Hard A-Level Hypothesis testing (normal) Questions

Challenging, exam-style A-Level Hypothesis testing (normal) questions with worked solutions. Stretch yourself on the hardest z-test, test-statistic, critical-region, p-value problems.

z-testtest-statisticcritical-regionp-valuecritical-valueconclusion
A-Level34 questionsStep-by-step solutions
Question 1
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?
Show worked solution

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
Question 2
8 markschallenging
A pharmaceutical batch should have a mean active-ingredient content of 200200 mg. A regulator wants to detect any deviation from this target in either direction. Which hypotheses should be used?
Show worked solution

Worked solution

  1. Identify the population parameter under test

    μ=200 (claimed mean)\mu=200\ (\text{claimed mean})

    The null hypothesis fixes the value of the population mean.

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

    two-tailed test\text{two-tailed test}

    The wording tells us the direction of the alternative hypothesis.

  3. Write the null hypothesis

    H0 ⁣:μ=200H_0\!:\mu=200

    The null hypothesis always states equality with the claimed value.

  4. Write the alternative hypothesis

    H1 ⁣:μ≠200H_1\!:\mu\neq200

    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=200, H1 ⁣:μ≠200H_0\!:\mu=200,\ H_1\!:\mu\neq200

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

Answer
H0 ⁣:μ=200, H1 ⁣:μ≠200H_0\!:\mu=200,\ H_1\!:\mu\neq200
Question 3
8 markschallenging
The population standard deviation is σ=1.2\sigma=1.2 and is known. Under H0:μ=7.5H_0:\mu=7.5 a random sample of size 4848 has sample mean xˉ=7.05\bar{x}=7.05. Calculate the test statistic z=xˉ−μ0σ/nz=\dfrac{\bar{x}-\mu_0}{\sigma/\sqrt{n}}, giving your answer to 3 decimal places.
Show worked solution

Worked solution

  1. State the distribution of the sample mean under H0

    Xˉ∼N ⁣(7.5, 1.4448)\bar{X}\sim N\!\left(7.5,\ \dfrac{1.44}{48}\right)

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

  2. Evaluate the standard error

    σn=1.248=0.1732\dfrac{\sigma}{\sqrt{n}}=\dfrac{1.2}{\sqrt{48}}=0.1732

    The denominator of the test statistic is the standard error.

  3. Write the standardising formula

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

    Standardise the observed sample mean under \(H_0\).

  4. Substitute the values

    z=7.05−7.51.2/48z=\dfrac{7.05-7.5}{1.2/\sqrt{48}}

    Insert the sample mean, hypothesised mean, \(\sigma\) and \(n\).

  5. Compute the numerator

    xˉ−μ0=−0.45\bar{x}-\mu_0=-0.45

    This measures how far the sample mean is from \(\mu_0\).

  6. Divide by the standard error

    −0.450.1732=−2.598\dfrac{-0.45}{0.1732}=-2.598

    Dividing the difference by the standard error gives \(z\).

  7. 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 \(\sigma^2/n\).

  8. Recall the mean of the sample mean

    E(Xˉ)=μE(\bar{X})=\mu

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

  9. 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.

  10. 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.

  11. 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.

  12. 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.

  13. 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).

  14. 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.

  15. State the value of the test statistic

    z=7.05−7.50.1732=−2.598z=\dfrac{7.05-7.5}{0.1732}=-2.598

    This is the test statistic to 3 decimal places.

Answer
−2.598-2.598
Question 4
8 markschallenging
The population standard deviation is σ=250\sigma=250 and is known. Under H0:μ=2000H_0:\mu=2000 a random sample of size 5050 has sample mean xˉ=2085\bar{x}=2085. Calculate the test statistic z=xˉ−μ0σ/nz=\dfrac{\bar{x}-\mu_0}{\sigma/\sqrt{n}}, giving your answer to 3 decimal places.
Show worked solution

Worked solution

  1. State the distribution of the sample mean under H0

    Xˉ∼N ⁣(2000, 6250050)\bar{X}\sim N\!\left(2000,\ \dfrac{62500}{50}\right)

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

  2. Evaluate the standard error

    σn=25050=35.3553\dfrac{\sigma}{\sqrt{n}}=\dfrac{250}{\sqrt{50}}=35.3553

    The denominator of the test statistic is the standard error.

  3. Write the standardising formula

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

    Standardise the observed sample mean under \(H_0\).

  4. Substitute the values

    z=2085−2000250/50z=\dfrac{2085-2000}{250/\sqrt{50}}

    Insert the sample mean, hypothesised mean, \(\sigma\) and \(n\).

  5. Compute the numerator

    xˉ−μ0=85\bar{x}-\mu_0=85

    This measures how far the sample mean is from \(\mu_0\).

  6. Divide by the standard error

    8535.3553=2.404\dfrac{85}{35.3553}=2.404

    Dividing the difference by the standard error gives \(z\).

  7. 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 \(\sigma^2/n\).

  8. Recall the mean of the sample mean

    E(Xˉ)=μE(\bar{X})=\mu

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

  9. 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.

  10. 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.

  11. 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.

  12. 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.

  13. 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).

  14. 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.

  15. State the value of the test statistic

    z=2085−200035.3553=2.404z=\dfrac{2085-2000}{35.3553}=2.404

    This is the test statistic to 3 decimal places.

Answer
2.4042.404
Question 5
8 markschallenging
A normal population has known standard deviation σ=12\sigma=12. To test H0:μ=60H_0:\mu=60 against H1:μ≠60H_1:\mu\neq60 at the 2%2\% level, a random sample of size 2525 is used. Find the critical region for the sample mean xˉ\bar{x}, giving boundaries to 4 decimal places.
Show worked solution

Worked solution

  1. State the distribution of the sample mean under H0

    Xˉ∼N ⁣(60, 14425)\bar{X}\sim N\!\left(60,\ \dfrac{144}{25}\right)

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

  2. Evaluate the standard error

    σn=1225=2.4000\dfrac{\sigma}{\sqrt{n}}=\dfrac{12}{\sqrt{25}}=2.4000

    This scales z-values onto the sample-mean axis.

  3. Find the critical value of the test statistic

    zc=±2.3263z_c=\pm 2.3263

    This is the boundary z-value for the chosen significance level.

  4. Convert the critical z back to the sample-mean scale

    xˉc=μ0+zc σn\bar{x}_c=\mu_0+z_c\,\dfrac{\sigma}{\sqrt{n}}

    Rearranging the standardising formula gives the boundary.

  5. Substitute to find the boundary

    xˉc=60±2.3263×2.4000\bar{x}_c=60\pm 2.3263\times 2.4000

    This gives the boundary of the critical region.

  6. 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 \(\sigma^2/n\).

  7. Recall the mean of the sample mean

    E(Xˉ)=μE(\bar{X})=\mu

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

  8. 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.

  9. 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.

  10. 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.

  11. 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.

  12. 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).

  13. 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.

  14. 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.

  15. State the critical region for the sample mean

    xˉ<54.4168 or xˉ>65.5832\bar{x} < 54.4168 \ \text{or}\ \bar{x} > 65.5832

    The sample mean leads to rejection of \(H_0\) inside this region.

Answer
xˉ<54.4168 or xˉ>65.5832\bar{x} < 54.4168 \ \text{or}\ \bar{x} > 65.5832

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