A-Level Hypothesis testing Practice Questions

Free A-Level Hypothesis testing practice questions with full step-by-step worked solutions. Covers hypothesis testing, binomial distribution, p-value, null and alternative hypotheses. Practise exam-style problems and check your method.

hypothesis testingbinomial distributionp-valuenull and alternative hypothesesconclusion in contextcritical region
A-Level70 questionsStep-by-step solutions
Question 1
2 markseasy
A seed supplier states that 40% of its seeds germinate. A random sample of 20 seeds is taken; under H0H_0 the number that germinated is modelled by X∼B(20,0.4)X \sim B(20, 0.4). Find P(X≥12)P(X \ge 12), giving your answer to 4 significant figures.
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Worked solution

  1. Model the test statistic under H_0

    X∼B(20,0.4)X \sim B(20, 0.4)

    If H_0 is true, X, the number of seeds that germinated, is binomial.

  2. Express the required tail probability as a sum

    P(X≥12)=∑x=1220(20x)(0.4)x(0.6)20−xP(X \ge 12) = \sum_{x=12}^{20} \binom{20}{x} (0.4)^{x} (0.6)^{20-x}

    The p-value is the probability of the observed result or something more extreme.

  3. Evaluate the tail probability

    P(X≥12)=0.05653P(X \ge 12) = 0.05653

    Summing the binomial terms gives the p-value.

Answer
P(X≥12)=0.05653P(X \ge 12) = 0.05653
Question 2
2 markseasy
A poll suggests that 45% of residents support a plan. A test is carried out to see whether the proportion has increased. Taking pp as the proportion who support the plan, which is the correct alternative hypothesis?
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Worked solution

  1. Define the parameter under test

    p=the proportion who support the planp = \text{the proportion who support the plan}

    Let p be the proportion who support the plan.

  2. State the null hypothesis

    H0:p=0.45H_0: p = 0.45

    Under the null hypothesis the proportion is unchanged from the stated value.

  3. State the alternative hypothesis

    H1:p>0.45H_1: p > 0.45

    We test whether the proportion has increased.

Answer
H1:p>0.45H_1: p > 0.45
Question 3
3 marksintermediate
A basketball player historically scores 60% of free throws. A test at the 5% significance level uses a sample of 25 throws of which 20 were scored. Under H0H_0, X∼B(25,0.6)X \sim B(25, 0.6) and P(X≥20)=0.02936P(X \ge 20) = 0.02936. Which conclusion is correct?
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Worked solution

  1. State the null hypothesis

    H0:p=0.6H_0: p = 0.6

    Under the null hypothesis the proportion is unchanged from the stated value.

  2. State the alternative hypothesis

    H1:p>0.6H_1: p > 0.6

    We test whether the proportion has increased.

  3. Model the test statistic under H_0

    X∼B(25,0.6)X \sim B(25, 0.6)

    If H_0 is true, X, the number of throws that were scored, is binomial.

  4. Evaluate the tail probability

    P(X≥20)=0.02936P(X \ge 20) = 0.02936

    Summing the binomial terms gives the p-value.

  5. Compare the p-value with the significance level

    0.02936≤0.050.02936 \le 0.05

    The p-value is at most alpha.

  6. State the statistical decision

    reject H0\text{reject } H_0

    Given the comparison, we reject the null hypothesis.

Answer
Reject H0H_0: there is significant evidence at the 5% level that the proportion has increased.
Question 4
5 markshard
Records show that 30% of website visitors sign up for a newsletter. A one-tailed test at the 5% level has critical region X≥10X \ge 10, where X∼B(20,0.3)X \sim B(20, 0.3) under H0H_0. What does this critical region represent?
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Worked solution

  1. Define the parameter under test

    p=the proportion of visitors who sign upp = \text{the proportion of visitors who sign up}

    Let p be the proportion of visitors who sign up.

  2. State the null hypothesis

    H0:p=0.3H_0: p = 0.3

    Under the null hypothesis the proportion is unchanged from the stated value.

