Hypothesis Testing

A binomial test is carried out to investigate whether a coin is biased towards heads, where p is the probability of obtaining a head. Which of the following correctly states the hypotheses?

• AH₀: p = 0.5, H₁: p > 0.5✓
• BH₀: p > 0.5, H₁: p = 0.5
• CH₀: p = 0.5, H₁: p ≠ 0.5
• DH₀: p = 0.5, H₁: p < 0.5

Answer: The null hypothesis assumes no bias, p = 0.5. 'Biased towards heads' is a one-tailed alternative in the upper direction, so H₁: p > 0.5. Setting H₁: p ≠ 0.5 makes the test two-tailed; H₁: p < 0.5 tests for bias towards tails; and putting the inequality into H₀ reverses the roles of the two hypotheses.

A teacher claims a six-sided die is biased, but does not say in which direction. A binomial test on the number of sixes is planned. Which type of test is appropriate, and what is H₁?

• AOne-tailed, H₁: p > 1/6
• BTwo-tailed, H₁: p ≠ 1/6✓
• COne-tailed, H₁: p < 1/6
• DTwo-tailed, H₁: p = 1/6

Answer: Because the claim of bias gives no direction, the test must allow deviation either way, so it is two-tailed with H₁: p ≠ 1/6. A one-tailed test would only detect bias in a single direction.

A two-tailed binomial hypothesis test is to be conducted at the 5% significance level. How is the significance level applied to the two tails of the distribution?

• A5% is placed in each tail
• B5% is placed in the upper tail only
• C10% is split as 5% in each tail
• D2.5% is placed in each tail✓

Answer: For a two-tailed test the significance level is split equally between the two tails, so each tail contains 5% ÷ 2 = 2.5%. Placing 5% in each tail double-counts and gives a 10% test overall; placing 5% in the upper tail only describes a one-tailed test; and splitting 10% as 5% per tail applies the wrong overall level.

For X ~ B(10, 0.25) a one-tailed test of H₀: p = 0.25 against H₁: p > 0.25 is carried out at the 5% level. The observed value is x = 6, and P(X ≥ 6) = 0.0197. What is the conclusion?

• AAccept H₁ is impossible to determine
• BDo not reject H₀; the result is not significant
• CReject H₀; there is no evidence p has increased
• DReject H₀; the result is significant, so there is evidence p > 0.25✓

Answer: Compare the p-value with the significance level: 0.0197 < 0.05, so the result lies in the critical region. We reject H₀ and conclude there is evidence that p > 0.25. Choosing not to reject H₀ ignores that the p-value is below the level, and rejecting H₀ while stating there is no evidence of an increase contradicts itself.

A random variable X ~ B(20, 0.3) is used to test H₀: p = 0.3 against H₁: p < 0.3 at the 5% significance level. Given P(X ≤ 2) = 0.0355 and P(X ≤ 3) = 0.1071, state the critical region.

• AX ≤ 2✓
• BX ≤ 3
• CX ≥ 2
• DX ≤ 1

Answer: For a lower-tailed test we need the largest region with probability ≤ 5%. Since P(X ≤ 2) = 0.0355 < 0.05 but P(X ≤ 3) = 0.1071 > 0.05, the critical region is X ≤ 2. Taking X ≤ 3 gives a region of probability 0.1071, above the 5% level; X ≤ 1 is smaller than it needs to be; and X ≥ 2 lies in the upper tail, the wrong direction for H₁: p < 0.3.

In a binomial hypothesis test of H₀: p = 0.4 against H₁: p > 0.4, the calculated p-value is 0.08 and the significance level is 5%. Which conclusion is correct?

