Answer to Question #160738 in Statistics and Probability for amna

Question

A study examined rates of posttraumatic stress disorder (PTSD) in mothers and fathers. Parents were interviewed 5 to 6 weeks after an accident or a new diagnosis of cancer or diabetes mellitus type I for their child. Twenty- eight of the 175 fathers interviewed and 43 of the 180 mothers interviewed met the criteria for current PTSD. Is there sufficient evidence for us to conclude that fathers are less likely to develop PTSD than mothers when a child is traumatized by an accident, cancer diagnosis, or diabetes diagnosis? Let α=5%.

Answer 1

Fathers:

$n_1 = 175 \\ x_1 = 28$

Mothers:

$n_2 = 180 \\ x_2 = 43 \\ α = 0.05$

$H_0 = p_1 = p_2 \\ H_1 = p_1 < p_2 \\ \hat{p_1} = \frac{x_1}{n_1} \\ = \frac{28}{175} \\ = 0.16 \\ \hat{p_2} = \frac{x_2}{n_2} \\ = \frac{43}{180} \\ = 0.2389 \\ \bar{p} = \frac{x_1+x_2}{n_1+n_2} \\ = \frac{28+43}{175+180} \\ = 0.2$

Test-statistic:

$Z = \frac{\hat{p_1}- \hat{p_2}}{\bar{p}(1-\bar{p})( \frac{1}{n_1} + \frac{1}{n_2} )} \\ = \frac{0.16- 0.2389}{0.2(1-0.2)( \frac{1}{175} + \frac{1}{180} )} \\ = -1.858$

Z-critical value = -1.645

Since, Z-calc. < Z-crit. Hence reject the null hypothesis.

There is enough evidence to conclude that the rate of PTSD in fathers is less than its rate in mothers.