(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.