Further Maths Further hypothesis tests Practice Questions

Free Further Maths Further hypothesis tests practice questions with full step-by-step worked solutions. Covers choosing-a-test, paired-data, paired-t-test, pooled-variance. Practise exam-style problems and check your method.

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Further Maths70 questionsStep-by-step solutions
Question 1
2 markseasy
Twelve members of a slimming club are each weighed at the start of a six-week course and again at the end of it: the same twelve people are weighed on both occasions. The club wishes to test whether the course changes the mean mass. Each observation in one sample is matched with exactly one observation in the other sample. The claim concerns the means of the two populations. The population variances are unknown. Both populations are assumed to be normal. Which of the following tests is appropriate?
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Worked solution

  1. Decide whether the population variances are known

    the population variances are unknown\text{the population variances are unknown}

    A known population variance allows a zz-test; an unknown one forces a tt-test.

  2. Match the situation to the standard tests

    a matched-pairs design with an unknown variance\text{a matched-pairs design with an unknown variance}

    Each of the standard two-sample tests answers exactly one kind of question.

  3. State the appropriate test and its test statistic

    d=x−y,t=dˉ−μdsd/n,ν=n−1d=x-y,\quad t=\frac{\bar{d}-\mu_d}{s_d/\sqrt{n}},\quad \nu=n-1

    Each 'before' reading is tied to exactly one 'after' reading on the same person, so the two samples are not independent.

Answer
A paired tt-test on the differences d=x−yd=x-y, with ν=n−1\nu=n-1 degrees of freedom
Question 2
2 markseasy
Ten cars are each fitted with one tyre of type AA and one tyre of type BB. After a fixed distance the tread depth, in tenths of a millimetre, of the type AA tyre is recorded as xx and that of the type BB tyre on the same car as yy. The values of xx are x: 33.3, 33.7, 32.3, 32.9, 31.8, 32.8, 32.3, 31.6, 30.9, 36.8x:\ 33.3,\ 33.7,\ 32.3,\ 32.9,\ 31.8,\ 32.8,\ 32.3,\ 31.6,\ 30.9,\ 36.8 and the corresponding values of yy are y: 34.5, 32.8, 33.1, 32.3, 30.3, 32.6, 31.2, 32.7, 31.5, 37.2y:\ 34.5,\ 32.8,\ 33.1,\ 32.3,\ 30.3,\ 32.6,\ 31.2,\ 32.7,\ 31.5,\ 37.2; the iith value of xx and the iith value of yy come from the same car, so that n=10n=10. The observations are naturally paired, so the test is carried out on the differences d=x−yd=x-y. The test statistic is t=dˉ−μdsd/nt=\dfrac{\bar{d}-\mu_d}{s_d/\sqrt{n}}, where sd2s_d^2 is the unbiased variance of the differences, and it has a tt distribution with ν=n−1\nu=n-1 degrees of freedom when H0H_0 is true. The hypotheses are H0:μd=0againstH1:μd≠0H_0:\mu_d=0\quad\text{against}\quad H_1:\mu_d\neq0, where μd\mu_d is the population mean of the differences. The test is carried out at the 5%5\% significance level. This is a two-tailed test, so the critical value is the upper 2.5%2.5\% point tcritt_{\text{crit}} of the null distribution and H0H_0 is rejected if and only if ∣t∣>tcrit|t|>t_{\text{crit}}. Do not round any intermediate value, and give every numerical answer correct to 33 decimal places. Calculate the mean dˉ\bar{d} of the differences, giving your answer to 33 decimal places.
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Worked solution

  1. Form the differences d=x−yd=x-y

    d: −1.2, 0.9, −0.8, 0.6, 1.5, 0.2, 1.1, −1.1, −0.6, −0.4d:\ -1.2,\ 0.9,\ -0.8,\ 0.6,\ 1.5,\ 0.2,\ 1.1,\ -1.1,\ -0.6,\ -0.4

    Pairing removes the between-unit variation, which is exactly why a paired test is more powerful here.

  2. Add the differences

    ∑d=0.2\sum d=0.2

    Differences of opposite sign partly cancel, which is exactly what the test is looking at.

  3. Divide by the number of pairs

    dˉ=0.210\bar{d}=\frac{0.2}{10}

    There are n=10n=10 pairs, not 2n2n observations.

  4. State the mean difference

    dˉ=0.020\bar{d}=0.020

    This estimates the population mean difference μd\mu_d.

