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The proportion [tex]\( p \)[/tex] of residents in a community who recycle has traditionally been [tex]\( 70\% \)[/tex]. A policy maker claims that the proportion is less than [tex]\( 70\% \)[/tex] now that one of the recycling centers has been relocated. If 128 out of a random sample of 210 residents in the community said they recycle, is there enough evidence to support the policy maker's claim at the 0.01 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 [tex]\( H_0 \)[/tex] and the alternative hypothesis [tex]\( H_1 \)[/tex].

[tex]\[
\begin{array}{l}
H_0: p = 0.70 \\
H_1: p \ \textless \ 0.70
\end{array}
\][/tex]

(b) Determine the type of test statistic to use.

[tex]\[
\begin{array}{l}
\square \\
Z
\end{array}
\][/tex]

(c) Find the value of the test statistic. (Round to three or more decimal places.)

[tex]\[
-2.861
\][/tex]

(d) Find the [tex]\( p \)[/tex]-value. (Round to three or more decimal places.)

[tex]\[
0.002
\][/tex]

(e) Is there enough evidence to support the policy maker's claim that the proportion of residents who recycle is less than [tex]\( 70\% \)[/tex]?



Answer :

GinnyAnswer
To determine if there is enough evidence to support the policy maker's claim that the proportion of residents who recycle is less than 70%, we will perform a one-tailed hypothesis test. Below are the steps in a detailed, step-by-step solution:

### (a) State the null hypothesis [tex]\(H_0\)[/tex] and the alternative hypothesis [tex]\(H_1\)[/tex].

The null hypothesis ([tex]\(H_0\)[/tex]) and the alternative hypothesis ([tex]\(H_1\)[/tex]) are stated as follows:
[tex]\[ \begin{array}{l} H_0: p = 0.70 \\ H_1: p < 0.70 \end{array} \][/tex]

### (b) Determine the type of test statistic to use.

Since we are testing a hypothesis about a population proportion, we use the Z-test for proportions.

[tex]\[ \square \quad Z \][/tex]

### (c) Find the value of the test statistic. (Round to three or more decimal places.)

To find the value of the test statistic, we will use the formula for the Z-score in hypothesis testing for proportions:
[tex]\[ Z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1 - p_0)}{n}}} \][/tex]

Where:
- [tex]\(\hat{p}\)[/tex] is the sample proportion
- [tex]\(p_0\)[/tex] is the population proportion under the null hypothesis
- [tex]\(n\)[/tex] is the sample size

Given data:
- The number of residents who recycle ([tex]\(x\)[/tex]) is 128.
- The sample size ([tex]\(n\)[/tex]) is 210.
- The population proportion ([tex]\(p_0\)[/tex]) is 0.70.

First, we compute the sample proportion ([tex]\(\hat{p}\)[/tex]):
[tex]\[ \hat{p} = \frac{x}{n} = \frac{128}{210} = 0.6095238095238096 \][/tex]

Next, we will calculate the standard error (SE) of the sample proportion:
[tex]\[ SE = \sqrt{\frac{p_0(1 - p_0)}{n}} = \sqrt{\frac{0.70(1 - 0.70)}{210}} = 0.03162277660168379 \][/tex]

Now, we can find the Z-value:
[tex]\[ Z = \frac{0.6095238095238096 - 0.70}{0.03162277660168379} = -2.861 \][/tex]

### (d) Find the [tex]\(p\)[/tex]-value. (Round to three or more decimal places.)

The [tex]\(p\)[/tex]-value is the probability of observing a test statistic as extreme as, or more extreme than, the observed value under the null hypothesis. For a one-tailed test, we use the cumulative distribution function (CDF) of the standard normal distribution.

[tex]\[ p\text{-value} = P(Z < -2.861) = 0.002 \][/tex]

### (e) Is there enough evidence to support the policy maker's claim that the proportion of residents who recycle is less than 70%?

We compare the [tex]\(p\)[/tex]-value to the significance level [tex]\(\alpha\)[/tex]:
[tex]\[ p\text{-value} = 0.002 < \alpha = 0.01 \][/tex]

Since the [tex]\(p\)[/tex]-value is less than the significance level, we reject the null hypothesis [tex]\(H_0\)[/tex].

### Conclusion

There is enough evidence to support the policy maker's claim that the proportion of residents who recycle is now less than 70%.