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# If $800$ people were asked whether they wear glasses for reading with following results. Compute the ${\chi ^2}$ square statistics.

Last updated date: 13th Jun 2024
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Hint: Here we will use the concept of expected frequencies to calculate the ${\chi ^2}$ square statistics. We will first find the sum of the rows and columns. Then we will calculate the value of the expected frequencies of each square. We will use the formula for the ${\chi ^2}$ square to get its value.

Formula used:
We will use the formula ${\chi ^2} = \sum {\left[ {\dfrac{{{{\left( {{O_{ij}} - {E_{ij}}} \right)}^2}}}{{{E_{ij}}}}} \right]}$where, $O$ is denoted for original value and $E$ is denoted for expected value.

So firstly we will find the sum of the rows and the columns, we get

We know that the expected frequency is generally given by ${E_{ij}} = \dfrac{{{R_i} \times {C_j}}}{N}$ .
So we will find the expected frequency for every square using the formula . Therefore, we get
${E_{11}} = \dfrac{{{R_1} \times {C_1}}}{N}$
Now substituting ${R_1} = 400$, ${C_1} = 600$ and $N = 800$ in above equation, we get
$\Rightarrow {E_{11}} = \dfrac{{400 \times 600}}{{800}} = 300$
Now substituting ${R_1} = 400$, ${C_1} = 200$ and $N = 800$ in the formula, we get

$\begin{array}{l}{E_{12}} = \dfrac{{{R_1} \times {C_2}}}{N}\\ \Rightarrow {E_{12}} = \dfrac{{400 \times 200}}{{800}} = 100\end{array}$
Now substituting ${R_2} = 400$, ${C_1} = 600$ and $N = 800$ in formula, we get
$\begin{array}{l}{E_{21}} = \dfrac{{{R_2} \times {C_1}}}{N}\\ \Rightarrow {E_{21}} = \dfrac{{400 \times 600}}{{800}} = 300\end{array}$
Now substituting ${R_2} = 400$, ${C_2} = 200$ and $N = 800$ in formula, we get
$\begin{array}{l}{E_{22}} = \dfrac{{{R_2} \times {C_2}}}{N}\\ \Rightarrow {E_{22}} = \dfrac{{400 \times 200}}{{800}} = 100\end{array}$
Now we will use the formula of the ${\chi ^2}$ square statistics to get its value.
Therefore, substituting the value in formula ${\chi ^2} = \sum {\left[ {\dfrac{{{{\left( {{O_{ij}} - {E_{ij}}} \right)}^2}}}{{{E_{ij}}}}} \right]}$, we get
${\chi ^2} = \dfrac{{{{\left( {310 - 300} \right)}^2}}}{{300}} + \dfrac{{{{\left( {90 - 100} \right)}^2}}}{{100}} + \dfrac{{{{\left( {290 - 300} \right)}^2}}}{{300}} + \dfrac{{{{\left( {110 - 100} \right)}^2}}}{{100}}$
By solving this, we get
$\Rightarrow {\chi ^2} = \dfrac{{{{\left( {10} \right)}^2}}}{{300}} + \dfrac{{{{\left( { - 10} \right)}^2}}}{{100}} + \dfrac{{{{\left( { - 10} \right)}^2}}}{{300}} + \dfrac{{{{\left( {10} \right)}^2}}}{{100}}$
Simplifying the terms, we get
$\Rightarrow {\chi ^2} = 0.33 + 1 + 0.33 + 1$
$\Rightarrow {\chi ^2} = 2.66$
Hence, the value of ${\chi ^2}$ square statistics is $2.66$.