Further Maths Chi-squared tests Practice Questions

Free Further Maths Chi-squared tests practice questions with full step-by-step worked solutions. Covers chi-squared, hypothesis-testing, goodness-of-fit, uniform-model. Practise exam-style problems and check your method.

chi-squaredhypothesis-testinggoodness-of-fituniform-modelexpected-frequenciesgiven-ratio
Further Maths70 questionsStep-by-step solutions
Question 1
2 markseasy
A six-sided die is rolled 120120 times and the score on the uppermost face is recorded each time. The observed frequencies are given in the table. Score123456Observed frequency182316212418\begin{array}{c|cccccc}\text{Score} & 1 & 2 & 3 & 4 & 5 & 6\\\hline\text{Observed frequency} & 18 & 23 & 16 & 21 & 24 & 18\end{array} Under H0H_0 the six outcomes are equally likely (a uniform model); under H1H_1 they are not all equally likely. No parameter is estimated from the data. Any class whose expected frequency is less than 55 must be pooled with the adjacent class nearer the tail of the distribution, repeatedly, until every expected frequency is at least 55; the tail is pooled first. The number of degrees of freedom is ν=(number of classes after pooling)−1−(number of parameters estimated from the data)\nu=(\text{number of classes after pooling})-1-(\text{number of parameters estimated from the data}). Expected frequencies are to be used exactly (they must not be rounded before being used) and every numerical answer is to be given to 33 decimal places. The test is carried out at the 5%5\% significance level and H0H_0 is rejected if and only if χcalc2>χcrit2\chi^2_{\text{calc}}>\chi^2_{\text{crit}} (a strict inequality). Calculate the expected frequency for the class 11 (before any pooling), giving your answer to 33 decimal places.
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Worked solution

  1. Find the model probability of the class 11

    p=0.16667p=0.16667

    This is the probability the model assigns to that class.

  2. Multiply the probability by the sample size

    E=120×0.16667E=120\times0.16667

    The expected frequency is E=N×pE=N\times p.

  3. State the expected frequency

    E=20.000E=20.000

    This is the expected frequency to 33 decimal places.

Answer
E=20.000E=20.000
Question 2
2 markseasy
The number of flaws in each of 200200 metre-lengths of cloth is recorded. The observed frequencies are given in the table. Number of flaws0123≥4Observed frequency4070551421\begin{array}{c|ccccc}\text{Number of flaws} & 0 & 1 & 2 & 3 & \ge 4\\\hline\text{Observed frequency} & 40 & 70 & 55 & 14 & 21\end{array} Under H0H_0 the number of events XX follows the distribution Po(1.5)\mathrm{Po}(1.5). The value λ=1.5\lambda=1.5 is given in this question and is not estimated from the data, so no degree of freedom is lost for parameter estimation. Any class whose expected frequency is less than 55 must be pooled with the adjacent class nearer the tail of the distribution, repeatedly, until every expected frequency is at least 55; the tail is pooled first. The number of degrees of freedom is ν=(number of classes after pooling)−1−(number of parameters estimated from the data)\nu=(\text{number of classes after pooling})-1-(\text{number of parameters estimated from the data}). Expected frequencies are to be used exactly (they must not be rounded before being used) and every numerical answer is to be given to 33 decimal places. The test is carried out at the 5%5\% significance level and H0H_0 is rejected if and only if χcalc2>χcrit2\chi^2_{\text{calc}}>\chi^2_{\text{crit}} (a strict inequality). Find the critical value χcrit2\chi^2_{\text{crit}} for this test, giving your answer to 33 decimal places.
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Worked solution

  1. Check the pooling rule

    min⁡iEi=13.1285≥5  ⇒  no pooling\min_i E_i=13.1285\ge5\;\Rightarrow\;\text{no pooling}

    Every expected frequency is at least 55, so no classes are combined.

  2. Find the number of degrees of freedom

    ν=5−1−0=4\nu=5-1-0=4

    There are 5 classes after pooling and 0 parameters estimated from the data.

  3. State the significance level

    α=5%\alpha=5\%

    The chi-squared test is one-tailed, so the whole of α\alpha is in the upper tail.

  4. Read the critical value from the chi-squared tables

    χcrit2=χ0.052(4)=9.488\chi^2_{\text{crit}}=\chi^2_{0.05}(4)=9.488

    This is the upper 5\% point with ν=4\nu=4.

