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?

Question

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?

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:$$	ext{Expected frequency} = \frac{(	ext{row total})(	ext{column total})}{	ext{grand total}}$$The grand total is 110, the row totals are 80 and 30, and the column totals are 60 and 50.The expected frequencies are:$$	ext{Expected frequency (men, health insurance)} = \frac{(80)(60)}{110} = 43.64$$$$	ext{Expected frequency (men, no health insurance)} = \frac{(80)(50)}{110} = 36.36$$$$	ext{Expected frequency (women, health insurance)} = \frac{(30)(60)}{110} = 16.36$$$$	ext{Expected frequency (women, no health insurance)} = \frac{(30)(50)}{110} = 13.64$$The chi-square test statistic is:$$\chi^2 = \frac{(50 - 43.64)^2}{43.64} + \frac{(30 - 36.36)^2}{36.36} + \frac{(20 - 16.36)^2}{16.36} + \frac{(10 - 13.64)^2}{13.64} = 0.1637$$Therefore, the value of the test statistic is 0.1637.

A Survey Conducted in a Small Business Yielded the Results