Answer
dˉ=0.020\bar{d}=0.020
Question 3
4 marksintermediate
A psychologist recruits 2424 volunteers and allocates them at random, 1212 to a quiet room and 1212 to a noisy room; each volunteer then completes one puzzle and the completion time is recorded, and no volunteer appears in both rooms. The psychologist wishes to compare the mean completion times. No observation in one sample is matched with any particular observation in the other sample. The claim concerns the means of the two populations. The population variances are unknown. Both populations are normal and may be assumed to have a common variance, which must therefore be estimated from the two samples. Which of the following tests is appropriate?
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Worked solution

  1. Identify what is being compared

    the difference of two population MEANS\text{the difference of two population MEANS}

    The parameter in the alternative hypothesis decides the family of test.

  2. Decide whether the observations are paired

    each volunteer contributes exactly one time to exactly one room, so the samples are independent, NOT paired\text{each volunteer contributes exactly one time to exactly one room, so the samples are independent, NOT paired}

    Two samples are paired only if each observation in one is matched with exactly one in the other on the same unit.

  3. Decide whether the population variances are known

    the common variance is unknown and must be estimated by pooling\text{the common variance is unknown and must be estimated by pooling}

    A known population variance allows a zz-test; an unknown one forces a tt-test.

  4. Match the situation to the standard tests

    two independent groups of equal size with an unknown common variance\text{two independent groups of equal size with an unknown common variance}

    Each of the standard two-sample tests answers exactly one kind of question.

  5. Recall the test statistic for a difference of means with known variances

    z=(xˉ1−xˉ2)−(μ1−μ2)σ12n1+σ22n2z=\frac{(\bar{x}_1-\bar{x}_2)-(\mu_1-\mu_2)}{\sqrt{\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}}}

    The variances of independent sample means add, which is why the two terms under the root are added.

  6. Recall the pooled estimate of the common variance

    sp2=(n1−1)s12+(n2−1)s22n1+n2−2s_p^2=\frac{(n_1-1)s_1^2+(n_2-1)s_2^2}{n_1+n_2-2}

    It is the weighted mean of the two unbiased sample variances, each weighted by its own degrees of freedom.

  7. State the appropriate test and its test statistic

    sp2=(n1−1)s12+(n2−1)s22n1+n2−2,ν=n1+n2−2s_p^2=\frac{(n_1-1)s_1^2+(n_2-1)s_2^2}{n_1+n_2-2},\quad \nu=n_1+n_2-2

    Equal sample sizes do NOT make data paired: pairing requires each observation in one sample to be matched to one in the other.

Answer
A two-sample (unpaired) tt-test for the difference of the means, using the pooled variance estimate and ν=n1+n2−2\nu=n_1+n_2-2 degrees of freedom
Question 4
6 markshard
Two varieties of wheat, AA and BB, are grown on similar plots. The yield, in kilograms per plot, of variety AA is denoted by xx and the yield of variety BB is denoted by yy. An agronomist wishes to know whether the two varieties have the same mean yield. The observations from the first sample are x: 20.8, 21.5, 22.9, 20.9, 20.8, 21.2, 21.7, 19.2x:\ 20.8,\ 21.5,\ 22.9,\ 20.9,\ 20.8,\ 21.2,\ 21.7,\ 19.2 and those from the second sample are y: 19.6, 21.7, 21.5, 22.5, 22.4, 23.5, 19.6, 19.1, 21.3, 20.4y:\ 19.6,\ 21.7,\ 21.5,\ 22.5,\ 22.4,\ 23.5,\ 19.6,\ 19.1,\ 21.3,\ 20.4, so that n1=8n_1=8 and n2=10n_2=10. The two samples are independent random samples. The two populations are normally distributed with the same unknown variance, which is estimated by the pooled estimate sp2=(n1−1)s12+(n2−1)s22n1+n2−2s_p^2=\dfrac{(n_1-1)s_1^2+(n_2-1)s_2^2}{n_1+n_2-2}, where s12s_1^2 and s22s_2^2 are the unbiased sample variances. The test statistic is t=(xˉ−yˉ)−(μ1−μ2)sp1n1+1n2t=\dfrac{(\bar{x}-\bar{y})-(\mu_1-\mu_2)}{s_p\sqrt{\dfrac{1}{n_1}+\dfrac{1}{n_2}}}, which has a tt distribution with ν=n1+n2−2\nu=n_1+n_2-2 degrees of freedom when H0H_0 is true. The hypotheses are H0:μ1−μ2=0againstH1:μ1−μ2≠0H_0:\mu_1-\mu_2=0\quad\text{against}\quad H_1:\mu_1-\mu_2\neq0. The test is carried out at the 5%5\% significance level. This is a two-tailed test, so the critical value is the upper 2.5%2.5\% point tcritt_{\text{crit}} of the null distribution and H0H_0 is rejected if and only if ∣t∣>tcrit|t|>t_{\text{crit}}. Do not round any intermediate value, and give every numerical answer correct to 33 decimal places. Which of the following correctly states the value of the test statistic, the critical value and the conclusion of this test in context?
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Worked solution

  1. Find both sample means

    xˉ=21.125,yˉ=21.16\bar{x}=21.125,\qquad \bar{y}=21.16

    Each mean is the sum of that sample's values divided by its size.

