A survey conducted in a small business yielded the results shown in the table. $\begin{array} { l c c } & \text { Men } & \text { Women } \ \text { Health insurance } & 50 & 20 \ \text { No health insurance } & 30 & 10 \end{array}$
Test the claim that health care coverage is independent of gender. Use a $0.05$ significance level. What is the value of the test statistic?
A) $\chi ^ { 2 } = 0.1637$
B) $\chi ^ { 2 } = 3.841$
C) $\chi ^ { 2 } = 8.263$
D) $\chi ^ { 2 } = 5.821$

Question

Quizplus · Accepted Answer

The chi-square test statistic is calculated as:$\chi ^ { 2 } = \sum \frac{(O - E) ^ { 2 }} { E }$where O is the observed frequency and E is the expected frequency.The expected frequencies are calculated as:$E = \frac { (row \ total) (column \ total) } { grand \ total }$The grand total is 110, the row totals are 80 and 30, and the column totals are 60 and 50.The expected frequencies are:$E (men, health insurance) = \frac { (80) (60) } { 110 } = 43.64$$E (men, no health insurance) = \frac { (80) (50) } { 110 } = 36.36$$E (women, health insurance) = \frac { (30) (60) } { 110 } = 16.36$$E (women, no health insurance) = \frac { (30) (50) } { 110 } = 13.64$The chi-square test statistic is:$\chi ^ { 2 } = \frac { (50 - 43.64) ^ { 2 }} { 43.64 } + \frac { (30 - 36.36) ^ { 2 }} { 36.36 } + \frac { (20 - 16.36) ^ { 2 }} { 16.36 } + \frac { (10 - 13.64) ^ { 2 }} { 13.64 } = 0.1637$The p-value for this test statistic is 0.921, which is greater than the significance level of 0.05. Therefore, we fail to reject the claim that health care coverage is independent of gender.

A Survey Conducted in a Small Business Yielded the Results