A poll asked random samples of adults in 2016 and 2018 if they were satisfied with the quality of the environment. In 2016, 533 were satisfied with the quality of the environment and 455 were dissatisfied. In 2018, 470 were satisfied and 515 were dissatisfied. Determine whether the proportion of adults who are satisfied with the quality of the environment has declined. Use a 0.10 significance level.
Let $p_{1}$ represent the population proportion of adults in 2016 who were satisfied with the quality of the environment and let $p_{2}$ represent the population proportion of adults in 2018 who were satisfied with the quality of the environment. Write the hypotheses for the test.
\[
\begin{array}{l}
H_{0}: p_{1}=p_{2} \\
H_{a}: p_{1}> p_{2}
\end{array}
\]
Find the test statistic for this test.
$z=\square$ (Round to two decimal places as needed.)
Answer
\(\boxed{z = 2.82}\) is the final answer.
Step 1 ：Calculate the sample proportions for 2016 and 2018. For 2016, the sample proportion (p1) is \(\frac{533}{533 + 455} = 0.539\). For 2018, the sample proportion (p2) is \(\frac{470}{470 + 515} = 0.477\).
Step 2 ：Calculate the pooled proportion (p), which is \(\frac{533 + 470}{533 + 455 + 470 + 515} = 0.508\).
Step 3 ：Calculate the test statistic (z) using the formula for the z-score of a proportion: \(z = \frac{p1 - p2}{\sqrt{p * (1 - p) * [(1/n1) + (1/n2)]}}\), where n1 and n2 are the total number of adults surveyed in 2016 and 2018, respectively.
Step 4 ：Substitute the values into the formula: \(z = \frac{0.539 - 0.477}{\sqrt{0.508 * (1 - 0.508) * [(1/988) + (1/985)]}}\).
Step 5 ：Calculate the above expression to get the test statistic: \(z = \frac{0.062}{0.022} = 2.82\).
Step 6 ：\(\boxed{z = 2.82}\) is the final answer.
