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 123 out of a random sample of 240 residents in the community said they recycle, is there enough evidence to support the policy makers claim at the 0.10 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 h0 and the alternative hypothesis h1. h0:□ h1:□ (b) determine the type of test statistic to use. select (c) find the value of the test statistic. (round to three or more decimal places.) □ (d) find the p - value. (round to three or more decimal places.) □ (e) is there enough evidence to support the policy makers claim that the proportion of residents who recycle is less than 60%? yes no

Explanation:

Step1: State hypotheses

The null hypothesis $H_0$ is that the proportion $p$ is equal to the traditional proportion. The alternative hypothesis $H_1$ is the policy - maker's claim.
$H_0:p = 0.6$
$H_1:p<0.6$

Step2: Determine test - statistic type

We are dealing with a hypothesis test for a proportion, so we use the z - test statistic for a proportion.
The formula for the z - test statistic 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 proportion, and $n$ is the sample size.

Step3: Calculate sample proportion

The sample size $n = 240$ and the number of residents who recycle in the sample $x = 123$. The sample proportion $\hat{p}=\frac{x}{n}=\frac{123}{240}=0.5125$.

Step4: Calculate test - statistic value

$p_0 = 0.6$, $n = 240$, $\hat{p}=0.5125$.
$z=\frac{0.5125 - 0.6}{\sqrt{\frac{0.6\times(1 - 0.6)}{240}}}=\frac{- 0.0875}{\sqrt{\frac{0.6\times0.4}{240}}}=\frac{-0.0875}{\sqrt{\frac{0.24}{240}}}=\frac{-0.0875}{\sqrt{0.001}}\approx\frac{-0.0875}{0.0316}\approx - 2.769$.

Step5: Calculate p - value

Since this is a left - tailed test, the p - value is the probability of getting a z - score less than the calculated z - value. Using a standard normal distribution table or calculator, the p - value for $z=-2.769$ is $P(Z < - 2.769)\approx0.003$.

Step6: Make a decision

The level of significance $\alpha = 0.10$. Since the p - value ($0.003$) is less than $\alpha(0.10)$, we reject the null hypothesis.

Answer:

(a) $H_0:p = 0.6$, $H_1:p<0.6$
(b) z - test statistic for a proportion
(c) $-2.769$
(d) $0.003$
(e) Yes