• AReject H₀: there is strong evidence that p > 0.4
• BDo not reject H₀: there is insufficient evidence that p > 0.4✓
• CAccept H₁: the proportion has definitely increased
• DReject both H₀ and H₁ as inconclusive

Answer: Since the p-value 0.08 > 0.05, the result is not in the critical region, so we do not reject H₀. We conclude there is insufficient evidence to support p > 0.4. Rejecting H₀ despite the p-value exceeding the significance level is the classic error, and claiming the proportion has definitely increased overstates what a hypothesis test can show.

In a hypothesis test at the 5% significance level, the p-value is 0.032. What is the conclusion?

• AThe test is inconclusive because 0.032 is close to 0.05
• BAccept H₀, because 0.032 is small
• CReject H₁ in favour of H₀
• DReject H₀ — there is significant evidence at the 5% level✓

Answer: Compare the p-value with the significance level: 0.032 < 0.05, so the result falls in the critical region and H₀ is rejected. Note the wording — you reject H₀ or fail to reject it; you never 'accept' H₀, because failing to find evidence is not the same as proving something true.

A one-tailed binomial test uses H₀: p = 0.4 and H₁: p > 0.4. Where is the critical region?

• AIn the lower tail, at low values
• BIn the upper tail, at high values of the test statistic✓
• CSplit equally between both tails
• DCentred on the mean of the distribution

Answer: H₁ points upwards (p > 0.4), so only unusually LARGE observed values count as evidence against H₀ — the critical region sits entirely in the upper tail. A two-tailed test would split the significance level between both tails, which is why the alternative hypothesis has to be settled before testing.

What does the significance level of a hypothesis test represent?

• AThe proportion of the sample that must disagree with H₀
• BThe probability that H₀ is true
• CThe probability of accepting H₁ when it is true
• DThe probability of rejecting H₀ when it is actually true✓

Answer: The significance level is the probability of a Type I error — rejecting a true null hypothesis. It is emphatically not the probability that H₀ is true; H₀ is either true or it is not, and the test gives no probability for that.

What is the null hypothesis in a hypothesis test?

• AThe hypothesis that is automatically accepted once the p-value is calculated
• BThe particular outcome that the researcher is actively hoping to demonstrate
• CThe default assumption of no effect or no change✓
• DA statement about the sample rather than about the wider population

Answer: H₀ states the status quo, and the test asks whether the evidence is strong enough to reject it. Hypotheses are always statements about the POPULATION parameter, never about the sample.

A manufacturer states that battery lifetimes are normally distributed with mean 20 hours and standard deviation 1.5 hours. A random sample of 25 batteries has mean lifetime 20.6 hours. Assuming the standard deviation is unchanged, test at the 1% significance level whether the mean lifetime has changed (H₀: μ = 20, H₁: μ ≠ 20).

Which is the correct calculation and conclusion?

• Az = 2.00 and 2.00 > 1.960, so reject H₀: there is evidence at the 1% level that the mean lifetime has changed
• Bz = 0.40 and 0.40 < 2.576, so do not reject H₀: there is insufficient evidence at the 1% level that the mean lifetime has changed
• Cz = 2.00 and 2.00 > 1.645, so reject H₀: there is evidence at the 1% level that the mean lifetime has changed
• Dz = 2.00 and 2.00 < 2.576, so do not reject H₀: there is insufficient evidence at the 1% level that the mean lifetime has changed✓

Answer: Under H₀ the sample mean is N(20, 1.5²/25), with standard error 0.3. z = (20.6 − 20)/0.3 = 2.00. A two-tailed 1% test puts 0.5% in each tail, so the critical value is 2.576. Since 2.00 < 2.576, H₀ is not rejected. 1.960 is the two-tailed 5% value, so it does not apply here; z = 0.40 forgets to divide σ by √25.

Over 12 days a café records the daily maximum temperature and the number of hot drinks sold. The product moment correlation coefficient is r = −0.61. The owner tests H₀: ρ = 0 against H₁: ρ < 0 at the 5% level; the critical value is −0.4973.

Which conclusion is correct?

