13. Find the value of the test statistic [tex]z[/tex] using

[tex]z=\frac{\hat{p} - p}{\sqrt{\frac{pq}{n}}}[/tex].

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 [tex]n=800[/tex] deaths of the elderly with [tex]15\%[/tex] of them attributable to residential falls.

A. 3.96
B. -3.96
C. 4.71
D. -4.71



Answer :

To find the value of the test statistic [tex]\( z \)[/tex] using the given formula [tex]\( z = \frac{\hat{p} - p}{\sqrt{\frac{p q}{n}}} \)[/tex], we need to follow these steps methodically:

1. Identify the given values:
- [tex]\( p \)[/tex] (the population proportion claim) = 0.10
- [tex]\( \hat{p} \)[/tex] (the sample proportion) = 0.15
- [tex]\( n \)[/tex] (the sample size) = 800

2. Calculate [tex]\( q \)[/tex], which is [tex]\( 1 - p \)[/tex]:
[tex]\[ q = 1 - p = 1 - 0.10 = 0.90 \][/tex]

3. Find the standard error of the proportion:
[tex]\[ \text{Standard Error} = \sqrt{\frac{p \cdot q}{n}} = \sqrt{\frac{0.10 \cdot 0.90}{800}} \][/tex]
Plugging in the values:
[tex]\[ \sqrt{\frac{0.09}{800}} = \sqrt{0.0001125} \approx 0.01061 \][/tex]

4. Calculate the test statistic [tex]\( z \)[/tex]:
[tex]\[ z = \frac{\hat{p} - p}{\text{Standard Error}} = \frac{0.15 - 0.10}{0.01061} \approx \frac{0.05}{0.01061} \approx 4.71 \][/tex]

So, the value of the test statistic [tex]\( z \)[/tex] is approximately [tex]\( 4.71 \)[/tex].

Therefore, the correct answer is:
[tex]\[ \boxed{4.71} \][/tex]