The null hypothesis states that there is no difference in the distribution of the categorical variable for exch of the populations/treatments. The alternative hypothesis states that there is a difference.

\(H_0\): The distribution of candy chosen is the same for each survey type.

\(H_a\): The distribution of candy chosen is not the same for each survey type.

The expected frequencies are the product of the column and row total,divided by the table total.

\(E_{11}=\frac{r_1 \times c_1}{n}=\frac{442 \times 319}{1207} \approx 116.82\)

\(E_{12}=\frac{r_1 \times c_2}{n}=\frac{442 \times 261}{1207} \approx 95.58\)

\(E_{13}=\frac{r_1 \times c_3}{n}=\frac{442 \times 627}{1207} \approx 229.61\)

\(E_{21}=\frac{r_2 \times c_1}{n}=\frac{765 \times 319}{1207} \approx 202.18\)

\(E_{22}=\frac{r_2 \times c_2}{n}=\frac{765 \times 261}{1207} \approx 165.42\)

\(E_{23}=\frac{r_2 \times c_3}{n}=\frac{765 \times 627}{1207} \approx 397.39\) Conditions

The conditions for performing a chi-square test of homogeneity /independence are: Random, Independent: (10%), Large counts.

Random: Not satisfied, because the passengers of the titanic were not randomly selected..

Independent: Satisfied, because the 1207 people are less than 10% of all people (since there are more than 12070 people).

Large counts: Satisfied, because all expected counts are at least 5.

Since the random condition is not satisfied, it is not appropriate to carry out a test of homogeneity â€˜independence.

All conditions are satisfied.