  2. Find both unbiased sample variances

    s12=1.0850,s22=2.1027s_1^2=1.0850,\qquad s_2^2=2.1027

    Each is ∑x2−nxˉ2n−1\frac{\sum x^2-n\bar{x}^2}{n-1}, computed from the raw data.

  3. Find the pooled estimate of the common variance

    sp2=(8−1)×1.0850+(10−1)×2.10278+10−2=1.6574s_p^2=\frac{(8-1)\times1.0850+(10-1)\times2.1027}{8+10-2}=1.6574

    The two sample variances are combined, each weighted by its own degrees of freedom.

  4. Find the standard error of xˉ−yˉ\bar{x}-\bar{y}

    sp1n1+1n2=1.6574×18+110=0.6107s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}=\sqrt{1.6574}\times\sqrt{\frac{1}{8}+\frac{1}{10}}=0.6107

    This is the estimated standard deviation of the difference of the two sample means.

  5. Substitute into the test statistic

    t=(21.125−21.16)−00.6107t=\frac{(21.125-21.16)-0}{0.6107}

    The hypothesised difference μ1−μ2\mu_1-\mu_2 is subtracted before dividing by the standard error.

  6. Evaluate the test statistic

    t=−0.057t=-0.057

    This is the observed value of tt.

  7. Find the critical value

    tcrit=t2.5%(16)=2.120t_{\text{crit}}=t_{2.5\%}(16)=2.120

    This is the upper 2.5%2.5\% point of the tt distribution with ν=16\nu=16.

  8. Compare the statistic with the critical value

    ∣t∣=0.057<2.120=tcrit|t|=0.057<2.120=t_{\text{crit}}

    For a two-tailed test the MODULUS of the statistic is compared with the critical value.

  9. Recall the test statistic for a difference of means with known variances

    z=(xˉ1−xˉ2)−(μ1−μ2)σ12n1+σ22n2z=\frac{(\bar{x}_1-\bar{x}_2)-(\mu_1-\mu_2)}{\sqrt{\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}}}

    The variances of independent sample means add, which is why the two terms under the root are added.

  10. Recall the pooled estimate of the common variance

    sp2=(n1−1)s12+(n2−1)s22n1+n2−2s_p^2=\frac{(n_1-1)s_1^2+(n_2-1)s_2^2}{n_1+n_2-2}

    It is the weighted mean of the two unbiased sample variances, each weighted by its own degrees of freedom.

  11. State the conclusion in context

    Do not reject H0: there is insufficient evidence at the 5% level of a difference between the mean yields of the two varieties\text{Do not reject }H_0\text{: there is insufficient evidence at the 5\% level of a difference between the mean yields of the two varieties}

    A conclusion is worth nothing unless it is expressed in the language of the original question.

Answer
ν=16,  t=−0.057,  tcrit=2.120,  so ∣t∣<tcrit: do not reject H0; there is insufficient evidence at the 5% level of a difference between the mean yields of the two varieties\nu=16,\;t=-0.057,\;t_{\text{crit}}=2.120,\;\text{so }|t|<t_{\text{crit}}\text{: do not reject }H_0\text{; there is insufficient evidence at the 5\% level of a difference between the mean yields of the two varieties}
Question 5
9 markschallenging
Ten examination scripts are each marked independently by two examiners. The mark given by examiner AA is xx and the mark given by examiner BB to the same script is yy. The values of xx are x: 61.7, 60.5, 57.9, 56.2, 61.8, 66.7, 59.2, 66.1, 59.0, 61.8x:\ 61.7,\ 60.5,\ 57.9,\ 56.2,\ 61.8,\ 66.7,\ 59.2,\ 66.1,\ 59.0,\ 61.8 and the corresponding values of yy are y: 63.3, 58.5, 57.4, 54.3, 62.6, 65.6, 58.1, 66.1, 58.0, 61.4y:\ 63.3,\ 58.5,\ 57.4,\ 54.3,\ 62.6,\ 65.6,\ 58.1,\ 66.1,\ 58.0,\ 61.4; the iith value of xx and the iith value of yy come from the same script, so that n=10n=10. The observations are naturally paired, so the test is carried out on the differences d=x−yd=x-y. The test statistic is t=dˉ−μdsd/nt=\dfrac{\bar{d}-\mu_d}{s_d/\sqrt{n}}, where sd2s_d^2 is the unbiased variance of the differences, and it has a tt distribution with ν=n−1\nu=n-1 degrees of freedom when H0H_0 is true. The hypotheses are H0:μd=0againstH1:μd≠0H_0:\mu_d=0\quad\text{against}\quad H_1:\mu_d\neq0, where μd\mu_d is the population mean of the differences. The test is carried out at the 10%10\% significance level. This is a two-tailed test, so the critical value is the upper 5%5\% point tcritt_{\text{crit}} of the null distribution and H0H_0 is rejected if and only if ∣t∣>tcrit|t|>t_{\text{crit}}. Do not round any intermediate value, and give every numerical answer correct to 33 decimal places. Calculate the value of the paired tt test statistic, giving your answer to 33 decimal places.
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Worked solution

