Chi-squared tests Worked Solutions — Further Maths Maths

Fully worked, step-by-step solutions to Further Maths Chi-squared tests questions. See exactly how to solve problems on chi-squared, hypothesis-testing, goodness-of-fit, uniform-model.

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.

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
A spinner has five sectors of equal size, numbered 11 to 55. It is spun 200200 times and the sector it lands on is recorded. The observed frequencies are given in the table. Sector12345Observed frequency4835404235\begin{array}{c|ccccc}\text{Sector} & 1 & 2 & 3 & 4 & 5\\\hline\text{Observed frequency} & 48 & 35 & 40 & 42 & 35\end{array} Under H0H_0 the five 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 10%10\% 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 33 (before any pooling), giving your answer to 33 decimal places.

Worked solution

  1. Find the model probability of the class 33

    p=0.20000p=0.20000

    This is the probability the model assigns to that class.

  2. Multiply the probability by the sample size

    E=200×0.20000E=200\times0.20000

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

  3. State the expected frequency

    E=40.000E=40.000

    This is the expected frequency to 33 decimal places.

Answer
E=40.000E=40.000
Question 3
2 markseasy
A packet of 160160 sweets contains sweets of four colours. A manufacturer claims that the four colours occur in equal proportions. The observed frequencies are given in the table. ColourRedGreenBlueYellowObserved frequency56323834\begin{array}{c|cccc}\text{Colour} & \text{Red} & \text{Green} & \text{Blue} & \text{Yellow}\\\hline\text{Observed frequency} & 56 & 32 & 38 & 34\end{array} Under H0H_0 the four 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 Green\text{Green} (before any pooling), giving your answer to 33 decimal places.

Worked solution

  1. State the model probability of each class

    pi=14(i=1,…,4)p_i=\frac{1}{4}\quad(i=1,\dots,4)

    A uniform model gives every class the same probability.

  2. Find the model probability of the class Green\text{Green}

    p=0.25000p=0.25000

    This is the probability the model assigns to that class.

  3. Multiply the probability by the sample size

    E=160×0.25000E=160\times0.25000

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

  4. State the expected frequency

    E=40.000E=40.000

    This is the expected frequency to 33 decimal places.

Answer
E=40.000E=40.000
Question 4
2 markseasy
A genetic theory predicts that four phenotypes A, B, C and D of a plant occur in the ratio 9:3:3:19:3:3:1. In a sample of 240240 plants the observed frequencies are given in the table. PhenotypeABCDObserved frequency128524416\begin{array}{c|cccc}\text{Phenotype} & \text{A} & \text{B} & \text{C} & \text{D}\\\hline\text{Observed frequency} & 128 & 52 & 44 & 16\end{array} Under H0H_0 the categories occur in the ratio 9:3:3:19:3:3:1; under H1H_1 they do not occur in this ratio. 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 A\text{A} (before any pooling), giving your answer to 33 decimal places.

Worked solution

  1. Find the model probability of the class A\text{A}

    p=0.56250p=0.56250

    This is the probability the model assigns to that class.

  2. Multiply the probability by the sample size

    E=240×0.56250E=240\times0.56250

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

  3. State the expected frequency

    E=135.000E=135.000

    This is the expected frequency to 33 decimal places.

Answer
E=135.000E=135.000
Question 5
2 markseasy
A company claims that its gold, silver and bronze packs are sold in the ratio 2:3:52:3:5. In a sample of 300300 packs the observed frequencies are given in the table. GradeGoldSilverBronzeObserved frequency7580145\begin{array}{c|ccc}\text{Grade} & \text{Gold} & \text{Silver} & \text{Bronze}\\\hline\text{Observed frequency} & 75 & 80 & 145\end{array} Under H0H_0 the categories occur in the ratio 2:3:52:3:5; under H1H_1 they do not occur in this ratio. 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 10%10\% 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 Bronze\text{Bronze} (before any pooling), giving your answer to 33 decimal places.

Worked solution

  1. Find the model probability of the class Bronze\text{Bronze}

    p=0.50000p=0.50000

    This is the probability the model assigns to that class.

  2. Multiply the probability by the sample size

    E=300×0.50000E=300\times0.50000

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

  3. State the expected frequency

    E=150.000E=150.000

    This is the expected frequency to 33 decimal places.

Answer
E=150.000E=150.000

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