Question

A sample of boys and girls were asked to choose their favorite sport, with the following results:

	Football	Cricket	Hockey	Basketball
Boys	$86$	$60$	$44$	$10$
Girls	$40$	$30$	$25$	$5$

Find the value of ${\chi ^2}$ statistics.

Answer 1

Hint: In this question, we are given data of boys’ and girls’ favorite sports. We have been asked to find the ${\chi ^2}$. At first, calculate expected frequencies using the formula $\dfrac{{{T_i} \times {T_j}}}{N}$. After this, subtract the expected frequency from the corresponding actual observation and square the difference. Then, divide the square by their corresponding expected frequency and add all of them. You will have the final answer.

Formula used: 1) ${E_{ij}} = \dfrac{{{T_i} \times {T_j}}}{N}$, where ${E_{ij}} = $Expected frequency for i(th) row and j(th) column, ${T_i} = $Total of i(th) row, ${T_j} = $Total of j(th) column, $N = $Total cells in table.
2) ${\chi ^2} = \dfrac{{\sum {{{\left( {{O_{ij}} - {E_{ij}}} \right)}^2}} }}{{{E_{ij}}}}$

Complete step-by-step solution:
We will start by finding the expected frequencies. We have to calculate the expected frequencies of each and every cell in the given table. But first we will find the total of each row and columns.

	Football	Cricket	Hockey	Basketball	Row total $\left( {{T_i}} \right)$
Boys	$86$	$60$	$44$	$10$	$200$
Girls	$40$	$30$	$25$	$5$	$100$
Column Total $\left( {{T_j}} \right)$	$126$	$90$	$69$	$15$	$300$

Now, we will find the expected frequencies using the formula ${E_{ij}} = \dfrac{{{T_i} \times {T_j}}}{N}$.
Expected frequency for $i = 1$ and $j = 1$,
${E_{11}} = \dfrac{{200 \times 126}}{{300}} = 84$
Expected frequency for $i = 1$ and $j = 2$,
${E_{12}} = \dfrac{{200 \times 90}}{{300}} = 60$
Expected frequency for $i = 1$ and $j = 3$,
${E_{13}} = \dfrac{{200 \times 69}}{{300}} = 46$
Expected frequency for $i = 1$ and $j = 4$,
${E_{14}} = \dfrac{{200 \times 15}}{{300}} = 10$
Expected frequency for $i = 2$ and $j = 1$,
${E_{21}} = \dfrac{{100 \times 126}}{{300}} = 42$
Expected frequency for $i = 2$ and $j = 2$,
${E_{22}} = \dfrac{{100 \times 90}}{{300}} = 30$
Expected frequency for $i = 2$ and $j = 3$,
${E_{23}} = \dfrac{{100 \times 69}}{{300}} = 23$
Expected frequency for $i = 2$ and $j = 4$,
${E_{24}} = \dfrac{{100 \times 15}}{{300}} = 5$
The table of expected frequencies is below:

	Football	Cricket	Hockey	Basketball	Row total $\left( {{T_i}} \right)$
Boys	$84$	$60$	$46$	$10$	$200$
Girls	$42$	$30$	$23$	$5$	$100$
Column Total $\left( {{T_j}} \right)$	$126$	$90$	$69$	$15$	$300$

 Next step is to subtract the expected frequency from the corresponding actual observation and then we have to square the difference. The final square of the difference is divided by the expected frequency of that cell. Each cell’s final answer is added to each other.
It is done like below:
$ \Rightarrow \dfrac{{{{\left( {86 - 84} \right)}^2}}}{{84}} + \dfrac{{{{\left( {60 - 60} \right)}^2}}}{{60}} + \dfrac{{{{\left( {44 - 46} \right)}^2}}}{{46}} + \dfrac{{{{\left( {10 - 10} \right)}^2}}}{{10}} + \dfrac{{{{\left( {40 - 42} \right)}^2}}}{{42}} + \dfrac{{{{\left( {30 - 30} \right)}^2}}}{{30}} + \dfrac{{{{\left( {25 - 23} \right)}^2}}}{{23}} + \dfrac{{{{\left( {5 - 5} \right)}^2}}}{5}$
On simplifying the above equation, we will get –
$ \Rightarrow \dfrac{4}{{84}} + 0 + \dfrac{4}{{46}} + 0 + \dfrac{4}{{42}} + 0 + \dfrac{4}{{23}} + 0$
Hence,
$ \Rightarrow 0.048 + 0.087 + 0.095 + 0.174$
On adding,
$ \Rightarrow {\chi ^2} = 0.404$

The required value of ${\chi ^2}$ is 0.404.

Note: We can observe that the expected frequency is a probability count that appears in contingency table calculations including the Chi-square test. Expected frequencies also used to calculate standardized residuals, where the expected count is subtracted from the observed count in the numerator. An observed frequency are counts made from experimental data. In other words, you actually observe the data happening and take measurements. An expected frequencies are counts calculated using probability theory.