To find the probability that a student does not have an A given that the student is female, we use the formula:
\[
P(\text{Does not have an A} | \text{Female}) = \frac{P(\text{Does not have an A and Female})}{P(\text{Female})}
\]
From the data, we have:
- Number of females who do not have an A: \(6\)
- Total number of females: \(2 + 6 = 8\)
Thus, the probability is calculated as:
\[
P(\text{Does not have an A} | \text{Female}) = \frac{6}{8} = 0.7500
\]
We perform a Chi-Square Test of Independence to determine if gender and having an A are independent. The observed frequencies are:
\[
\text{Observed} = \begin{bmatrix}
2 & 3 \\
6 & 16
\end{bmatrix}
\]
The expected frequencies for each cell are calculated using the formula:
\[
E = \frac{R_i \times C_j}{N}
\]
Where \(R_i\) is the total for row \(i\), \(C_j\) is the total for column \(j\), and \(N\) is the total number of observations.
For cell (1, 1):
\[
E = \frac{5 \times 8}{27} = 1.4815
\]
For cell (1, 2):
\[
E = \frac{5 \times 19}{27} = 3.5185
\]
For cell (2, 1):
\[
E = \frac{22 \times 8}{27} = 6.5185
\]
For cell (2, 2):
\[
E = \frac{22 \times 19}{27} = 15.4815
\]
Thus, the expected frequencies are:
\[
\text{Expected} = \begin{bmatrix}
1.4815 & 3.5185 \\
6.5185 & 15.4815
\end{bmatrix}
\]
The Chi-Square statistic is calculated as:
\[
\chi^2 = \sum \frac{(O - E)^2}{E}
\]
Calculating for each cell:
For cell (1, 1):
\[
\frac{(2 - 1.4815)^2}{1.4815} = 0.1815
\]
For cell (1, 2):
\[
\frac{(3 - 3.5185)^2}{3.5185} = 0.0764
\]
For cell (2, 1):
\[
\frac{(6 - 6.5185)^2}{6.5185} = 0.0412
\]
For cell (2, 2):
\[
\frac{(16 - 15.4815)^2}{15.4815} = 0.0174
\]
Summing these values gives:
\[
\chi^2 = 0.1815 + 0.0764 + 0.0412 + 0.0174 = 0.3165
\]
The critical value at \(\alpha = 0.05\) for a Chi-Square distribution with \(1\) degree of freedom is:
\[
\chi^2_{\alpha, df} = 3.8415
\]
The p-value associated with the Chi-Square statistic is:
\[
P = P(\chi^2 > 0.3165) = 0.984
\]
The probability that a student does not have an A given that the student is female is:
\[
\boxed{0.7500}
\]
The results of the Chi-Square Test are:
- Chi-Square Statistic: \(0.3165\)
- Critical Value: \(3.8415\)
- Degrees of Freedom: \(1\)
- P-Value: \(0.984\)
Answered
