Question

In a survey of 200 boys, 75 were intelligent and out of these intelligent boys, 40 had an education from the government schools. Out of not intelligent boys, 85 had an education form the private schools. Then, the value of the test statistic, to test the hypothesis that there is no association between the education from the schools and intelligence of boys, is:

• A. 7.80
• B. 6.28
• C. 4.80
• D. 8.89

Hint:

For a 2x2 contingency table, a faster formula for the Chi-squared statistic is: \[ \chi^2 = \frac{N(ad-bc)^2}{(a+b)(c+d)(a+c)(b+d)} \] Where a, b, c, d are the cell frequencies and N is the grand total. In this case: \(a=40, b=35, c=40, d=85\). \[ \chi^2 = \frac{200(40 . 85 - 35 . 40)^2}{(75)(125)(80)(120)} = \frac{200(3400 - 1400)^2}{90000000} = \frac{200(2000)^2}{90000000} = \frac{800000000}{90000000} = \frac{80}{9} \approx 8.89 \] This avoids calculating expected values separately and can be quicker.

Answer 1

The Correct Option is D

Step 1: Concept Overview:
This problem involves a Chi-squared (\(\chi^2\)) test for independence to assess the relationship between two categorical variables: 'intelligence' (intelligent/not intelligent) and 'school type' (government/private). The null hypothesis assumes no association between these variables.

Step 2: Core Formula:
The Chi-squared test statistic is computed as follows: \[ \chi^2 = \sum \frac{(O - E)^2}{E} \] Where: - \(O\) represents the observed frequency in each contingency table cell. - \(E\) represents the expected frequency in each cell, assuming independence. The expected frequency is calculated using: \[ E = \frac{(\text{Row Total}) \times (\text{Column Total})}{\text{Grand Total}} \]
Step 3: Detailed Calculation:
First, create a 2x2 contingency table using the observed frequencies (O) from the provided data.- Total boys: 200.- Intelligent boys: 75. Thus, Not Intelligent boys: 200 - 75 = 125.- Intelligent boys from Government schools: 40.- Intelligent boys from Private schools: 75 - 40 = 35.- Not Intelligent boys from Private schools: 85.- Not Intelligent boys from Government schools: 125 - 85 = 40.Observed Frequencies (O):\begin{center}\begin{tabular}{|l|c|c|c|}\hline & Government & Private & Row Total
\hlineIntelligent & 40 & 35 & 75
\hlineNot Intelligent & 40 & 85 & 125
\hlineColumn Total & 80 & 120 & 200
\hline\end{tabular}\end{center}Next, determine the expected frequencies (E) for each cell:- E(Int, Gov) = \(\frac{75 \times 80}{200} = 30\)- E(Int, Pri) = \(\frac{75 \times 120}{200} = 45\)- E(Not Int, Gov) = \(\frac{125 \times 80}{200} = 50\)- E(Not Int, Pri) = \(\frac{125 \times 120}{200} = 75\)Now, calculate the \(\chi^2\) statistic:\[ \chi^2 = \frac{(40 - 30)^2}{30} + \frac{(35 - 45)^2}{45} + \frac{(40 - 50)^2}{50} + \frac{(85 - 75)^2}{75} \]\[ \chi^2 = \frac{10^2}{30} + \frac{(-10)^2}{45} + \frac{(-10)^2}{50} + \frac{10^2}{75} \]\[ \chi^2 = \frac{100}{30} + \frac{100}{45} + \frac{100}{50} + \frac{100}{75} \]\[ \chi^2 = \frac{10}{3} + \frac{20}{9} + 2 + \frac{4}{3} \]\[ \chi^2 = \frac{30}{9} + \frac{20}{9} + \frac{18}{9} + \frac{12}{9} = \frac{30+20+18+12}{9} = \frac{80}{9} \]\[ \chi^2 \approx 8.888... \]
Step 4: Result:
The Chi-squared test statistic is approximately 8.89.