Questions on hypothesis testing with the binomial

Question 1

Define the null hypothesis, the alternative hypothesis and the significance level of a hypothesis test.

Accepted Answer

The null hypothesis H₀ states the assumed value of the population parameter, the value taken as true while the test runs. The alternative H₁ states the change being tested for. The significance level is the probability threshold, so evidence rarer than that threshold under H₀ counts as grounds for rejection. B1 B1 B1, one for each of the three definitions. Both hypotheses must name a parameter, never an observed count.

Question 2

A die is suspected of being biased towards sixes. It is rolled 30 times. Write down suitable hypotheses in terms of p, and state the distribution of the number of sixes X under H₀.

Accepted Answer

H₀: p = 1/6 and H₁: p > 1/6, where p is the probability that a single roll gives a six. Under H₀, X ~ B(30, 1/6). B1 B1 for the two hypotheses and B1 for the distribution. 'Biased towards sixes' points one way only, so the alternative is one-tailed with a strict inequality.

Question 3

A manufacturer claims that at most 5% of its bulbs fail within a year. A consumer group suspects the failure rate is higher and tests a random sample of 40 bulbs. Write down suitable hypotheses for the test, defining the parameter you use.

Accepted Answer

H₀: p = 0.05 and H₁: p > 0.05, where p is the probability that a randomly chosen bulb fails within a year. B1 for H₀ written as an equality at the claimed value, B1 for the one-sided alternative with p defined. The hypotheses name the population proportion p and never the count of failures among the 40; that count is the test statistic used to judge them. A claim of 'at most 5%' is tested at its boundary, because 0.05 is the value in the claim that is hardest to reject.

Question 4

A coin is tested for bias by tossing it 50 times and counting the number of heads, X. Write down suitable hypotheses, state the distribution of X under H₀, and state whether the test is one-tailed or two-tailed, with a reason.

Accepted Answer

H₀: p = 0.5 and H₁: p ≠ 0.5, where p is the probability of a head on a single toss. B1 for the pair. Under H₀, X ~ B(50, 0.5). B1 for the distribution with both parameters. The test is two-tailed, because the word 'biased' names no direction: a coin favouring tails is as much biased as one favouring heads, so both ends of the distribution count as evidence. B1 for two-tailed with that reason. Compare 'biased towards heads', which points one way and would give H₁: p > 0.5.

Question 5

A spinner is claimed to land on red with probability 0.3. In 25 independent spins it lands on red 12 times. Test, at the 5% significance level, whether the spinner favours red more often than claimed. State your hypotheses clearly.

Accepted Answer

H₀: p = 0.3 and H₁: p > 0.3, where p is the probability of red on one spin. Under H₀, X ~ B(25, 0.3). The probability of a result at least as extreme as the one observed is P(X ≥ 12) = 1 − P(X ≤ 11) = 0.0442. Since 0.0442 < 0.05, reject H₀. There is significant evidence at the 5% level that the spinner lands on red more often than claimed. B1 hypotheses, M1 the binomial model, M1 the correct tail, A1 0.0442, A1 comparison and conclusion in context. Working out P(X ≤ 12) on the calculator instead of 1 − P(X ≤ 11) gives 0.9825 and a conclusion pointing the wrong way.

Question 6

Under H₀, X ~ B(20, 0.4), and the alternative hypothesis is H₁: p > 0.4. Find the critical region for a test at the 5% significance level, and state the actual significance level of the test.

Accepted Answer

Work down from the top of the distribution. P(X ≥ 12) = 0.0565, which is bigger than 0.05, so 12 cannot be in the region. P(X ≥ 13) = 0.0210, which is smaller, so 13 can. The critical region is X ≥ 13, and the actual significance level is 0.0210, or 2.10%. M1 for attempting a tail, A1 for 0.0565, A1 for the region, B1 for the actual level. Because X is discrete the advertised 5% is almost never achievable exactly, and the actual level is the true tail probability of whatever region you end up with.

Question 7

For a test of H₀: p = 0.4 against H₁: p > 0.4 based on X ~ B(20, 0.4), the critical region is X ≥ 13. State the conclusion when a value of 12 is observed, and explain why this conclusion is not the same as proving that p = 0.4.