• ASince −0.61 < −0.4973, do not reject H₀: there is insufficient evidence of negative correlation
• BSince −0.61 < −0.4973, reject H₀: there is evidence of negative correlation in the population✓
• CSince 0.61 > 0.4973, reject H₀: there is evidence that higher temperatures cause fewer hot-drink sales
• DSince −0.61 < −0.4973, reject H₀: the population correlation coefficient is shown to be −0.61

Answer: For H₁: ρ < 0 the critical region is r ≤ −0.4973, and −0.61 lies in it, so H₀ is rejected: there is evidence of negative correlation between temperature and sales. A significant correlation does not show causation, and r = −0.61 estimates ρ from a sample; it does not establish its value.

Historically 35% of patients at a clinic miss appointments. After a text-reminder scheme, 4 of a random sample of 25 patients miss appointments. The clinic tests H₀: p = 0.35 against H₁: p < 0.35 at the 5% level, using X ~ B(25, 0.35).

Which is the correct calculation and conclusion?

• AP(X ≤ 4) = 0.0320 > 0.025, so do not reject H₀: there is insufficient evidence at the 5% level that the proportion has decreased
• BP(X ≤ 4) = 0.0320 < 0.05, so reject H₀: there is evidence at the 5% level that the proportion has decreased✓
• CP(X ≤ 4) = 0.0320 < 0.05, so reject H₀: this proves that the proportion missing appointments is now 16%
• DP(X ≥ 4) = 0.9903 > 0.05, so do not reject H₀: there is insufficient evidence at the 5% level that the proportion has decreased

Answer: For a lower-tail test the p-value is P(X ≤ 4) under B(25, 0.35), which is 0.0320. This is below 0.05, so H₀ is rejected. Comparing with 0.025 halves the level as if the test were two-tailed. A significant result is evidence in context, not proof, and 4/25 = 16% is only a sample estimate.

A seed supplier states that 15% of its seeds fail to germinate. A gardener suspects the proportion is higher and plants 20 seeds, testing H₀: p = 0.15 against H₁: p > 0.15 at the 5% level, with X ~ B(20, 0.15) the number that fail.

State the critical region, and the probability of rejecting H₀ when H₀ is true.

• AX ≥ 6; actual significance level 6.73%
• BX ≥ 8; actual significance level 0.59%
• CX ≥ 7; actual significance level 6.73%
• DX ≥ 7; actual significance level 2.19%✓

Answer: Under H₀, P(X ≥ 6) = 0.0673, which exceeds 0.05, and P(X ≥ 7) = 0.0219, which does not. The critical region is the largest upper tail with probability at most 5%: X ≥ 7, with actual significance level 2.19%. X ≥ 6 would reject H₀ with probability 6.73%, more than the 5% allowed.

A spinner is claimed to land on red with probability 0.25. In 50 spins it lands on red 20 times. A test of H₀: p = 0.25 against H₁: p ≠ 0.25 is carried out at the 5% level, with X ~ B(50, 0.25).

Which is the correct calculation and conclusion?

• AP(X ≥ 20) = 0.0139 < 0.025, so reject H₀: there is evidence at the 5% level that P(red) is not 0.25✓
• B2 × P(X ≥ 20) = 0.0278 > 0.025, so do not reject H₀: there is insufficient evidence at the 5% level that P(red) is not 0.25
• CP(X ≤ 20) = 0.9937 > 0.025, so do not reject H₀: there is insufficient evidence at the 5% level that P(red) is not 0.25
• DP(X ≥ 20) = 0.0139 < 0.025, so reject H₀: there is evidence at the 5% level that P(red) is below 0.25

Answer: The expected number is 12.5, so 20 is in the upper tail. For a two-tailed 5% test, compare P(X ≥ 20) = 0.0139 with 0.025: it is smaller, so H₀ is rejected and the conclusion is non-directional. Doubling the tail probability and then also halving the level applies the two-tail adjustment twice.