Hard Further Maths Further hypothesis tests Questions

Challenging, exam-style Further Maths Further hypothesis tests questions with worked solutions. Stretch yourself on the hardest two-sample-t-test, difference-of-means, pooled-variance, test-statistic problems.

two-sample-t-testdifference-of-meanspooled-variancetest-statisticpaired-t-testpaired-data
Further Maths34 questionsStep-by-step solutions
Question 1
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.
Show worked solution

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
Question 2
9 markschallenging
A company claims that its warehouse staff work, on average, exactly 22 hours per week more than its office staff. The weekly hours worked by a warehouse worker are normally distributed with mean μ1\mu_1 and those worked by an office worker are normally distributed with mean μ2\mu_2; both variances are known from many years of payroll records. The population variances are known from long experience. The summary statistics of the two independent random samples are n1=60, xˉ1=32.4, σ12=25;n2=60, xˉ2=28.9, σ22=20.25n_1=60,\ \bar{x}_1=32.4,\ \sigma_1^2=25;\qquad n_2=60,\ \bar{x}_2=28.9,\ \sigma_2^2=20.25. Because both population variances are known, the test statistic is z=(xˉ1−xˉ2)−(μ1−μ2)σ12n1+σ22n2z=\dfrac{(\bar{x}_1-\bar{x}_2)-(\mu_1-\mu_2)}{\sqrt{\dfrac{\sigma_1^2}{n_1}+\dfrac{\sigma_2^2}{n_2}}}, which has the standard normal distribution when H0H_0 is true. The hypotheses are H0:μ1−μ2=2againstH1:μ1−μ2≠2H_0:\mu_1-\mu_2=2\quad\text{against}\quad H_1:\mu_1-\mu_2\neq2. 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 zcritz_{\text{crit}} of the null distribution and H0H_0 is rejected if and only if ∣z∣>zcrit|z|>z_{\text{crit}}. Do not round any intermediate value, and give every numerical answer correct to 33 decimal places. Calculate the value of the test statistic zz, giving your answer to 33 decimal places.
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Worked solution

  1. State the hypotheses

    H0:μ1−μ2=2againstH1:μ1−μ2≠2H_0:\mu_1-\mu_2=2\quad\text{against}\quad H_1:\mu_1-\mu_2\neq2

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

  2. State the modelling assumptions

    Xˉ1−Xˉ2∼N ⁣(μ1−μ2,σ12n1+σ22n2)\bar{X}_1-\bar{X}_2\sim N\!\left(\mu_1-\mu_2,\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}\right)

    Independent normal samples make the difference of the sample means exactly normal.

  3. Find the standard error of xˉ1−xˉ2\bar{x}_1-\bar{x}_2

    2560+20.2560=0.7542=0.8684\sqrt{\frac{25}{60}+\frac{20.25}{60}}=\sqrt{0.7542}=0.8684

    The variances of the two independent sample means are added, then square rooted.

  4. Substitute into the test statistic

    z=(32.4−28.9)−20.8684z=\frac{(32.4-28.9)-2}{0.8684}

    The hypothesised difference is subtracted before dividing by the standard error.

  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. 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.

  8. 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.

  9. Recall the paired tt statistic

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

    A paired test reduces the two samples to the single sample of differences.

  10. Recall the unbiased sample variance

    s2=Sxxn−1,Sxx=∑x2−nxˉ2s^2=\frac{S_{xx}}{n-1},\qquad S_{xx}=\sum x^2-n\bar{x}^2

    Dividing by n−1n-1 rather than nn makes the estimator unbiased.

  11. Recall the FF statistic for comparing two variances

    F=larger s2smaller s2≥1F=\frac{\text{larger }s^2}{\text{smaller }s^2}\ge1

    Putting the larger variance on top means only the upper tail of the FF distribution is ever needed.

  12. Recall the degrees of freedom of the FF statistic

    F∼F(ν1,ν2),νi=ni−1F\sim F(\nu_1,\nu_2),\qquad \nu_i=n_i-1

    The numerator degrees of freedom belong to the sample with the larger variance.