Accepted Answer

12 lies outside the critical region, so do not reject H₀. There is insufficient evidence at the 5% level that p has risen above 0.4. That is a failure to disprove, not a proof. The data are consistent with p = 0.4 and equally consistent with p = 0.45 or p = 0.5, none of which the test has ruled out. M1 for comparing with the region, A1 for the conclusion in context, B1 for the explanation.

Question 8

A seed supplier claims that 80% of its seeds germinate. A gardener plants a random sample of 20 of the seeds and 13 of them germinate. Test, at the 5% significance level, whether the germination rate is lower than the supplier claims. State your hypotheses clearly.

Accepted Answer

H₀: p = 0.8 and H₁: p < 0.8, where p is the probability that a seed germinates. B1 for the hypotheses. Under H₀, X ~ B(20, 0.8). M1 for the binomial model. The alternative points downwards, so the tail to take is the lower one: P(X ≤ 13) = 0.0867. M1 for the correct tail, A1 for 0.0867. Since 0.0867 > 0.05, do not reject H₀. There is insufficient evidence at the 5% level that the germination rate is below the 80% claimed. A1 for the comparison and a conclusion written about the seeds. Thirteen out of twenty is 65% against a claimed 80%, which looks far short, and yet on only twenty seeds a result at least this low turns up about once in every twelve plantings even when the claim is true.

Question 9

Under H₀, X ~ B(30, 0.4), and the alternative hypothesis is H₁: p ≠ 0.4. A value of 17 is observed, and P(X ≥ 17) = 0.0481. Test at the 5% significance level.

Accepted Answer

The alternative names no direction, so the 5% is split between the two ends and each tail is judged against 0.025. B1 for halving the significance level. The observed 17 sits above the mean of 30 × 0.4 = 12, so the upper tail is the relevant one: P(X ≥ 17) = 0.0481. M1 for choosing that tail. Since 0.0481 > 0.025, the value 17 is not in the critical region, so do not reject H₀. A1. There is insufficient evidence at the 5% level that p differs from 0.4. A1 for the conclusion. Had the alternative been H₁: p > 0.4, the same 0.0481 would have been set against 0.05 and H₀ would have been rejected.

Question 10

A politician claims that 15% of the voters in a constituency support a proposal. A campaigner believes the proportion is higher and asks a random sample of 40 voters, of whom 11 support it. Given that P(X ≥ 11) = 0.0299 for X ~ B(40, 0.15), test the campaigner's belief at the 5% significance level.

Accepted Answer

H₀: p = 0.15 and H₁: p > 0.15, where p is the proportion of voters in the constituency who support the proposal. B1 for the hypotheses. Under H₀, X ~ B(40, 0.15), and the alternative points upwards, so the tail runs from the observed value: P(X ≥ 11) = 0.0299. M1 for the upper tail taken from 11. Since 0.0299 < 0.05, reject H₀. A1. There is significant evidence at the 5% level that more than 15% of the voters in the constituency support the proposal. A1 for the conclusion in context. On a calculator this tail is 1 − P(X ≤ 10), with the boundary dropped by one. Entering 1 − P(X ≤ 11) leaves the observed value out of its own tail and returns 0.0120, a figure that overstates how unusual the result is.

Question 11

Explain why the actual significance level of a test based on a binomial distribution is rarely equal to the level stated in the question, and state how the two compare.

Accepted Answer

The test statistic is a whole number of successes, so the tail probability takes only the values the distribution offers and cannot be tuned to 0.05 exactly. B1 for the discreteness of X. The critical region is built by taking the largest tail whose probability does not exceed the stated level, so the region stops one value short of that level rather than reaching it. B1 for how the region is chosen. The actual significance level is that region's own tail probability, which is therefore at or below the stated level, and in practice below it. B1 for the comparison. The same number is the probability of rejecting H₀ when H₀ is true.

Question 12

A researcher believes the proportion p of a population holding a certain view differs from 0.25. A random sample of 30 people is taken and X, the number holding the view, is recorded. Find the critical region for a two-tailed test of H₀: p = 0.25 at the 5% significance level, and find the actual significance level of the test.

