Find the value of the test statistic z using z $z = \frac { \hat { p } - p } { \sqrt { \frac { p q } { n } } }$
-The claim is that the proportion of accidental deaths of the elderly attributable to residential falls is more than 0.10, and the sample statistics include n = 800 deaths of the elderly with 15% of them attributable to residential
Falls.
A) -3.96
B) -4.71
C) 3.96
D) 4.71

Question

Quizplus · Accepted Answer

Here,

n = 800 (sample size)
p = 0.10 (null hypothesis proportion of accidental deaths of the elderly attributable to residential falls) 
q = 1 - p = 0.9 
$\hat{p}$ = 0.15 (sample proportion of accidental deaths of the elderly attributable to residential falls)

Now,

$z = \frac {\hat{p} - p}{\sqrt{\frac{pq}{n}}} = \frac{0.15 - 0.10}{\sqrt{\frac{0.1 \cdot 0.9}{800}}} \approx 4.71$

Since the alternative hypothesis is that the proportion is more than 0.10, this is a right-tailed test. The area to the right of $z = 4.71$ is very close to 0, so we can conclude that the sample provides very strong evidence to support the alternative hypothesis. Therefore, the best choice is to select option D, 4.71.

Find the Value of the Test Statistic Z Using Z z=p^−ppqnz = \frac { \hat { p } - p } { \sqrt { \frac { p q } { n } } }z=npq​​p^​−p​

Find the Value of the Test Statistic Z Using Z $z = \frac { \hat { p } - p } { \sqrt { \frac { p q } { n } } }$