  13. Note the modelling assumptions

    X∼N(μ1,σ12),Y∼N(μ2,σ22)X\sim N(\mu_1,\sigma_1^2),\qquad Y\sim N(\mu_2,\sigma_2^2)

    Both populations are normal and the samples are independent random samples.

  14. Note the effect of a two-tailed test on the critical value

    12α in each tail\tfrac{1}{2}\alpha\text{ in each tail}

    A two-tailed test splits the significance level equally between the two tails.

  15. Note why the difference of the sample means is used

    E(Xˉ−Yˉ)=μ1−μ2E(\bar{X}-\bar{Y})=\mu_1-\mu_2

    The difference of the sample means is an unbiased estimator of the difference of the population means.

  16. Note the variance of the difference of two independent sample means

    Var⁡(Xˉ−Yˉ)=σ12n1+σ22n2\operatorname{Var}(\bar{X}-\bar{Y})=\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}

    Independence lets the two variances be added, even though the means are subtracted.

  17. Evaluate the test statistic

    z=1.727z=1.727

    This is zz correct to 33 decimal places.

Answer
z=1.727z=1.727
Question 3
9 markschallenging
A meteorologist records the daily rainfall, in millimetres, at station AA (values xx) and at station BB (values yy) on independent randomly chosen days. It is claimed that the rainfall at station AA is more variable than the rainfall at station BB. The observations from the first sample are x: 26.6, 18.6, 28.2, 19.5, 22.2, 21.5, 22.2, 31.7, 14.7, 26.4x:\ 26.6,\ 18.6,\ 28.2,\ 19.5,\ 22.2,\ 21.5,\ 22.2,\ 31.7,\ 14.7,\ 26.4 and those from the second sample are y: 23.0, 23.0, 19.3, 23.6, 24.4, 22.7, 20.6, 22.7, 18.5, 24.9y:\ 23.0,\ 23.0,\ 19.3,\ 23.6,\ 24.4,\ 22.7,\ 20.6,\ 22.7,\ 18.5,\ 24.9, so that n1=10n_1=10 and n2=10n_2=10. Both populations are assumed to be normally distributed and the samples are independent. The test statistic is F=larger sample variancesmaller sample varianceF=\dfrac{\text{larger sample variance}}{\text{smaller sample variance}}, with ν1\nu_1 one less than the size of the sample having the larger variance and ν2\nu_2 one less than the size of the other sample. The hypotheses are H0:σ12=σ22againstH1:σ12>σ22H_0:\sigma_1^2=\sigma_2^2\quad\text{against}\quad H_1:\sigma_1^2>\sigma_2^2. The test is carried out at the 5%5\% significance level. This is a one-tailed test. The larger sample variance is placed on top, so the critical value is the upper 5%5\% point Fcrit=F5%(ν1,ν2)F_{\text{crit}}=F_{5\%}(\nu_1,\nu_2) and H0H_0 is rejected if and only if F>FcritF>F_{\text{crit}}. Do not round any intermediate value, and give every numerical answer correct to 33 decimal places. Calculate the value of the test statistic FF, giving your answer to 33 decimal places.
Show worked solution

Worked solution

  1. State the hypotheses

    H0:σ12=σ22againstH1:σ12>σ22H_0:\sigma_1^2=\sigma_2^2\quad\text{against}\quad H_1:\sigma_1^2>\sigma_2^2

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

  2. Find ∑x\sum x for the sample of size 1010

    ∑x=231.6\sum x=231.6

    Adding the raw values is the first step towards the sample mean.

  3. Find ∑x2\sum x^2 for that sample

    ∑x2=5594.88\sum x^2=5594.88

    The sum of squares is needed for the unbiased variance.

  4. Find the mean of the xx sample

    xˉ=231.610=23.16\bar{x}=\frac{231.6}{10}=23.16

    The sample mean is the sum of the values divided by the sample size.

  5. Find SxxS_{xx}

    Sxx=∑x2−nxˉ2=5594.88−10×(23.16)2=231.0240S_{xx}=\sum x^2-n\bar{x}^2=5594.88-10\times(23.16)^2=231.0240

    Using ∑x2−nxˉ2\sum x^2-n\bar{x}^2 avoids computing every deviation separately.

