14.1 The language of hypothesis testing — Practise with help

1. $P(X = 0) = 0.75^{20} = 0.0032$ and $P(X = 1) = \binom{20}{1}(0.25)(0.75)^{19} = 0.0211$, each to 4 decimal places.
2. $\dfrac{30}{200} = 0.15$.
3. $0.05 = \dfrac{5}{100} = \dfrac{1}{20}$.

If question 1 was hard, revise the binomial probability formula before you start.

The claim supplies the value inside $H_0$. The suspicion supplies the direction of $H_1$.

Write $H_0: p = 0.05$. Ask which direction worries the inspector, then choose $<$ or $>$ for $H_1$.

The claim is that the fault rate is 5%, so

The inspector suspects a higher rate, so

This is a one-tailed test on the upper tail. Under $H_0$ the number of faulty bolts in 40, written $X$, has the model $X \sim \mathrm{B}(40, 0.05)$.

Ask whether one direction is the only concern, or whether both directions would matter.

For part (a), the words "too often" name one direction. For part (b), the words "either direction" name two.

(a) One-tailed, upper tail, because the suspicion points to too many sixes and too few is not the concern.

(b) Two-tailed, because a drift up and a drift down are both problems.

(c) One-tailed, lower tail, because the customer suspects fewer on-time deliveries than the claim.

The lower tail is the sum of the probabilities for $X = 0$, $X = 1$ and $X = 2$.

Use the binomial formula for each of the three values, then add the three probabilities. Compare the total with $0.05$ and with $0.01$ in turn.

to 4 decimal places. At the 5% level, $0.0913 > 0.05$, so the test retains $H_0$. At the 1% level, $0.0913 > 0.01$, so the test also retains $H_0$. The same data can retain the claim at both levels, because the evidence is not strong at either one.

Compare the p-value with each significance level separately.

Notice that $0.032$ is below $0.05$ but above $0.01$. Write one comparison for each level.

At the 5% level, $0.032 < 0.05$, so the test rejects $H_0$. At the 1% level, $0.032 > 0.01$, so the test retains $H_0$.

The decisions differ because a smaller significance level demands stronger evidence. The 1% level accepts a smaller risk of wrongly rejecting a true $H_0$, so the result must be more extreme before it counts.

A fair coin has $p = 0.5$, so the hypotheses use $\ne$ for $H_1$. The test has two tails.

Use symmetry: with $p = 0.5$, $P(X \ge 17)$ equals $P(X \le 7)$. Double the smaller tail before comparing with $0.05$.

The hypotheses are

The value 7 is below the centre of the distribution, so the smaller tail is the lower one:

The two-tailed p-value is

Since $0.0640 > 0.05$, the test retains $H_0$. There is not sufficient evidence at the 5% level to suggest the coin is not fair.

A p-value above the significance level retains the claim. Then think about what the evidence can and cannot show.

Decide the test first. Then write a sentence that starts "There is not sufficient evidence…" for the context.

Since $0.21 > 0.05$, the test retains $H_0$ at the 5% level. There is not sufficient evidence to suggest that the proportion differs from $0.4$.

The conclusion does not mean that the proportion is exactly $0.4$. It means that this sample did not provide strong enough evidence against $0.4$. A larger sample might give a different verdict.