Upper & Lower Bound on P-Value using printed t table

Question

Suppose we have a sample size of $n=25,$ and want to test $H_0 :\mu = 0$ against $H_A : \mu > 0$ using a t test at the 5% level of significance.

The first question was

Suppose the t-statistic were calculated to be $t = 1.972.$ What is the greatest upper & lower bound for the corresponding p-value that is implied by your critical value table?

And my answer was $0.025 <$ p value $< 0.050.$

However, the next part ask

Using a normal approximation to the null distribution of the t statistic, what is the greatest upper & lower bound for the p-value that is implied?

I'm not sure how to do this? Do I use the z table but not quite sure how.

Answer 1

For the first part, the answer depends on the format of the printed t table you are supposed to use. For a one-sample t test with a sample of size $n = 25,$ the degrees of freedom are $\nu = n - 1 = 25.$ Reading across the row for $\nu = 24$ in the t table I have at hand, the statistic $t = 1.972$ is bracketed by 1.711 [which cuts probability $0.05$ from the upper tail of $\mathsf{T}(\nu=24)]$ and 2.064 [which cuts probability 0.025 from the upper tail]. So I get the same answer for the p-value from my table that you got from yours.

Using software, you can get the exact p-value by finding what probability 1.972 cuts from the upper tail of this t distribution. In R statistical software, pt is the CDF of a t distribution, so the exact p-value is 0.0301 (to four places). P-values were not widely used in statistical testing until computers and statistical calculators became widely available.

1 - pt(1.972, 24)
[1]  0.03012028

For the second part, you are right in saying that you would use a z table. Again, the answer depends on the format of the z table in your book. Typical z tables give a lot more detail about the standard normal distribution than typical t tables give about any one t distribution.

In my z table, the statistic $t = 1.972$ is bracketed by values $1.97$ and $1.98$ in the margins of the table which cut respective probabilities 0.0244 and 0.0239 from the upper tail of $\mathsf{Norm}(0,1)$. So the approximated p-value is between 0.0239 and 0.0244.

Notice that the distributions $\mathsf{T}(24)$ and $\mathsf{Norm}(0,1)$ are roughly the same, but not exactly. So in practice, it would not be a good idea to use the z table for this purpose. [The value $1.972$ happens to cut very nearly probability 0.0243 from the upper tail of the standard normal distribution, but this is still only a rough approximation of the actual p-value.]

1 - pnorm(1.972)
[1] 0.0243048

In summary, I'd like to mention that this problem draws attention to an important difference between t tables and z tables: t tables have probabilities in the margin and cut-off values in the body of the table. By contrast, z tables have probabilities in the body of the table and cut-off value in the margins.