Question

a medical researcher says that less than 72% of adults in a certain country think that healthy children should be required to be vaccinated. in a random sample of 500 adults in that country, 69% think that healthy children should be required to be vaccinated. at \\(\alpha = 0.05\\), is there enough evidence to support the researchers claim? complete parts (a) through (d) below.

a. the rejection region is \\(z < -1.65\\).
b. the rejection regions are \\(z < \square\\) and \\(z > \square\\).
c. the rejection region is \\(z > \square\\).
d. the rejection region is \\(\square < z < \square\\).

(c) find the standardized test statistic \\(z\\).
\\(z = \square\\) (round to two decimal places as needed.)

Explanation:

Step1: Identify the test type

This is a one - tailed (left - tailed) hypothesis test for a proportion. The formula for the standardized test statistic \(z\) for a proportion is \(z=\frac{\hat{p}-p_0}{\sqrt{\frac{p_0(1 - p_0)}{n}}}\), where \(\hat{p}\) is the sample proportion, \(p_0\) is the hypothesized population proportion, and \(n\) is the sample size.

Step2: Define the values

We are given that \(p_0 = 0.72\), \(\hat{p}=0.69\), and \(n = 500\).

Step3: Calculate the standard error

First, calculate the standard error \(SE=\sqrt{\frac{p_0(1 - p_0)}{n}}\). Substitute the values:
\(SE=\sqrt{\frac{0.72\times(1 - 0.72)}{500}}=\sqrt{\frac{0.72\times0.28}{500}}=\sqrt{\frac{0.2016}{500}}=\sqrt{0.0004032}\approx0.0201\)

Step4: Calculate the z - statistic

Now, use the formula for \(z\):
\(z=\frac{\hat{p}-p_0}{SE}=\frac{0.69 - 0.72}{0.0201}=\frac{- 0.03}{0.0201}\approx - 1.49\)

Answer:

\(z\approx - 1.49\)