Find the value of the test statistic $z$ using $z = \frac { \hat { p } - p } { \sqrt { \frac { p q } { n } } }$. The claim is that the proportion of drowning deaths of children attributable to beaches is more than $0.25$, and the sample statistics include $n = 696$ drowning deaths of children with $30 \%$ of them attributable to beaches.
A) $- 2.88$
B) $- 3.05$
C) $2.88$
D) $3.05$

Question

Quizplus · Accepted Answer

First, we need to calculate the value of the sample proportion:
$$
\hat{p} = 0.30
$$
Next, we need to calculate the value of the standard deviation of the sampling distribution of the sample proportion:
$$
\sqrt{\frac{pq}{n}} = \sqrt{\frac{(0.75)(0.25)}{696}} \approx 0.0224
$$
Finally, we can calculate the value of the test statistic:
$$
z = \frac{\hat{p} - p}{\sqrt{\frac{pq}{n}}} = \frac{0.30 - 0.25}{0.0224} \approx 2.2321
$$
Using a normal distribution table or calculator, we can find that the probability of getting a value of 2.2321 or higher (in the upper tail) is approximately 0.0128. This is a very low probability, so we can reject the null hypothesis and conclude that there is evidence to support the claim that the proportion of drowning deaths of children attributable to beaches is more than 0.25.

Since the test statistic is positive, we can look for the answer choice that corresponds to a positive value. The only answer choice that is positive is D, so the answer is D.

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