Questions: Question 5 (2 points) Use chi-square testing to solve the following problem: A survey conducted in a small business yielded the results shown in the table. Men Women ----------------- Health Insurance 50 20 No Health Insurance 30 10 Test the claim that health care coverage is independent of gender. Use a 0.05 significance level.

Solution Steps

The observed frequencies from the survey are represented in the contingency table as follows:

\[ \begin{array}{|c|c|c|} \hline & \text{Men} & \text{Women} \\ \hline \text{Health Insurance} & 50 & 20 \\ \hline \text{No Health Insurance} & 30 & 10 \\ \hline \end{array} \]

To calculate the expected frequencies \(E\) for each cell in the table, we use 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 grand total.

Calculating the expected frequencies:

• For cell (1, 1): \[ E = \frac{70 \times 80}{110} = 50.9091 \]

• For cell (1, 2): \[ E = \frac{70 \times 30}{110} = 19.0909 \]

• For cell (2, 1): \[ E = \frac{40 \times 80}{110} = 29.0909 \]

• For cell (2, 2): \[ E = \frac{40 \times 30}{110} = 10.9091 \]

Thus, the expected frequencies are:

\[ \begin{array}{|c|c|c|} \hline & \text{Men} & \text{Women} \\ \hline \text{Health Insurance} & 50.9091 & 19.0909 \\ \hline \text{No Health Insurance} & 29.0909 & 10.9091 \\ \hline \end{array} \]

The Chi-Square test statistic \(\chi^2\) is calculated using the formula:

\[ \chi^2 = \sum \frac{(O - E)^2}{E} \]

where \(O\) is the observed frequency and \(E\) is the expected frequency.

Calculating for each cell:

• For cell (1, 1): \[ \frac{(50 - 50.9091)^2}{50.9091} = 0.0162 \]

• For cell (1, 2): \[ \frac{(20 - 19.0909)^2}{19.0909} = 0.0433 \]

• For cell (2, 1): \[ \frac{(30 - 29.0909)^2}{29.0909} = 0.0284 \]

• For cell (2, 2): \[ \frac{(10 - 10.9091)^2}{10.9091} = 0.0758 \]

Summing these values gives:

\[ \chi^2 = 0.0162 + 0.0433 + 0.0284 + 0.0758 = 0.1637 \]

The critical value for a Chi-Square distribution with 1 degree of freedom at \(\alpha = 0.05\) is:

\[ \chi^2_{\alpha, df} = 3.8415 \]

The p-value associated with the calculated Chi-Square statistic is:

\[ P = P(\chi^2 > 0.1637) = 0.8555 \]

Since the p-value \(0.8555\) is greater than the significance level \(\alpha = 0.05\), we fail to reject the null hypothesis. This indicates that there is not enough evidence to suggest that health care coverage is dependent on gender.

Final Answer