Accepted Answer

Under H₀, X ~ B(30, 0.25), and a two-tailed 5% test puts at most 2.5% in each tail. Lower end: P(X ≤ 2) = 0.0106, which is under 0.025, while P(X ≤ 3) = 0.0374, which is over, so the lower region is X ≤ 2. Upper end: P(X ≥ 13) = 0.0216, under 0.025, while P(X ≥ 12) = 0.0507, over, so the upper region is X ≥ 13. The critical region is X ≤ 2 or X ≥ 13, and the actual significance level is 0.0106 + 0.0216 = 0.0322, or 3.22%. B1 for splitting the 5% into two 2.5% halves, M1 A1 for the lower tail, M1 A1 for the upper tail, A1 for the actual level. Comparing each tail with 0.05 rather than 0.025 makes the region far too big, and 5% is not the actual level.

Question 13

A machine is claimed to produce faulty items with probability 0.3. An engineer believes the proportion has fallen. A random sample of 30 items is inspected. Find the critical region for a test of H₀: p = 0.3 against H₁: p < 0.3 at the 1% significance level, and write down the probability of a Type I error for this test.

Accepted Answer

Under H₀, X ~ B(30, 0.3), and the alternative points downwards so the region sits in the lower tail. P(X ≤ 3) = 0.00932, which is under 0.01, while P(X ≤ 4) = 0.0302, which is over. The critical region is therefore X ≤ 3. A Type I error is rejecting H₀ when it is true, which happens exactly when X lands in the critical region while p really is 0.3, so its probability is 0.00932. M1 for working in the lower tail, A1 for 0.00932, A1 for 0.0302, A1 for the region, B1 for the Type I probability. The probability of a Type I error and the actual significance level are the same number, and questions ask for it under either name.

Question 14

A spinner is claimed to land on red with probability 0.2. A test of H₀: p = 0.2 against H₁: p > 0.2 at the 5% significance level uses X, the number of reds in 25 spins. Find the critical region and the actual significance level. The whole test is then run a second time on a fresh, independent set of 25 spins. Find the probability that at least one of the two tests rejects H₀ when p really is 0.2.

Accepted Answer

Under H₀, X ~ B(25, 0.2), and the alternative points upwards, so work down from the top. P(X ≥ 8) = 0.1091, which is above 0.05, so 8 cannot be in the region. P(X ≥ 9) = 0.0468, which is below it, so 9 can. M1 for attempting a tail, A1 for 0.1091, A1 for the critical region X ≥ 9. The actual significance level is that region's own probability, 0.0468. B1. Each run rejects a true H₀ with probability 0.0468, and the two sets of spins are independent, so the probability that neither rejects is 0.9532² = 0.9086 and the probability that at least one does is 1 − 0.9086 = 0.0914. M1 for the complement of both failing to reject, A1 for 0.0914. Running the same 5% test twice therefore rejects a true H₀ about 9% of the time rather than 5%: a significance level is a property of one test, and repeating a test raises the chance of a false rejection somewhere among the runs.

Question 15

A grower claims that 35% of a batch of bulbs are yellow. A buyer believes the proportion is lower and inspects a random sample of 20 bulbs. Find the critical region for a test of H₀: p = 0.35 against H₁: p < 0.35 at the 10% significance level, state the actual significance level, and explain why an observed value of 3 leads to rejection while 4 does not.

Accepted Answer

Under H₀, X ~ B(20, 0.35), and the alternative points downwards, so work up from the bottom. P(X ≤ 4) = 0.1182, above 0.10, so 4 cannot be in the region. P(X ≤ 3) = 0.0444, below it, so 3 can. M1 for working in the lower tail, A1 for 0.1182, A1 for the critical region X ≤ 3. The actual significance level is 0.0444, or 4.44%. B1. So 3 rejects H₀ and 4 does not, though the two counts differ by a single bulb and their tail probabilities differ by 0.074. B1 for the explanation. The boundary has to land between two whole numbers because X is discrete, and the rule that fixes where is that the region's own probability must not exceed the stated 10%: letting 4 in would take the region to 11.8%, above the level the test was set at.