Question

the proportion p of residents in a community who recycle has traditionally been 70%. a policy maker claims that the proportion is less than 70% now that one of the recycling centers has been relocated. if 138 out of a random sample of 210 residents in the community said they recycle, is there enough evidence to support the policy makers 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 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 70%? oyes ono

Explanation:

Step1: State the hypotheses

The null hypothesis $H_0$ is that the proportion $p$ is equal to the traditional proportion. The alternative hypothesis $H_1$ is the claim of the policy - maker.
$H_0:p = 0.7$
$H_1:p<0.7$

Step2: Determine the test - statistic type

For testing a proportion, 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 the sample proportion $\hat{p}$

The sample size $n = 210$ and the number of successes $x = 138$. The sample proportion $\hat{p}=\frac{x}{n}=\frac{138}{210}\approx0.65714$.

Step4: Calculate the z - test statistic

We have $p_0 = 0.7$, $n = 210$, and $\hat{p}=0.65714$.
$z=\frac{\hat{p}-p_0}{\sqrt{\frac{p_0(1 - p_0)}{n}}}=\frac{0.65714 - 0.7}{\sqrt{\frac{0.7\times(1 - 0.7)}{210}}}$
$=\frac{- 0.04286}{\sqrt{\frac{0.7\times0.3}{210}}}=\frac{-0.04286}{\sqrt{\frac{0.21}{210}}}=\frac{-0.04286}{\sqrt{0.001}}=\frac{-0.04286}{0.03162}\approx - 1.356$.

Step5: Calculate the 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 $z=-1.356$ is $P(Z < - 1.356)\approx0.087$.

Step6: Make a decision

The significance level $\alpha = 0.05$. Since the p - value ($0.087$) is greater than $\alpha(0.05)$, we fail to reject the null hypothesis.

Answer:

(a) $H_0:p = 0.7$, $H_1:p<0.7$
(b) z - test statistic for a proportion
(c) $-1.356$
(d) $0.087$
(e) No