Answer
χcrit2=9.488\chi^2_{\text{crit}}=9.488
Question 3
4 marksintermediate
The number of emails received by an office in each of 100100 five-minute intervals is recorded. The observed frequencies are given in the table. Number of emails01234≥5Observed frequency2132261560\begin{array}{c|cccccc}\text{Number of emails} & 0 & 1 & 2 & 3 & 4 & \ge 5\\\hline\text{Observed frequency} & 21 & 32 & 26 & 15 & 6 & 0\end{array} Under H0H_0 the number of events XX follows a Poisson distribution Po(λ)\mathrm{Po}(\lambda), where λ\lambda is estimated from the sample by the sample mean λ^=xˉ\hat{\lambda}=\bar{x}; no observation falls in the class ≥5\ge 5, so xˉ\bar{x} is found exactly from the table. One degree of freedom is therefore lost for this estimated parameter. Any class whose expected frequency is less than 55 must be pooled with the adjacent class nearer the tail of the distribution, repeatedly, until every expected frequency is at least 55; the tail is pooled first. The number of degrees of freedom is ν=(number of classes after pooling)−1−(number of parameters estimated from the data)\nu=(\text{number of classes after pooling})-1-(\text{number of parameters estimated from the data}). Expected frequencies are to be used exactly (they must not be rounded before being used) and every numerical answer is to be given to 33 decimal places. The test is carried out at the 5%5\% significance level and H0H_0 is rejected if and only if χcalc2>χcrit2\chi^2_{\text{calc}}>\chi^2_{\text{crit}} (a strict inequality). State the number of degrees of freedom ν\nu for this test.
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Worked solution

  1. Estimate λ\lambda from the sample

    λ^=xˉ=153100=1.53\hat{\lambda}=\bar{x}=\frac{153}{100}=1.53

    The parameter is estimated from the data, so one degree of freedom is lost.

  2. Find every expected frequency

    E1=21.6536,  E2=33.1300,  E3=25.3444,  E4=12.9257,  E5=4.9441,  E6=2.0023E_{1}=21.6536,\;E_{2}=33.1300,\;E_{3}=25.3444,\;E_{4}=12.9257,\;E_{5}=4.9441,\;E_{6}=2.0023

    Each expected frequency is the sample size multiplied by the model probability.

  3. Pool the classes whose expected frequency is below 55

    classes 0,  1,  2,  3,  4,  ≥5  ⇒  0,  1,  2,  3,  ≥4\text{classes }0,\;1,\;2,\;3,\;4,\;\ge 5\;\Rightarrow\;0,\;1,\;2,\;3,\;\ge 4

    Merging from the tail inwards leaves 5 classes, each with E≥5E\ge5.

  4. Count the classes used in the test

    k=5k=5

    This is the number of classes remaining after any pooling.

  5. Count the parameters estimated from the data

    m=1m=1

    A parameter that is given in the question does not cost a degree of freedom.

  6. State the number of degrees of freedom

    ν=3\nu=3

    This is the value of ν\nu used to find the critical value.

Answer
ν=3\nu=3
Question 4
6 markshard
A poll of 200200 voters records their age band and whether they agree with a statement. The results are given in the contingency table.  AgreeDisagreeAge 16-253120Age 26-402525Age 41-602030Age over 601435\begin{array}{l|cc}\text{ } & \text{Agree} & \text{Disagree}\\\hline\text{Age 16-25} & 31 & 20\\\text{Age 26-40} & 25 & 25\\\text{Age 41-60} & 20 & 30\\\text{Age over 60} & 14 & 35\end{array} The data form a contingency table. Under H0H_0 the two factors are independent; under H1H_1 they are not independent. Expected frequencies are found from E=(row total)×(column total)grand totalE=\frac{(\text{row total})\times(\text{column total})}{\text{grand total}} and the number of degrees of freedom is ν=(r−1)(c−1)\nu=(r-1)(c-1), where rr and cc are the numbers of rows and columns. No classes are to be pooled: every expected frequency in this table is at least 55. Yates' continuity correction does not apply to a table of this size, so the test statistic is χcalc2=∑(O−E)2E\chi^2_{\text{calc}}=\sum\frac{(O-E)^2}{E}. Expected frequencies are to be used exactly (they must not be rounded before being used) and every numerical answer is to be given to 33 decimal places. The test is carried out at the 1%1\% significance level and H0H_0 is rejected if and only if χcalc2>χcrit2\chi^2_{\text{calc}}>\chi^2_{\text{crit}} (a strict inequality). Calculate the value of the test statistic χcalc2\chi^2_{\text{calc}}, giving your answer to 33 decimal places.
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Worked solution

  1. State the hypotheses

    H0:the two factors are independentH1:the two factors are not independentH_0:\text{the two factors are independent}\quad H_1:\text{the two factors are not independent}

    A chi-squared test always tests a model against the alternative that the model is wrong.

