Find the value of the test statistic z using $z = \frac { \hat { p } - p } { \sqrt { \frac { p q } { n } } }$ A claim is made that the proportion of children who play sports is less than 0.5, and the sample statistics include n = 1320 subjects with 30% saying that they play a sport.
A) -29.66
B) 29.66
C) 14.53
D) -14.53

Question

Quizplus · Accepted Answer

Given:
Number of subjects, n = 1320
Sample proportion, $\hat p$ = 0.30
Claimed proportion, $p_0$ = 0.5

Calculate:
$q = 1-p = 1-0.5 = 0.5$
$ \frac { p q } { n } = \frac { 0.5 * 0.5 } { 1320 } = 0.000188 $
$z = \frac { \hat { p } - p } { \sqrt { \frac { p q } { n } } } = \frac {0.30 - 0.5} { \sqrt {0.000188} } \approx -14.53$

Since the claimed proportion is greater than the sample proportion, we expect the test statistic to be negative. The closest answer choice to -14.53 is D, so the answer is D.

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