  6. Find the unbiased variance s12s_1^2

    s12=Sxx9=25.6693s_1^2=\frac{S_{xx}}{9}=25.6693

    Dividing the sum of squares by n−1n-1 gives the unbiased estimate of the population variance.

  7. Find ∑y\sum y for the sample of size 1010

    ∑y=222.7\sum y=222.7

    Adding the raw values is the first step towards the sample mean.

  8. Find ∑y2\sum y^2 for that sample

    ∑y2=5000.01\sum y^2=5000.01

    The sum of squares is needed for the unbiased variance.

  9. Find the mean of the yy sample

    yˉ=222.710=22.27\bar{y}=\frac{222.7}{10}=22.27

    The sample mean is the sum of the values divided by the sample size.

  10. Find SyyS_{yy}

    Syy=∑y2−nyˉ2=5000.01−10×(22.27)2=40.4810S_{yy}=\sum y^2-n\bar{y}^2=5000.01-10\times(22.27)^2=40.4810

    Using ∑y2−nyˉ2\sum y^2-n\bar{y}^2 avoids computing every deviation separately.

  11. Find the unbiased variance s22s_2^2

    s22=Syy9=4.4979s_2^2=\frac{S_{yy}}{9}=4.4979

    Dividing the sum of squares by n−1n-1 gives the unbiased estimate of the population variance.

  12. Identify the larger and the smaller sample variance

    larger=25.6693 (sample A),smaller=4.4979\text{larger}=25.6693\ (\text{sample }A),\qquad \text{smaller}=4.4979

    The larger variance always goes on top, so that F≥1F\ge1 and only the upper tail is needed.

  13. Find the degrees of freedom

    ν1=10−1=9,ν2=10−1=9\nu_1=10-1=9,\qquad \nu_2=10-1=9

    The numerator degrees of freedom belong to sample A, which has the larger variance.

  14. Substitute into the FF statistic

    F=25.66934.4979F=\frac{25.6693}{4.4979}

    The statistic is the ratio of the two unbiased variance estimates.

  15. 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.

  16. Evaluate the FF statistic

    F=5.707F=5.707

    This is FF correct to 33 decimal places, on (9,9)(9,9) degrees of freedom.

Answer
F=5.707F=5.707
Question 4
9 markschallenging
A drug company claims that its new tablet lowers systolic blood pressure, in millimetres of mercury, by exactly 22 units on average. Ten patients each have their blood pressure measured before treatment (xx) and after treatment (yy), and the claim is tested using the differences d=x−yd=x-y. The values of xx are x: 156.0, 141.2, 160.0, 161.2, 155.5, 151.1, 164.2, 157.7, 158.0, 140.0x:\ 156.0,\ 141.2,\ 160.0,\ 161.2,\ 155.5,\ 151.1,\ 164.2,\ 157.7,\ 158.0,\ 140.0 and the corresponding values of yy are y: 155.0, 139.0, 158.6, 157.7, 155.0, 147.9, 160.7, 155.0, 157.5, 139.5y:\ 155.0,\ 139.0,\ 158.6,\ 157.7,\ 155.0,\ 147.9,\ 160.7,\ 155.0,\ 157.5,\ 139.5; the iith value of xx and the iith value of yy come from the same patient, 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=2againstH1:μd≠2H_0:\mu_d=2\quad\text{against}\quad H_1:\mu_d\neq2, 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 value of the paired tt test statistic, giving your answer to 33 decimal places.
Show worked solution

Worked solution

  1. State the hypotheses

    H0:μd=2againstH1:μd≠2H_0:\mu_d=2\quad\text{against}\quad H_1:\mu_d\neq2

    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: 1.0, 2.2, 1.4, 3.5, 0.5, 3.2, 3.5, 2.7, 0.5, 0.5d:\ 1.0,\ 2.2,\ 1.4,\ 3.5,\ 0.5,\ 3.2,\ 3.5,\ 2.7,\ 0.5,\ 0.5

    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=19.0,∑d2=50.58\sum d=19.0,\qquad \sum d^2=50.58

    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=19.010=1.9\bar{d}=\frac{19.0}{10}=1.9

    This estimates the population mean difference μd\mu_d.

