.965 | 4 | 0.389 | 2.60000 | −4.8810 | 10.0810 |
We note from the above output:
Our obtained t‐statistic is equal to 0.965 and is evaluated on four degrees of freedom (i.e., n − 1 = 5 − 1 = 4). We lose a degree of freedom because recall that in estimating the population variance σ2 with s2, we had to compute a sample mean and hence this value is regarded as “fixed” as we carry on with our t‐test. Hence, we lose a single degree of freedom.
The