  2. State the model of independence

    Eij=Ri×CjTE_{ij}=\frac{R_i\times C_j}{T}

    Under H0H_0 the two factors are independent, which fixes every expected frequency.

  3. Find every expected frequency

    E11=22.9500,  E12=28.0500,  E21=22.5000,  E22=27.5000,  E31=22.5000,  E32=27.5000,  E41=22.0500,  E42=26.9500E_{11}=22.9500,\;E_{12}=28.0500,\;E_{21}=22.5000,\;E_{22}=27.5000,\;E_{31}=22.5000,\;E_{32}=27.5000,\;E_{41}=22.0500,\;E_{42}=26.9500

    Each expected frequency is (row total)(column total)/(grand total).

  4. Check that the expected frequencies sum to the observed total

    ∑E=200.000=200=∑O\sum E=200.000=200=\sum O

    This confirms that no arithmetic slip has been made.

  5. Check the pooling rule

    min⁡iEi=22.0500≥5  ⇒  no pooling\min_i E_i=22.0500\ge5\;\Rightarrow\;\text{no pooling}

    Every expected frequency is at least 55, so no classes are combined.

  6. Find the number of degrees of freedom

    ν=(4−1)(2−1)=3\nu=(4-1)(2-1)=3

    For a contingency table ν=(r−1)(c−1)\nu=(r-1)(c-1).

  7. Find the contribution of every cell

    2.8236+2.3102+0.2778+0.2273+0.2778+0.2273+2.9389+2.4045  ⇒  χcalc2=11.4872.8236+2.3102+0.2778+0.2273+0.2778+0.2273+2.9389+2.4045\;\Rightarrow\;\chi^2_{\text{calc}}=11.487

    The test statistic is the sum of the contributions of all the cells.

  8. Recall the chi-squared test statistic

    χcalc2=∑(O−E)2E\chi^2_{\text{calc}}=\sum\frac{(O-E)^2}{E}

    Each class contributes (O−E)2E\frac{(O-E)^2}{E}, so the statistic is never negative.

  9. Recall the pooling rule

    Ei≥5 for every class used in the testE_i\ge5\text{ for every class used in the test}

    Classes are combined with their neighbour towards the tail until this holds.

  10. State the value of the test statistic

    χcalc2=11.487\chi^2_{\text{calc}}=11.487

    This is χcalc2\chi^2_{\text{calc}} correct to 33 decimal places.

Answer
χcalc2=11.487\chi^2_{\text{calc}}=11.487
Question 5
9 markschallenging
The number of accidents on a stretch of road in each of 150150 weeks is recorded. The observed frequencies are given in the table. Number of accidents012345≥6Observed frequency8203532261613\begin{array}{c|ccccccc}\text{Number of accidents} & 0 & 1 & 2 & 3 & 4 & 5 & \ge 6\\\hline\text{Observed frequency} & 8 & 20 & 35 & 32 & 26 & 16 & 13\end{array} Under H0H_0 the number of events XX follows the distribution Po(3)\mathrm{Po}(3). The value λ=3\lambda=3 is given in this question and is not estimated from the data, so no degree of freedom is lost for parameter estimation. Any class whose expected frequency is less than 55 must be pooled with the adjacent class nearer the tail of the distribution, repeatedly, until every expected frequency is at least 55; the tail is pooled first. The number of degrees of freedom is ν=(number of classes after pooling)−1−(number of parameters estimated from the data)\nu=(\text{number of classes after pooling})-1-(\text{number of parameters estimated from the data}). Expected frequencies are to be used exactly (they must not be rounded before being used) and every numerical answer is to be given to 33 decimal places. The test is carried out at the 1%1\% significance level and H0H_0 is rejected if and only if χcalc2>χcrit2\chi^2_{\text{calc}}>\chi^2_{\text{crit}} (a strict inequality). Which of the following correctly states the number of degrees of freedom, the value of the test statistic, the critical value and the conclusion of this test in context?
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Worked solution

  1. State the hypotheses

    H0:the model fits the dataH1:the model does not fit the dataH_0:\text{the model fits the data}\quad H_1:\text{the model does not fit the data}

    The stated model is the null hypothesis for a goodness-of-fit test.

