EXM.1.SL.TZ0.10

pestleMathematicsAISLPaper 1EXM· sl-4-11-expected-observed-hypotheses-chi-squared-gof-t-testsource ↗

Give your answers to four significant figures.

A die is thrown 120 times with the following results.

Showing all steps clearly, test whether the die is fair

(i) at the 5% level of significance;

(ii) at the 1% level of significance.

[7]

Explain what is meant by “level of significance” in part (a).

[3]

Markscheme / solution

The data can be described by the following table

(A1)

$\left. {\begin{array}{*{20}{c}} {{{\text{H}}_0}\,{\text{: the die is fair}}\,\,\,\,\,\,\,\,\,{\text{ }}} \\ {{{\text{H}}_1}\,{\text{: the die is not fair}}} \end{array}} \right\}$ (C1)

$\chi _{{\text{calc}}}^2 = \frac{{{{\left( {20 - 27} \right)}^2}}}{{20}} + \frac{{{{\left( {12 - 20} \right)}^2}}}{{20}} + \frac{{{{\left( {16 - 20} \right)}^2}}}{{20}} + \frac{{{{\left( {25 - 20} \right)}^2}}}{{20}} + \frac{{{{\left( {26 - 20} \right)}^2}}}{{20}} + \frac{{{{\left( {14 - 20} \right)}^2}}}{{20}}$

$= 11.3$ (A1)

(i) $\chi _{0.95{\text{;}}\,\,5}^2 = 11.07$ (A1)

Hence, since 11.3 > 11.07 at the 5% level we must accept H₁. (R1)

(ii) $\chi _{0.99{\text{;}}\,\,5}^2 = 15.086$ (A1)

Hence, since 11.3 < 15.086, at the 1% level, there is not enough evidence to conclude that the die is not fair (and hence we accept H₀). (R1)

[7 marks]

Let $\alpha$ denote the significance level. If $\chi _{{\text{calc}}}^2$ is greater than $\chi _{\alpha ,\,\,n - 1}^2$ then it means that the probability of obtaining the results obtained is less than $\alpha$ if H₀ is correct. (R3)

Note: Award (R3) for any correct explanation. Use discretion to award (R2) or (R1).

[3 marks]

Examiners’ report

[N/A]