Question

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

Answer 1

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.

Complete step-by-step answer:
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
\[ \Rightarrow {\chi ^2} = 0.33 + 1 + 0.33 + 1\]
Adding the terms, we get
\[ \Rightarrow {\chi ^2} = 2.66\]
Hence, the value of \[{\chi ^2}\] square statistics is \[2.66\].
Note: Statistics is the science of collecting some data in the form of the number and studying it to forecast or predict its future possibility. An array is an arrangement of numbers or symbols in rows and columns. Frequency is defined as the number of times a value or some particular data is repeating itself. Frequency arrays help us a lot for the prediction and probability purpose and it also helps us to analyse real life data like to predict the sale of some specific product or to analyse the sales of the same type of products in the market. A