  2. State the Poisson model

    X∼Po(3)X\sim\mathrm{Po}(3)

    The value of λ\lambda is given, so no degree of freedom is lost for estimation.

  3. Find every expected frequency

    E1=7.4681,  E2=22.4042,  E3=33.6063,  E4=33.6063,  E5=25.2047,  E6=15.1228,  E7=12.5877E_{1}=7.4681,\;E_{2}=22.4042,\;E_{3}=33.6063,\;E_{4}=33.6063,\;E_{5}=25.2047,\;E_{6}=15.1228,\;E_{7}=12.5877

    Each expected frequency is the sample size multiplied by the model probability.

  4. Check that the expected frequencies sum to the observed total

    ∑E=150.000=150=∑O\sum E=150.000=150=\sum O

    This confirms that no arithmetic slip has been made.

  5. Check the pooling rule

    min⁡iEi=7.4681≥5  ⇒  no pooling\min_i E_i=7.4681\ge5\;\Rightarrow\;\text{no pooling}

    Every expected frequency is at least 55, so no classes are combined.

  6. Find the number of degrees of freedom

    ν=7−1−0=6\nu=7-1-0=6

    There are 7 classes after pooling and 0 parameters estimated from the data.

  7. Find the contribution of every class

    0.0379+0.2580+0.0578+0.0768+0.0251+0.0509+0.0135  ⇒  χcalc2=0.5200.0379+0.2580+0.0578+0.0768+0.0251+0.0509+0.0135\;\Rightarrow\;\chi^2_{\text{calc}}=0.520

    The test statistic is the sum of the contributions of the pooled classes.

  8. Add the contributions

    χcalc2=0.0379+0.2580+0.0578+0.0768+0.0251+0.0509+0.0135=0.520\chi^2_{\text{calc}}=0.0379+0.2580+0.0578+0.0768+0.0251+0.0509+0.0135=0.520

    This is the value of the test statistic, to 33 decimal places.

  9. Find the critical value

    χcrit2=χ0.012(6)=16.812\chi^2_{\text{crit}}=\chi^2_{0.01}(6)=16.812

    This is the upper 1\% point of the chi-squared distribution with ν=6\nu=6.

  10. Compare the statistic with the critical value

    0.520≤16.8120.520\le16.812

    H0H_0 is rejected if and only if χcalc2>χcrit2\chi^2_{\text{calc}}>\chi^2_{\text{crit}}.

  11. Recall the chi-squared test statistic

    χcalc2=∑(O−E)2E\chi^2_{\text{calc}}=\sum\frac{(O-E)^2}{E}

    Each class contributes (O−E)2E\frac{(O-E)^2}{E}, so the statistic is never negative.

  12. Recall the pooling rule

    Ei≥5 for every class used in the testE_i\ge5\text{ for every class used in the test}

    Classes are combined with their neighbour towards the tail until this holds.

  13. Recall the degrees-of-freedom rule for a goodness-of-fit test

    ν=k−1−m\nu=k-1-m

    Here kk is the number of classes AFTER pooling and mm the number of parameters estimated from the data.

  14. Recall the degrees-of-freedom rule for a contingency table

    ν=(r−1)(c−1)\nu=(r-1)(c-1)

    The row and column totals are fixed, which removes r+c−1r+c-1 degrees of freedom.

  15. Select the correct conclusion in context

    do not reject H0; there is insufficient evidence at the 1% level that Po(3) is a poor model for the number of accidents\text{do not reject }H_0\text{; there is insufficient evidence at the 1\% level that Po(3) is a poor model for the number of accidents}

    The conclusion must be stated in the context of the question.

Answer
ν=6,  χcalc2=0.520,  χcrit2=16.812,  so χcalc2≤χcrit2: do not reject H0; there is insufficient evidence at the 1% level that Po(3) is a poor model for the number of accidents\nu=6,\;\chi^2_{\text{calc}}=0.520,\;\chi^2_{\text{crit}}=16.812,\;\text{so }\chi^2_{\text{calc}}\le\chi^2_{\text{crit}}\text{: do not reject }H_0\text{; there is insufficient evidence at the 1\% level that Po(3) is a poor model for the number of accidents}

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