A-Level Hypothesis testing (normal) Practice Questions

Free A-Level Hypothesis testing (normal) practice questions with full step-by-step worked solutions. Covers sampling-distribution, sample-mean, variance, standard-error. Practise exam-style problems and check your method.

sampling-distributionsample-meanvariancestandard-errorz-testtest-statistic
A-Level70 questionsStep-by-step solutions
Question 1
2 markseasy
The random variable X∼N(50, 36)X\sim N(50,\ 36). A random sample of 44 observations is taken. Write down the distribution of the sample mean Xˉ\bar{X}.
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Worked solution

  1. Write down the population distribution

    X∼N(50, 36)X\sim N(50,\ 36)

    The individual observations are normally distributed.

  2. Apply the sample-mean result

    Xˉ∼N ⁣(μ, σ2n)\bar{X}\sim N\!\left(\mu,\ \dfrac{\sigma^2}{n}\right)

    The mean of a normal sample is normal with variance \(\sigma^2/n\).

  3. State the distribution of the sample mean

    Xˉ∼N ⁣(50, 9)\bar{X}\sim N\!\left(50,\ 9\right)

    Substituting \(\mu\), \(\sigma^2\) and \(n\) gives the distribution of \(\bar{X}\).

Answer
Xˉ∼N ⁣(50, 9)\bar{X}\sim N\!\left(50,\ 9\right)
Question 2
2 markseasy
A normal population has known standard deviation σ=4\sigma=4. A researcher tests H0:μ=20H_0:\mu=20 against H1:μ>20H_1:\mu>20 at the 5%5\% level. A random sample of size 1616 gives xˉ=20.5\bar{x}=20.5. State the correct conclusion.
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Worked solution

  1. State the hypotheses

    H0 ⁣:μ=20, H1 ⁣:μ>20H_0\!:\mu=20,\ H_1\!:\mu>20

    Set up the null and alternative hypotheses.

  2. Calculate the test statistic

    z=20.5−204/16=0.500z=\dfrac{20.5-20}{4/\sqrt{16}}=0.500

    Standardise the sample mean under \(H_0\).

  3. State the conclusion of the test

    Do not reject H0\text{Do not reject } H_0

    Do not reject \(H_0\) at the 5% level, based on the comparison above.

Answer
Thereisinsufficientevidenceatthe5There is insufficient evidence at the 5% level to reject H_0.
Question 3
3 marksintermediate
A normal population has known standard deviation σ=20\sigma=20. A test of H0:μ=100H_0:\mu=100 against H1:μ<100H_1:\mu<100 is carried out at the 5%5\% level. A random sample of size 2525 gives xˉ=92\bar{x}=92. State the correct conclusion.
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Worked solution

  1. State the hypotheses

    H0 ⁣:μ=100, H1 ⁣:μ<100H_0\!:\mu=100,\ H_1\!:\mu<100

    Set up the null and alternative hypotheses.

  2. Calculate the test statistic

    z=92−10020/25=−2.000z=\dfrac{92-100}{20/\sqrt{25}}=-2.000

    Standardise the sample mean under \(H_0\).

  3. State the significance level and tail

    α=0.05 (one-tailed (lower))\alpha=0.05\ (\text{one-tailed (lower)})

    The level and tail determine the critical region.

  4. Write the distribution of the sample mean under H0

    Xˉ∼N ⁣(100, 40025)\bar{X}\sim N\!\left(100,\ \dfrac{400}{25}\right)

    Under \(H_0\) the sample mean is normal about \(\mu_0\).

  5. Evaluate the standard error

    σn=2025=4.0000\dfrac{\sigma}{\sqrt{n}}=\dfrac{20}{\sqrt{25}}=4.0000

    This is the denominator of the test statistic.

  6. State the conclusion of the test

    Reject H0\text{Reject } H_0

    Reject \(H_0\) at the 5% level, based on the comparison above.

Answer
Thereissufficientevidenceatthe5There is sufficient evidence at the 5% level to reject H_0.
Question 4
5 markshard
A factory process targets a mean fill of 340340 g. An auditor wants to test whether the mean has increased. Which hypotheses are appropriate?
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Worked solution

  1. Identify the population parameter under test

    μ=340 (claimed mean)\mu=340\ (\text{claimed mean})

    The null hypothesis fixes the value of the population mean.

  2. Decide whether the test is one- or two-tailed

    one-tailed (upper) test\text{one-tailed (upper) test}

    The wording tells us the direction of the alternative hypothesis.

  3. Write the null hypothesis

    H0 ⁣:μ=340H_0\!:\mu=340

    The null hypothesis always states equality with the claimed value.

