Large df and the ∞ row

The rule

The course's t-table stops at df $=30$ and then gives one more row, $\infty$, holding the $z$ critical values. The large-df rule: when $\mathrm{df}>30$, use the $\infty$ row. In effect, $t$ in principle, $z$ in practice — because by df 30 the 0.025 critical value is 2.042, within 4 % of 1.960, and it keeps shrinking.

The sources disagree on the wording. IBS1 says a sample of 30 or more lets you use $z$ with $s$; BCcampus says always use $t$; the handouts and the paper sit between: the distribution is $t$, and the table row you read is $\infty$. State it that way and both markers are satisfied.

Worked example — forty-four timesheets

Cedar & Stone samples $n=44$ staff timesheets: mean weekly hours $\bar{x}=75.3$, $s=9.2$. A 95 % interval: $\sigma$ is unknown so this is a t-interval, df $=43>30$, so the multiplier is the $\infty$-row value 1.960:

$$\mathrm{SE}=\frac{9.2}{\sqrt{44}}=1.387,\text{margin}=1.96\times 1.387=2.718,(72.58,78.02)$$

What R does instead

R has no table and no $\infty$ row; it computes the exact $t$ quantile for df $=43$. The ladder below is a real console session:

> qt(0.975, c(10, 30, 43, 1000))
[1] 2.228139 2.042272 2.016692 1.962339
> qnorm(0.975)
[1] 1.959964

The values fall toward qnorm(0.975) as df grows — that is the convergence L9.3 promised. At df 43 R uses 2.0167, so its margin is $2.0167\times 1.387=2.797$, about 3 % wider than the hand margin, and the interval it prints is

> t.test(hours)$conf.int
[1] 72.50294 78.09706
attr(,"conf.level")
[1] 0.95

against the hand answer $(72.58,78.02)$. Both are accepted. The paper prints R output and asks you to reconcile it; the reconciliation is one sentence: R uses $t_{43}=2.017$ where the table's $\infty$ row gives 1.96, so its interval is slightly wider. Not "R uses $\sigma$", not "R divides by $n$" — the SE is identical, 1.387 in both.

Reading qt

qt(p, df) returns the $t$ with cumulative probability $p$ — the value with $p$ to its left. So qt(0.975, 43) is the value with 2.5 % in the upper tail, the 95 % two-sided multiplier; and qt(0.95, 19) is the value with 5 % in the upper tail, $1.729$, the table's $t_{0.05;19}$. The table is indexed by the upper tail; R by the lower. Translate: table column $\alpha$ = qt(1 - α, df). The mismatch is the single most common misreading of R on the paper.