  3. State the alternative hypothesis

    H1:p>0.3H_1: p > 0.3

    We test whether the proportion has increased.

  4. State the significance level

    α=0.05\alpha = 0.05

    The test is carried out at the 5% significance level.

  5. Model the test statistic under H_0

    X∼B(20,0.3)X \sim B(20, 0.3)

    If H_0 is true, X, the number of visitors that signed up, is binomial.

  6. Describe the critical region for the test

    Reject H0 if X≥c\text{Reject } H_0 \text{ if } X \ge c

    The critical region is the set of outcomes that would lead to rejecting H_0.

  7. Test the boundary at c = 10

    P(X≥10)=0.04796P(X \ge 10) = 0.04796

    Check whether this upper-tail probability is at most alpha.

  8. Identify the critical value

    c=10c = 10

    The critical value is the smallest c whose upper-tail probability is at most alpha.

  9. State the critical region

    X≥10X \ge 10

    Observations in this region lead to rejecting H_0.

  10. Find the actual significance level

    P(X≥10)=0.04796P(X \ge 10) = 0.04796

    The actual significance level is the probability of landing in the critical region when H_0 is true.

Answer
The set of values of the test statistic for which H0H_0 is rejected.
Question 5
8 markschallenging
A basketball player historically scores 60% of free throws. A one-tailed test at the 5% level uses critical region X≥20X \ge 20, where X∼B(25,0.6)X \sim B(25, 0.6) under H0H_0; the actual significance level is 0.029360.02936. A sample gives 21 throws that were scored. Which statement correctly interprets the test?
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Worked solution

  1. Define the parameter under test

    p=the player’s free-throw success proportionp = \text{the player's free-throw success proportion}

    Let p be the player's free-throw success proportion.

  2. State the null hypothesis

    H0:p=0.6H_0: p = 0.6

    Under the null hypothesis the proportion is unchanged from the stated value.

  3. State the alternative hypothesis

    H1:p>0.6H_1: p > 0.6

    We test whether the proportion has increased.

  4. State the significance level

    α=0.05\alpha = 0.05

    The test is carried out at the 5% significance level.

  5. Model the test statistic under H_0

    X∼B(25,0.6)X \sim B(25, 0.6)

    If H_0 is true, X, the number of throws that were scored, is binomial.

  6. Find the expected number under H_0

    E(X)=np=25×0.6=15E(X) = np = 25 \times 0.6 = 15

    This shows where the sample sits relative to expectation.

  7. Describe the critical region for the test

    Reject H0 if X≥c\text{Reject } H_0 \text{ if } X \ge c

    The critical region is the set of outcomes that would lead to rejecting H_0.

  8. Test the boundary at c = 19

    P(X≥19)=0.07357P(X \ge 19) = 0.07357

    Check whether this upper-tail probability is at most alpha.

  9. Test the boundary at c = 20

    P(X≥20)=0.02936P(X \ge 20) = 0.02936

    Check whether this upper-tail probability is at most alpha.

  10. Identify the critical value

    c=20c = 20

    The critical value is the smallest c whose upper-tail probability is at most alpha.

  11. State the critical region

    X≥20X \ge 20

    Observations in this region lead to rejecting H_0.

  12. Find the actual significance level

    P(X≥20)=0.02936P(X \ge 20) = 0.02936

    The actual significance level is the probability of landing in the critical region when H_0 is true.

  13. Check whether the observation lies in the critical region

    21≥2021 \ge 20

    The observed value lies in the critical region.

  14. State the statistical decision

    reject H0\text{reject } H_0

    Given the comparison, we reject the null hypothesis.

  15. Comment on the actual significance level

    0.02936≤0.050.02936 \le 0.05

    Because X is discrete the actual significance level is at most the nominal level.

Answer
The actual significance level is 0.029360.02936, the probability of rejecting H0H_0 when it is true, and H0H_0 is rejected only if XX lies in the critical region.

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