  4. Write the alternative hypothesis

    H1 ⁣:μ>340H_1\!:\mu>340

    The alternative captures the suspected change in the mean.

  5. Recall the distribution of the sample mean

    Xˉ∼N ⁣(μ, σ2n)\bar{X}\sim N\!\left(\mu,\ \dfrac{\sigma^2}{n}\right)

    For a normal population the sample mean is normal with variance \(\sigma^2/n\).

  6. Recall the mean of the sample mean

    E(Xˉ)=μE(\bar{X})=\mu

    The sample mean is an unbiased estimator of the population mean.

  7. Recall the standard error

    s.e.=σn\text{s.e.}=\dfrac{\sigma}{\sqrt{n}}

    The standard deviation of the sample mean is called the standard error.

  8. Recall the test statistic

    z=xˉ−μ0σ/nz=\dfrac{\bar{x}-\mu_0}{\sigma/\sqrt{n}}

    Standardising the sample mean under \(H_0\) gives a z statistic.

  9. Note the statistic is standard normal under the null hypothesis

    z∼N(0,1)z\sim N(0,1)

    Under \(H_0\) the standardised sample mean follows the standard normal distribution.

  10. State the correct hypotheses

    H0 ⁣:μ=340, H1 ⁣:μ>340H_0\!:\mu=340,\ H_1\!:\mu>340

    This pair matches the claim and the suspected departure from it.

Answer
H0 ⁣:μ=340, H1 ⁣:μ>340H_0\!:\mu=340,\ H_1\!:\mu>340
Question 5
8 markschallenging
A new teaching method is claimed to raise the mean test score above the historical mean of 6565 marks. A researcher wishes to test this claim. Which hypotheses should be used?
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Worked solution

  1. Identify the population parameter under test

    μ=65 (claimed mean)\mu=65\ (\text{claimed mean})

    The null hypothesis fixes the value of the population mean.

  2. Decide whether the test is one- or two-tailed

    one-tailed (upper) test\text{one-tailed (upper) test}

    The wording tells us the direction of the alternative hypothesis.

  3. Write the null hypothesis

    H0 ⁣:μ=65H_0\!:\mu=65

    The null hypothesis always states equality with the claimed value.

  4. Write the alternative hypothesis

    H1 ⁣:μ>65H_1\!:\mu>65

    The alternative captures the suspected change in the mean.

  5. Recall the distribution of the sample mean

    Xˉ∼N ⁣(μ, σ2n)\bar{X}\sim N\!\left(\mu,\ \dfrac{\sigma^2}{n}\right)

    For a normal population the sample mean is normal with variance \(\sigma^2/n\).

  6. Recall the mean of the sample mean

    E(Xˉ)=μE(\bar{X})=\mu

    The sample mean is an unbiased estimator of the population mean.

  7. Recall the standard error

    s.e.=σn\text{s.e.}=\dfrac{\sigma}{\sqrt{n}}

    The standard deviation of the sample mean is called the standard error.

  8. Recall the test statistic

    z=xˉ−μ0σ/nz=\dfrac{\bar{x}-\mu_0}{\sigma/\sqrt{n}}

    Standardising the sample mean under \(H_0\) gives a z statistic.

  9. Note the statistic is standard normal under the null hypothesis

    z∼N(0,1)z\sim N(0,1)

    Under \(H_0\) the standardised sample mean follows the standard normal distribution.

  10. Recall how to read a critical value

    P(Z>zc)=α ⇒ zc=Φ−1(1−α)P(Z>z_c)=\alpha\ \Rightarrow\ z_c=\Phi^{-1}(1-\alpha)

    Critical values come from the inverse normal function.

  11. Recall the decision rule

    reject H0 when the statistic lies in the critical region\text{reject } H_0 \text{ when the statistic lies in the critical region}

    We compare the test statistic (or p-value) with the critical value (or alpha).

  12. Recall the meaning of the p-value

    p=P(result as extreme as observed∣H0)p=P(\text{result as extreme as observed}\mid H_0)

    The p-value measures how surprising the data are if \(H_0\) is true.

  13. Recall the p-value decision rule

    reject H0  ⟺  p<α\text{reject } H_0 \iff p<\alpha

    A p-value below the significance level indicates a significant result.

  14. State the known-variance assumption

    σ is known\sigma \text{ is known}

    A z-test is appropriate because the population standard deviation is known.

  15. State the correct hypotheses

    H0 ⁣:μ=65, H1 ⁣:μ>65H_0\!:\mu=65,\ H_1\!:\mu>65

    This pair matches the claim and the suspected departure from it.

Answer
H0 ⁣:μ=65, H1 ⁣:μ>65H_0\!:\mu=65,\ H_1\!:\mu>65

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