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%.