  1. State the hypotheses

    H0:μd=0againstH1:μd≠0H_0:\mu_d=0\quad\text{against}\quad H_1:\mu_d\neq0

    The null hypothesis always states equality; the alternative encodes the claim being tested.

  2. State the modelling assumption

    D=X−Y∼N(μd,σd2)D=X-Y\sim N(\mu_d,\sigma_d^2)

    The differences are assumed to be a random sample from a normal population.

  3. Form the differences d=x−yd=x-y

    d: −1.6, 2.0, 0.5, 1.9, −0.8, 1.1, 1.1, 0.0, 1.0, 0.4d:\ -1.6,\ 2.0,\ 0.5,\ 1.9,\ -0.8,\ 1.1,\ 1.1,\ 0.0,\ 1.0,\ 0.4

    Pairing removes the between-unit variation, which is exactly why a paired test is more powerful here.

  4. Find ∑d\sum d and ∑d2\sum d^2

    ∑d=5.6,∑d2=14.64\sum d=5.6,\qquad \sum d^2=14.64

    The two sums are all that is needed for the mean and the variance of the differences.

  5. Find the mean of the differences

    dˉ=5.610=0.56\bar{d}=\frac{5.6}{10}=0.56

    This estimates the population mean difference μd\mu_d.

  6. Find SddS_{dd}

    Sdd=∑d2−ndˉ2=14.64−10×(0.56)2=11.5040S_{dd}=\sum d^2-n\bar{d}^2=14.64-10\times(0.56)^2=11.5040

    This is the sum of the squared deviations of the differences about their mean.

  7. Find the unbiased variance of the differences

    sd2=Sddn−1=11.50409=1.2782s_d^2=\frac{S_{dd}}{n-1}=\frac{11.5040}{9}=1.2782

    Only ONE sample (the differences) is used, so only one degree of freedom is lost.

  8. Find the standard deviation of the differences

    sd=1.2782=1.1306s_d=\sqrt{1.2782}=1.1306

    The square root of the unbiased variance estimates σd\sigma_d.

  9. Find the number of degrees of freedom

    ν=n−1=10−1=9\nu=n-1=10-1=9

    The paired test is a one-sample test on the nn differences, so ν=n−1\nu=n-1.

  10. Find the standard error of dˉ\bar{d}

    sdn=1.130610=0.3575\frac{s_d}{\sqrt{n}}=\frac{1.1306}{\sqrt{10}}=0.3575

    The mean of nn differences has standard error sd/ns_d/\sqrt{n}.

  11. Substitute into the test statistic

    t=0.56−00.3575t=\frac{0.56-0}{0.3575}

    The hypothesised mean difference is subtracted from dˉ\bar{d} before dividing by the standard error.

  12. Recall the test statistic for a difference of means with known variances

    z=(xˉ1−xˉ2)−(μ1−μ2)σ12n1+σ22n2z=\frac{(\bar{x}_1-\bar{x}_2)-(\mu_1-\mu_2)}{\sqrt{\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}}}

    The variances of independent sample means add, which is why the two terms under the root are added.

  13. Recall the pooled estimate of the common variance

    sp2=(n1−1)s12+(n2−1)s22n1+n2−2s_p^2=\frac{(n_1-1)s_1^2+(n_2-1)s_2^2}{n_1+n_2-2}

    It is the weighted mean of the two unbiased sample variances, each weighted by its own degrees of freedom.

  14. Recall the pooled two-sample tt statistic

    t=(xˉ−yˉ)−(μ1−μ2)sp1n1+1n2t=\frac{(\bar{x}-\bar{y})-(\mu_1-\mu_2)}{s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}

    The pooled standard deviation replaces the unknown common σ\sigma.

  15. Recall the degrees of freedom of a pooled two-sample tt-test

    ν=n1+n2−2\nu=n_1+n_2-2

    One degree of freedom is lost for each of the two sample means used in the variance estimate.

  16. Evaluate the test statistic

    t=1.566t=1.566

    This is tt correct to 33 decimal places, on ν=9\nu=9 degrees of freedom.

Answer
t=1.566t=1.566

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