Hypothesis testing Worked Solutions — A-Level Maths

Fully worked, step-by-step solutions to A-Level Hypothesis testing questions. See exactly how to solve problems on hypothesis testing, binomial distribution, p-value, null and alternative hypotheses.

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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.

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 machine historically produces bolts of which 20% are defective. A random sample of 25 bolts is taken; under H0H_0 the number that were defective is modelled by X∼B(25,0.2)X \sim B(25, 0.2). Find P(X≥9)P(X \ge 9), giving your answer to 4 significant figures.

Worked solution

  1. Model the test statistic under H_0

    X∼B(25,0.2)X \sim B(25, 0.2)

    If H_0 is true, X, the number of bolts that were defective, is binomial.

  2. Express the required tail probability as a sum

    P(X≥9)=∑x=925(25x)(0.2)x(0.8)25−xP(X \ge 9) = \sum_{x=9}^{25} \binom{25}{x} (0.2)^{x} (0.8)^{25-x}

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

  3. Evaluate the tail probability

    P(X≥9)=0.04677P(X \ge 9) = 0.04677

    Summing the binomial terms gives the p-value.

Answer
P(X≥9)=0.04677P(X \ge 9) = 0.04677
Question 3
2 markseasy
A coin is believed to be fair. A random sample of 20 tosss is taken; under H0H_0 the number that landed heads is modelled by X∼B(20,0.5)X \sim B(20, 0.5). Find P(X≥14)P(X \ge 14), giving your answer to 4 significant figures.

Worked solution

  1. Model the test statistic under H_0

    X∼B(20,0.5)X \sim B(20, 0.5)

    If H_0 is true, X, the number of tosss that landed heads, is binomial.

  2. Express the required tail probability as a sum

    P(X≥14)=∑x=1420(20x)(0.5)x(0.5)20−xP(X \ge 14) = \sum_{x=14}^{20} \binom{20}{x} (0.5)^{x} (0.5)^{20-x}

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

  3. Evaluate the tail probability

    P(X≥14)=0.05766P(X \ge 14) = 0.05766

    Summing the binomial terms gives the p-value.

Answer
P(X≥14)=0.05766P(X \ge 14) = 0.05766
Question 4
2 markseasy
A production line yields 10% faulty items on average. A random sample of 30 items is taken; under H0H_0 the number that were faulty is modelled by X∼B(30,0.1)X \sim B(30, 0.1). Find P(X≥6)P(X \ge 6), giving your answer to 4 significant figures.

Worked solution

  1. Model the test statistic under H_0

    X∼B(30,0.1)X \sim B(30, 0.1)

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

  2. Express the required tail probability as a sum

    P(X≥6)=∑x=630(30x)(0.1)x(0.9)30−xP(X \ge 6) = \sum_{x=6}^{30} \binom{30}{x} (0.1)^{x} (0.9)^{30-x}

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

  3. Evaluate the tail probability

    P(X≥6)=0.07319P(X \ge 6) = 0.07319

    Summing the binomial terms gives the p-value.

Answer
P(X≥6)=0.07319P(X \ge 6) = 0.07319
Question 5
2 markseasy
Records show that 30% of website visitors sign up for a newsletter. A random sample of 24 visitors is taken; under H0H_0 the number that signed up is modelled by X∼B(24,0.3)X \sim B(24, 0.3). Find P(X≥11)P(X \ge 11), giving your answer to 4 significant figures.

Worked solution

  1. 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.

  2. Express the required tail probability as a sum

    P(X≥11)=∑x=1124(24x)(0.3)x(0.7)24−xP(X \ge 11) = \sum_{x=11}^{24} \binom{24}{x} (0.3)^{x} (0.7)^{24-x}

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

  3. Evaluate the tail probability

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

    Summing the binomial terms gives the p-value.

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

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