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: Understanding the Concept:
The problem requires performing a Chi-squared (\(\chi^2\)) test for independence. This test is used to determine if there is a significant association between two categorical variables: in this case, 'intelligence' (intelligent/not intelligent) and 'type of school' (government/private). The null hypothesis is that there is no association between the two variables.

Step 2: Key Formula or Approach:
The Chi-squared test statistic is calculated as: \[ \chi^2 = \sum \frac{(O - E)^2}{E} \] Where: - \(O\) is the observed frequency in each cell of the contingency table. - \(E\) is the expected frequency in each cell, calculated under the assumption of independence. The formula for expected frequency is: \[ E = \frac{(\text{Row Total}) \times (\text{Column Total})}{\text{Grand Total}} \]
Step 3: Detailed Explanation:
First, construct the 2x2 contingency table of observed frequencies (O) from the given data. - Total boys = 200. - Intelligent boys = 75. So, Not Intelligent boys = 200 - 75 = 125. - Intelligent and from Government schools = 40. - Intelligent and from Private schools = 75 - 40 = 35. - Not Intelligent and from Private schools = 85. - Not Intelligent and from Government schools = 125 - 85 = 40. Observed Frequencies (O): \begin{center} \begin{tabular}{|l|c|c|c|} \hline & Government & Private & Row Total
\hline Intelligent & 40 & 35 & 75
\hline Not Intelligent & 40 & 85 & 125
\hline Column Total & 80 & 120 & 200
\hline \end{tabular} \end{center} Next, calculate 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: Final Answer:
The value of the test statistic is approximately 8.89.