Hard A-Level Hypothesis testing Questions

Challenging, exam-style A-Level Hypothesis testing questions with worked solutions. Stretch yourself on the hardest hypothesis testing, binomial distribution, p-value, critical region problems.

hypothesis testingbinomial distributionp-valuecritical regionconclusion in context
A-Level34 questionsStep-by-step solutions
Question 1
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?
Show worked solution

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.
Question 2
8 markschallenging
A production line yields 10% faulty items on average. A one-tailed test at the 5% level uses critical region X≥10X \ge 10, where X∼B(50,0.1)X \sim B(50, 0.1) under H0H_0; the actual significance level is 0.024540.02454. A sample gives 10 items that were faulty. Which statement correctly interprets the test?
Show worked solution

Worked solution

  1. Define the parameter under test

    p=the proportion of faulty itemsp = \text{the proportion of faulty items}

    Let p be the proportion of faulty items.

  2. State the null hypothesis

    H0:p=0.1H_0: p = 0.1

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

  3. State the alternative hypothesis

    H1:p>0.1H_1: p > 0.1

    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(50,0.1)X \sim B(50, 0.1)

    If H_0 is true, X, the number of items that were faulty, is binomial.

  6. Find the expected number under H_0

    E(X)=np=50×0.1=5E(X) = np = 50 \times 0.1 = 5

    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 = 9

    P(X≥9)=0.05787P(X \ge 9) = 0.05787

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

  9. Test the boundary at c = 10

    P(X≥10)=0.02454P(X \ge 10) = 0.02454

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

  10. Identify the critical value

    c=10c = 10

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

  11. State the critical region

    X≥10X \ge 10

    Observations in this region lead to rejecting H_0.

  12. Find the actual significance level

    P(X≥10)=0.02454P(X \ge 10) = 0.02454

    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

    10≥1010 \ge 10

    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

    0.02454≤0.050.02454 \le 0.05

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

Answer
The actual significance level is 0.024540.02454, the probability of rejecting H0H_0 when it is true, and H0H_0 is rejected only if XX lies in the critical region.
Question 3
8 markschallenging
A standard treatment cures 25% of patients. A one-tailed test at the 5% level uses critical region X≥13X \ge 13, where X∼B(30,0.25)X \sim B(30, 0.25) under H0H_0; the actual significance level is 0.021590.02159. A sample gives 13 patients that were cured. Which statement correctly interprets the test?
Show worked solution

Worked solution

  1. Define the parameter under test

    p=the proportion of patients curedp = \text{the proportion of patients cured}

    Let p be the proportion of patients cured.

  2. State the null hypothesis

    H0:p=0.25H_0: p = 0.25

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

  3. State the alternative hypothesis

    H1:p>0.25H_1: p > 0.25

    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(30,0.25)X \sim B(30, 0.25)

    If H_0 is true, X, the number of patients that were cured, is binomial.

  6. Find the expected number under H_0

    E(X)=np=30×0.25=7.5E(X) = np = 30 \times 0.25 = 7.5

    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 = 12

    P(X≥12)=0.05066P(X \ge 12) = 0.05066

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

  9. Test the boundary at c = 13

    P(X≥13)=0.02159P(X \ge 13) = 0.02159

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

  10. Identify the critical value

    c=13c = 13

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

  11. State the critical region

    X≥13X \ge 13

    Observations in this region lead to rejecting H_0.

  12. Find the actual significance level

    P(X≥13)=0.02159P(X \ge 13) = 0.02159

    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

    13≥1313 \ge 13

    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

    0.02159≤0.050.02159 \le 0.05

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

Answer
The actual significance level is 0.021590.02159, the probability of rejecting H0H_0 when it is true, and H0H_0 is rejected only if XX lies in the critical region.
Question 4
8 markschallenging
Records show that 30% of website visitors sign up for a newsletter. A one-tailed test at the 5% level uses critical region X≥12X \ge 12, where X∼B(24,0.3)X \sim B(24, 0.3) under H0H_0; the actual significance level is 0.031390.03139. A sample gives 13 visitors that signed up. Which statement correctly interprets the test?
Show worked solution

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(24,0.3)X \sim B(24, 0.3)

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

  6. Find the expected number under H_0

    E(X)=np=24×0.3=7.2E(X) = np = 24 \times 0.3 = 7.2

    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 = 11

    P(X≥11)=0.07424P(X \ge 11) = 0.07424

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

  9. Test the boundary at c = 12

    P(X≥12)=0.03139P(X \ge 12) = 0.03139

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

  10. Identify the critical value

    c=12c = 12

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

  11. State the critical region

    X≥12X \ge 12

    Observations in this region lead to rejecting H_0.

  12. Find the actual significance level

    P(X≥12)=0.03139P(X \ge 12) = 0.03139

    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

    13≥1213 \ge 12

    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

    0.03139≤0.050.03139 \le 0.05

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

Answer
The actual significance level is 0.031390.03139, the probability of rejecting H0H_0 when it is true, and H0H_0 is rejected only if XX lies in the critical region.
Question 5
8 markschallenging
A seed supplier states that 40% of its seeds germinate. A one-tailed test at the 5% level uses critical region X≥13X \ge 13, where X∼B(20,0.4)X \sim B(20, 0.4) under H0H_0; the actual significance level is 0.021030.02103. A sample gives 14 seeds that germinated. Which statement correctly interprets the test?
Show worked solution

Worked solution

  1. Define the parameter under test

    p=the proportion of seeds that germinatep = \text{the proportion of seeds that germinate}

    Let p be the proportion of seeds that germinate.

  2. State the null hypothesis

    H0:p=0.4H_0: p = 0.4

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

  3. State the alternative hypothesis

    H1:p>0.4H_1: p > 0.4

    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(20,0.4)X \sim B(20, 0.4)

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

  6. Find the expected number under H_0

    E(X)=np=20×0.4=8E(X) = np = 20 \times 0.4 = 8

    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 = 12

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

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

  9. Test the boundary at c = 13

    P(X≥13)=0.02103P(X \ge 13) = 0.02103

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

  10. Identify the critical value

    c=13c = 13

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

  11. State the critical region

    X≥13X \ge 13

    Observations in this region lead to rejecting H_0.

  12. Find the actual significance level

    P(X≥13)=0.02103P(X \ge 13) = 0.02103

    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

    14≥1314 \ge 13

    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

    0.02103≤0.050.02103 \le 0.05

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

Answer
The actual significance level is 0.021030.02103, 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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