- Check the conditions for a [latex]t[/latex]-distribution, then use a [latex]t[/latex]-distribution to calculate probabilities when appropriate.
[latex]z[/latex]-statistic vs. [latex]t[/latex]-statistic
[latex]z=\dfrac{\bar{x}-[\text{mean of } \bar{x}'s]}{\text{std. deviation of } \bar{x}'s}[/latex] [latex]= \dfrac{\bar{x}-\mu}{\frac{\sigma}{\sqrt{n}}}[/latex]
Suppose we do not know the value of the population standard deviation, [latex]\sigma[/latex]. If we want to calculate the value of a standardized sample mean, we need to know [latex]\sigma[/latex]. So, instead of [latex]\sigma[/latex], let’s substitute in our best estimate for [latex]\sigma[/latex]: the sample standard deviation, [latex]s[/latex].
Recall that an estimate of the standard deviation of a statistic is called the standard error of that statistic. Since we estimate the standard deviation of the sample mean [latex]\dfrac{\sigma}{\sqrt{n}}[/latex] by substituting in [latex]s[/latex] for [latex]\sigma[/latex], the standard error of the sample mean is [latex]SE(\bar{x})=\dfrac{s}{\sqrt{n}}[/latex].
[latex]t=\dfrac{\bar{x}-[\text{mean of } \bar{x}'s]}{\text{std. error of } \bar{x}'s}[/latex] [latex]= \dfrac{\bar{x}-\mu}{\frac{s}{\sqrt{n}}}[/latex]
The distribution of [latex]z[/latex]-scores is the standard normal curve, with a mean of 0 and a standard deviation of 1. The distribution of [latex]t[/latex]-scores depends on the sample size, [latex]n[/latex]. There is a different [latex]t[/latex]-model for every [latex]n[/latex]. Instead of referring to [latex]n[/latex] to specify which [latex]t[/latex]-model to use, we refer to the degrees of freedom, or [latex]df[/latex] for short. The number of degrees of freedom is 1 less than the sample size. That is, [latex]df = n – 1[/latex].
Let’s see how the degree of freedom impacts the [latex]t[/latex]-distribution curve.