  6. Find SddS_{dd}

    Sdd=∑d2−ndˉ2=50.58−10×(1.9)2=14.4800S_{dd}=\sum d^2-n\bar{d}^2=50.58-10\times(1.9)^2=14.4800

    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=14.48009=1.6089s_d^2=\frac{S_{dd}}{n-1}=\frac{14.4800}{9}=1.6089

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

  8. Find the standard deviation of the differences

    sd=1.6089=1.2684s_d=\sqrt{1.6089}=1.2684

    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=1.268410=0.4011\frac{s_d}{\sqrt{n}}=\frac{1.2684}{\sqrt{10}}=0.4011

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

  11. Substitute into the test statistic

    t=1.9−20.4011t=\frac{1.9-2}{0.4011}

    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. Recall the paired tt statistic

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

    A paired test reduces the two samples to the single sample of differences.

  17. Evaluate the test statistic

    t=−0.249t=-0.249

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

Answer
t=−0.249t=-0.249
Question 5
9 markschallenging
Twelve fields are each sown with the same crop in two successive years. The yield, in tonnes, of a field in the first year is xx and the yield of the same field in the second year, after the fertiliser was changed, is yy. It is claimed that the yields fell in the second year. The values of xx are x: 48.4, 53.5, 51.2, 44.2, 43.2, 42.0, 48.1, 41.2, 55.1, 46.8, 49.4, 44.7x:\ 48.4,\ 53.5,\ 51.2,\ 44.2,\ 43.2,\ 42.0,\ 48.1,\ 41.2,\ 55.1,\ 46.8,\ 49.4,\ 44.7 and the corresponding values of yy are y: 44.8, 50.2, 49.1, 43.3, 41.5, 40.7, 44.3, 40.2, 53.7, 43.3, 47.6, 41.5y:\ 44.8,\ 50.2,\ 49.1,\ 43.3,\ 41.5,\ 40.7,\ 44.3,\ 40.2,\ 53.7,\ 43.3,\ 47.6,\ 41.5; the iith value of xx and the iith value of yy come from the same plot, so that n=12n=12. 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>0, 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 one-tailed test with the critical region in the upper tail, 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>tcritt>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>0

    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: 3.6, 3.3, 2.1, 0.9, 1.7, 1.3, 3.8, 1.0, 1.4, 3.5, 1.8, 3.2d:\ 3.6,\ 3.3,\ 2.1,\ 0.9,\ 1.7,\ 1.3,\ 3.8,\ 1.0,\ 1.4,\ 3.5,\ 1.8,\ 3.2

    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=27.6,∑d2=76.78\sum d=27.6,\qquad \sum d^2=76.78

    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=27.612=2.3\bar{d}=\frac{27.6}{12}=2.3

    This estimates the population mean difference μd\mu_d.

  6. Find SddS_{dd}

    Sdd=∑d2−ndˉ2=76.78−12×(2.3)2=13.3000S_{dd}=\sum d^2-n\bar{d}^2=76.78-12\times(2.3)^2=13.3000

    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=13.300011=1.2091s_d^2=\frac{S_{dd}}{n-1}=\frac{13.3000}{11}=1.2091

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

  8. Find the standard deviation of the differences

    sd=1.2091=1.0996s_d=\sqrt{1.2091}=1.0996

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

  9. Find the number of degrees of freedom

    ν=n−1=12−1=11\nu=n-1=12-1=11

    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=1.099612=0.3174\frac{s_d}{\sqrt{n}}=\frac{1.0996}{\sqrt{12}}=0.3174

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

  11. Substitute into the test statistic

    t=2.3−00.3174t=\frac{2.3-0}{0.3174}

    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=7.246t=7.246

    This is tt correct to 33 decimal places, on ν=11\nu=11 degrees of freedom.

Answer
t=7.246t=7.246

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