Critical values/regions; tests for mean/proportion; Type I and II errors practice questions.

In a hypothesis test, a Type I error occurs when you:

• reject a true null hypothesis
• fail to reject a false null hypothesis
• reject a false null hypothesis
• fail to reject a true null hypothesis

Answer: reject a true null hypothesis

A Type I error is rejecting H0 when H0 is actually true; its probability equals the significance level. Failing to reject a false H0 is a Type II error.

A test is conducted at the 5% significance level and gives a p-value of 0.032. State the correct conclusion.

• Accept H0 because 0.032 > 0.05
• Reject H0 because 0.032 < 0.05
• Fail to reject H0 because 0.032 < 0.05
• Reject H1 because 0.032 < 0.05

Answer: Reject H0 because 0.032 < 0.05

When the p-value is less than the significance level, there is sufficient evidence to reject H0. We never 'accept' H0, only fail to reject it.

A sample of size n = 25 is drawn from a normal population with known variance to test H0: μ = 50 against H1: μ ≠ 50 at the 5% level. Which critical z-values bound the rejection region?

• z = ±1.645
• z = ±1.960
• z = ±2.326
• z = ±2.576

Answer: z = ±1.960

For a two-tailed z-test at 5%, each tail holds 2.5%, giving critical values of ±1.960. The value ±1.645 applies to a one-tailed 5% test.

A one-tailed t-test for a mean uses a sample of size n = 15. State the number of degrees of freedom.

• 15
• 16
• 14
• 13

Answer: 14

For a single-sample t-test the degrees of freedom are n − 1 = 15 − 1 = 14.

The power of a hypothesis test is best described as:

• the probability of making a Type I error
• the probability of correctly rejecting a false H₀
• the probability of failing to reject a true null hypothesis
• the significance level of the test

Answer: the probability of correctly rejecting a false H₀

Power = 1 − P(Type II error) = P(reject H0 | H0 false), i.e. the probability of correctly detecting a real effect.

A sample of 40 items has mean 51.5 and the population standard deviation is known to be 4. To test H0: μ = 50 against H1: μ > 50, calculate the test statistic z to 3 s.f.

• 2.37
• 1.50
• 0.375
• 9.49

Answer: 2.37

z = (x̄ − μ)/(σ/√n) = (51.5 − 50)/(4/√40) = 1.5/0.6325 = 2.37 (3 s.f.).