14.2 Binomial tests for a proportion — Practise with help

Attempt each question closed book first. Write the five steps down in order before you calculate.

Warm-up: remember

Write $H_0$ and $H_1$ for a claim that the proportion is $0.4$ , when any difference from $0.4$ matters.
$X \sim \mathrm{B}(20, 0.25)$ . Find $P(X \le 1)$ to 4 decimal places.
Explain, in one sentence, what the 5% significance level means.

Answers

$H_0: p = 0.4$ and $H_1: p \ne 0.4$ . The test has two tails.
$P(X \le 1) = 0.0032 + 0.0211 = 0.0243$ to 4 decimal places.
The 5% level is the probability of rejecting a true $H_0$ that you are willing to accept.

If question 1 mixed up the alternative, reread Section 14.1 before you start.

Practise with help

1. A delivery company claims that 20% of parcels arrive late. A shop records 30 parcels and finds 2 that arrive late. Test at the 5% level whether the late rate is lower than the claim.

Hint

The claim gives $p_0 = 0.2$ . The concern is "lower", so the alternative uses $<$ .

Next step

Write $H_0$ , $H_1$ and the model $X \sim \mathrm{B}(30, 0.2)$ . Then find the lower tail $P(X \le 2)$ .

Full working

H_0: p = 0.2, \qquad H_1: p < 0.2, \qquad \alpha = 0.05.

Under $H_0$ , $X \sim \mathrm{B}(30, 0.2)$ , and the sample gives $X = 2$ . The p-value is

P(X \le 2) = 0.0442 \quad (4 \text{ d.p.}).

Since $0.0442 < 0.05$ , the test rejects $H_0$ . There is sufficient evidence at the 5% level that fewer than 20% of parcels arrive late.

2. A manufacturer claims that 20% of components are faulty. A sample of 25 components contains 2 faulty ones. Test at the 5% level whether the fault rate is lower than the claim.

Hint

The model is $X \sim \mathrm{B}(25, 0.2)$ , not $\mathrm{B}(30, 0.2)$ . The sample size has changed.

Next step

Find the lower tail $P(X \le 2)$ for 25 components, then compare the value with $0.05$ .

Full working

H_0: p = 0.2, \qquad H_1: p < 0.2, \qquad \alpha = 0.05.

Under $H_0$ , $X \sim \mathrm{B}(25, 0.2)$ , and the sample gives $X = 2$ . The p-value is

P(X \le 2) = 0.0982 \quad (4 \text{ d.p.}).

Since $0.0982 > 0.05$ , the test retains $H_0$ . There is not sufficient evidence at the 5% level that fewer than 20% of components are faulty.

The count is the same as question 1, but the sample is smaller, so a count of 2 is less surprising.

3. A survey claims that 40% of shoppers prefer brand A. In a sample of 30 shoppers, 16 prefer brand A. Test at the 5% level whether the proportion is higher than the claim.

Hint

"Higher" gives a one-tailed upper test. The sample proportion is $\frac{16}{30} = 0.53$ , which is above the claim.

Next step

Under $H_0$ , $X \sim \mathrm{B}(30, 0.4)$ . Find the upper tail $P(X \ge 16)$ and compare it with $0.05$ .

Full working

H_0: p = 0.4, \qquad H_1: p > 0.4, \qquad \alpha = 0.05.

Under $H_0$ , $X \sim \mathrm{B}(30, 0.4)$ , and the sample gives $X = 16$ . The p-value is

P(X \ge 16) = 0.0971 \quad (4 \text{ d.p.}).

Since $0.0971 > 0.05$ , the test retains $H_0$ . There is not sufficient evidence at the 5% level that more than 40% of shoppers prefer brand A. The sample proportion is above the claim, but the sample is too small for the gap to count.

4. A coin is tossed 16 times and shows 12 heads. Test at the 5% level whether the coin is biased towards heads.

Hint

The suspicion points one way, so this is a one-tailed upper test.

Next step

Under $H_0$ , $X \sim \mathrm{B}(16, 0.5)$ . Find $P(X \ge 12)$ and compare with $0.05$ .

Full working

H_0: p = 0.5, \qquad H_1: p > 0.5, \qquad \alpha = 0.05.

Under $H_0$ , $X \sim \mathrm{B}(16, 0.5)$ , and the sample gives $X = 12$ . The p-value is

P(X \ge 12) = 0.0384 \quad (4 \text{ d.p.}).

Since $0.0384 < 0.05$ , the test rejects $H_0$ . There is sufficient evidence at the 5% level that the coin is biased towards heads.

5. A spinner is claimed to land on red 1 time in 4. In 30 spins, it lands on red 13 times. Test at the 5% level whether the spinner is unfair.

Hint

"Unfair" allows too many reds or too few, so the test has two tails.

Next step

Under $H_0$ , $X \sim \mathrm{B}(30, 0.25)$ . The value 13 is above the centre, so find $P(X \ge 13)$ , then double it.

Full working

H_0: p = 0.25, \qquad H_1: p \ne 0.25, \qquad \alpha = 0.05.

Under $H_0$ , $X \sim \mathrm{B}(30, 0.25)$ , and the sample gives $X = 13$ . The value is above the centre, so the smaller tail is the upper one:

P(X \ge 13) = 0.0216 \quad (4 \text{ d.p.}).

The two-tailed p-value is

2 \times 0.0216 = 0.0432.

Since $0.0432 < 0.05$ , the test rejects $H_0$ . There is sufficient evidence at the 5% level that the spinner is unfair.

Notice the two-stage decision. The one-sided tail of $0.0216$ and the two-sided $0.0432$ fall on opposite sides of $0.05$ only when the tail is close to the level. Always check which test the question asks for.

6. In question 1 the p-value was $0.0442$ . The same test is repeated at the 1% level. State the new decision and explain the difference.

Hint

The p-value has not changed, but the level has.

Next step

Compare $0.0442$ with $0.01$ . Then say what a smaller level does to the risk of wrongly rejecting a true claim.

Full working

At the 1% level, $0.0442 > 0.01$ , so the test retains $H_0$ . The evidence produced a rejection at the 5% level and a retention at the 1% level.

The 1% level accepts a smaller risk of wrongly rejecting a true claim, so a result must be more surprising before it counts. A conclusion must state the level, because the same data can give two different decisions.