14.1 The language of hypothesis testing — Practise

Work through these on your own. For each test, write $H_0$ , $H_1$ and the significance level before you calculate.

The questions and the answers below come from a set that is checked by computer every time the book is built.

A packet contains a voucher with probability $0.25$ . A sample of 20 packets is checked. Find $P(X \le 1)$ , where $X$ is the number of vouchers, to 4 decimal places.
A company claims that 1 packet in 4 contains a voucher. A customer buys 20 packets and finds 1 voucher. Test the claim at the 5% level against the alternative that fewer packets contain a voucher.
Use the same claim and sample of 20 packets, but suppose the customer finds 2 vouchers. Test the claim at the 5% level against the alternative that fewer packets contain a voucher.
A coin is tossed 20 times. Find $P(X \ge 15)$ for the number of heads $X$ when the coin is fair, to 4 decimal places.
A coin is tossed 20 times and shows 15 heads. Test at the 5% level whether the coin is fair, using a two-tailed test.
A coin is tossed 20 times and shows 5 heads. Test at the 5% level whether the coin is fair, using a two-tailed test.
A test gives a p-value of $0.032$ . State, with a symbol, how this p-value compares with the 5% significance level.
The same test gives a p-value of $0.032$ . State, with a symbol, how this p-value compares with the 1% significance level.
A test uses the 1% significance level. Write down the probability of incorrectly rejecting $H_0$ when $H_0$ is true.
A two-tailed test uses the 5% significance level. Find the probability placed in each tail.
A company claims that 30% of students walk to school. In a sample of 25 students, 12 walk. Find $P(X \ge 12)$ under the claim, to 4 decimal places.
Test the claim that 30% of students walk to school at the 5% level, against the alternative that the proportion is higher. Use the sample of 25 students in which 12 walk.

Answers and concise solutions

1. The lower tail probability

Answer: $P(X \le 1) = P(0) + P(1) = 0.0032 + 0.0211 = 0.0243$ to 4 decimal places.

2. The p-value and the decision

Answer: $H_0: p = 0.25$ and $H_1: p < 0.25$ . Under $H_0$ , $X \sim \mathrm{B}(20, 0.25)$ . The p-value is $P(X \le 1) = 0.0243 < 0.05$ , so reject $H_0$ . There is sufficient evidence at the 5% level that fewer than 1 packet in 4 contains a voucher.

3. The p-value and the decision

Answer: The p-value is $P(X \le 2) = 0.0913 > 0.05$ , so retain $H_0$ . Two vouchers is below the expected 5, but a result this low is not unusual when $p = 0.25$ .

4. The upper tail probability

Answer: $P(X \ge 15) = 0.0207$ to 4 decimal places. Only about 2% of fair-coin runs of 20 give 15 or more heads.

5. The two-tailed p-value and the decision

Answer: $H_0: p = 0.5$ and $H_1: p \ne 0.5$ . The two-tailed p-value is $2 \times 0.0207 = 0.0414 < 0.05$ , so reject $H_0$ . There is sufficient evidence that the coin is not fair.

6. The two-tailed p-value and the decision

Answer: The smaller tail is $P(X \le 5) = 0.0207$ , so the two-tailed p-value is $0.0414 < 0.05$ and the test rejects $H_0$ . By symmetry, 5 heads is as unlikely as 15 heads.

7. The comparison

Answer: $0.032 < 0.05$ , so the test rejects $H_0$ at the 5% level.

8. The comparison

Answer: $0.032 > 0.01$ , so the test retains $H_0$ at the 1% level. The smaller level demands stronger evidence.

9. The probability

Answer: The significance level is that probability, so it is $0.01$ .

10. The probability in each tail

Answer: The 5% is shared equally, so each tail holds $\dfrac{0.05}{2} = 0.025$ .

11. The upper tail probability

Answer: $P(X \ge 12) = 0.0442$ to 4 decimal places.

12. The p-value and the decision

Answer: $H_0: p = 0.3$ and $H_1: p > 0.3$ . The p-value is $0.0442 < 0.05$ , so reject $H_0$ . There is sufficient evidence at the 5% level that more than 30% of students walk to school.