Question

the proportion ( p ) of residents in a community who recycle has traditionally been 60%. a policy maker claims that the proportion is less than 60% now that one of the recycling centers has been relocated. if 136 out of a random sample of 235 residents in the community said they recycle, is there enough evidence to support the policy maker’s claim at the 0.05 level of significance?
perform a one - tailed test. then complete the parts below.
carry your intermediate computations to three or more decimal places. (if necessary, consult a list of formulas.)
(a) state the null hypothesis ( h_0 ) and the alternative hypothesis ( h_1 ).
( h_0: square )
( h_1: square )
(b) determine the type of test statistic to use.
(c) find the value of the test statistic. (round to three or more decimal places.)

Explanation:

Step1: State hypotheses

We test if proportion < 60%.
$H_0: p = 0.60$
$H_1: p < 0.60$

Step2: Choose test statistic

For population proportion, use z-test.
Test type: z-test for proportion

Step3: Calculate sample proportion

Sample proportion $\hat{p} = \frac{136}{235}$
$\hat{p} \approx 0.5787$

Step4: Compute test statistic

Use formula $z = \frac{\hat{p} - p}{\sqrt{\frac{p(1-p)}{n}}}$
$z = \frac{0.5787 - 0.60}{\sqrt{\frac{0.60(1-0.60)}{235}}}$
First compute denominator: $\sqrt{\frac{0.60 \times 0.40}{235}} = \sqrt{\frac{0.24}{235}} \approx \sqrt{0.001021} \approx 0.03195$
Then $z = \frac{-0.0213}{0.03195} \approx -0.667$

Answer:

(a) $H_0: p = 0.60$; $H_1: p < 0.60$
(b) z-test for a population proportion
(c) $-0.667$