Reading t.test output

◈ 7 cards

R’s t.test prints t, df, the p-value, the alternative, a confidence interval and the sample mean; alternative = sets the tail, and the one-sided p is half the two-sided one.

The printout, line by line

The paper prints R output and asks you to read it. For Cedar & Stone's sixteen files (mean 53, sd 6.4, stored in a vector hrs), the one-sided test of L10.5 is one call — mu = sets $μ_{0}$ and alternative = sets the tail:

> t.test(hrs, mu = 50, alternative = "greater")

	One Sample t-test

data:  hrs
t = 1.875, df = 15, p-value = 0.0402
alternative hypothesis: true mean is greater than 50
95 percent confidence interval:
 50.19512      Inf
sample estimates:
mean of x 
       53

Map each line to the hand computation:

t = 1.875, df = 15 — step 2, exactly: $(53 - 50) / (6.4/ 16)$ on $n - 1$ df.
p-value = 0.0402 — step 3, the exact upper-tail area beyond 1.875 that the table could only bracket as $0.025 < p < 0.05$ . It sits inside the bracket, as it must.
alternative hypothesis: true mean is greater than 50 — R restating $H_{a}$ ; check it against your step 1.
95 percent confidence interval: 50.19512 Inf — a one-sided interval, a lower bound only, matching the one-sided test: every $μ_{0}$ below 50.20 is rejected at 5 %. Not the two-sided interval of Module 9.
mean of x 53 — $\overset{x}{ˉ}$ .

The decision is yours to write; R never says reject. $0.0402 < 0.05$ : reject at 5 %, with the same sentence as L10.5.

The default call is two-sided

Drop alternative = and R tests $μ \neq = 50$ :

> t.test(hrs, mu = 50)
t = 1.875, df = 15, p-value = 0.0804
alternative hypothesis: true mean is not equal to 50
95 percent confidence interval:
 49.58968 56.41032

Same $t$ , same df; the p-value has doubled to 0.0804, and the interval is now the ordinary two-sided one — $53 \pm 2.131 \times 1.6 = (49.59, 56.41)$ , built with $t_{0.025; 15}$ . This is the one to compare with the hand bracket: doubling $0.025 < p < 0.05$ at both ends gives $0.05 < p < 0.10$ , and 0.0804 is inside. At 5 % two-sided, do not reject — 50 lies inside $(49.59, 56.41)$ , L10.7's duality once more.

If a question is one-sided and the paper prints the default output, halve the printed p (the estimate 53 lies in the predicted direction): $0.0804/2 = 0.0402$ .

The quantiles behind the table

> qt(0.95, 15); qt(0.975, 15)
[1] 1.75305
[1] 2.13145

qt(0.95, 15) is the table's $t_{0.05; 15} = 1.753$ (5 % above); qt(0.975, 15) is $t_{0.025; 15} = 2.131$ . R indexes by the cumulative probability, the table by the upper tail — the translation is $1 - α$ , as in L9.5. pt(1.875, 15, lower.tail = FALSE) returns the 0.0402 on the printout.

Reading checklist

Given any t.test printout: find $μ_{0}$ in mu =; find the tail in the alternative hypothesis line; read $t$ and df; compare p with $α$ — halving first if the call was two-sided and the question is not; write the decision and the sentence yourself; and know which interval the call produced.