written 4.3 years ago by
teamques10
★ 26k

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modified 4.1 years ago

(i) The null hypothesis $H_0$ : The die is unbiased.
Alternative hypothesis $H_a$ : The die is biased.
(ii) Calculation of test statistic : On the hypothesis, that the die is unbiased we should expect the frequency of each number to be = $\frac{total}{6}$ = $\frac{132}{6} = 22$
$\chi^2$ = $\sum$ $\frac{(O  E)^2}{E}$ = $\frac{(15  22)^2}{22}$ + $\frac{(20  22)^2}{22}$ + $\frac{(25  22)^2}{22}$ + $\frac{(15  22)^2}{22}$ + $\frac{(29  22)^2}{22}$ + $\frac{(28  22)^2}{22}$ = $\frac{196}{12}$ = $8.91$
(iii) Level of significance : $α = 0.05$
(iv) Critical value : the table value of χ$^2$ at 5% level of significance for $ν = 6  1 = 5$ degrees of freedom is $11.0$
(v) Decision : Since the calculated value of $t = 8.91$ is less than the table value $χ^2$ = $11.07$, the null hypothesis is accepted.
∴ The die is unbiased.