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 148 out of a random sample of 215 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.) -Find the value of the test statistic and round to 3 or more decimal places.
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 148 out of a random sample of 215 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.) -Find the value of the test statistic and round to 3 or more decimal places.
MATLAB: An Introduction with Applications
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ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
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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 148 out of a random sample of 215 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.)
-Find the value of the test statistic and round to 3 or more decimal places.
(I have posted a picture of an example problem and the equation to use, with the correct answer as every expert I have asked thus far has gotten this problem wrong.)
![### Finding the Value of the Test Statistic
The value of this test statistic is the z-value corresponding to the sample proportion under the assumption that \( H_0 \) is true. Here is the formula and calculation:
\[ z = \frac{\hat{p} - p}{\sqrt{\frac{p(1-p)}{n}}} = \frac{\frac{36}{43} - 0.80}{\sqrt{\frac{0.80(1 - 0.80)}{215}}} \approx 1.364 \]
#### Explanation:
- \( \hat{p} \) is the sample proportion, calculated as \( \frac{36}{43} \).
- \( p \) is the population proportion under the null hypothesis \( H_0 \), which is given as 0.80.
- \( n \) is the sample size, which is given as 215.
The z-value is calculated to be approximately 1.364. This value represents how many standard deviations the sample proportion is away from the population proportion under the null hypothesis.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F715ecfc9-db17-4203-b60f-304ee331059c%2F7e48103b-70a1-4a2f-833c-2866d0355e53%2Fpe8wvp_processed.png&w=3840&q=75)
Transcribed Image Text:### Finding the Value of the Test Statistic
The value of this test statistic is the z-value corresponding to the sample proportion under the assumption that \( H_0 \) is true. Here is the formula and calculation:
\[ z = \frac{\hat{p} - p}{\sqrt{\frac{p(1-p)}{n}}} = \frac{\frac{36}{43} - 0.80}{\sqrt{\frac{0.80(1 - 0.80)}{215}}} \approx 1.364 \]
#### Explanation:
- \( \hat{p} \) is the sample proportion, calculated as \( \frac{36}{43} \).
- \( p \) is the population proportion under the null hypothesis \( H_0 \), which is given as 0.80.
- \( n \) is the sample size, which is given as 215.
The z-value is calculated to be approximately 1.364. This value represents how many standard deviations the sample proportion is away from the population proportion under the null hypothesis.
 of 215 married couples who completed her program, 180 of them stayed together. Based on this sample, is there enough evidence to support the marriage counselor’s claim at the [0.05 level of significance](https://www.investopedia.com/terms/l/level-of-significance.asp)?
### Steps to Follow:
1. **Perform a [one-tailed test](https://stattrek.com/statistics/dictionary.aspx?definition=one-tailed%20test)**. Then complete the parts below.
2. Carry your intermediate computations to three or more decimal places. (If necessary, consult a [list of formulas](https://stattrek.com/hypothesis-test/formulas.aspx).)
### Key Concepts:
- **Proportion \( p \)**: Represents the traditional success rate of the program (0.80 or 80%).
- **Random Sample**: A subset of individuals chosen from a population, in which every individual has an equal probability of being selected.
- **Level of Significance (0.05)**: The probability threshold below which the null hypothesis will be rejected, indicating that the results are statistically significant.
### Sample Data and Analysis:
- **Sample Size (n)**: 215 couples
- **Number of Couples Stayed Together**: 180
Use this information to determine if the improvements in the program have significantly increased the success rate above 80%.
### Conclusion:
Determine if there is enough statistical evidence, based on the performed test, to support the marriage counselor's new claim that her improved program can prevent divorce in more than 80% of married couples.
For more detailed steps on calculating and interpreting the results, you can refer to the provided links on statistical concepts.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F715ecfc9-db17-4203-b60f-304ee331059c%2F7e48103b-70a1-4a2f-833c-2866d0355e53%2Frw3nwc_processed.png&w=3840&q=75)
Transcribed Image Text:### Evaluating the Effectiveness of a Communication Program in Preventing Divorce
A marriage counselor has traditionally seen that the proportion \( p \) of all married couples for whom her communication program can prevent divorce is 80%. After making some recent changes, the marriage counselor now claims that her program can prevent divorce in more than 80% of married couples.
In a [random sample](https://www.surveymonkey.com/mp/random-sample/) of 215 married couples who completed her program, 180 of them stayed together. Based on this sample, is there enough evidence to support the marriage counselor’s claim at the [0.05 level of significance](https://www.investopedia.com/terms/l/level-of-significance.asp)?
### Steps to Follow:
1. **Perform a [one-tailed test](https://stattrek.com/statistics/dictionary.aspx?definition=one-tailed%20test)**. Then complete the parts below.
2. Carry your intermediate computations to three or more decimal places. (If necessary, consult a [list of formulas](https://stattrek.com/hypothesis-test/formulas.aspx).)
### Key Concepts:
- **Proportion \( p \)**: Represents the traditional success rate of the program (0.80 or 80%).
- **Random Sample**: A subset of individuals chosen from a population, in which every individual has an equal probability of being selected.
- **Level of Significance (0.05)**: The probability threshold below which the null hypothesis will be rejected, indicating that the results are statistically significant.
### Sample Data and Analysis:
- **Sample Size (n)**: 215 couples
- **Number of Couples Stayed Together**: 180
Use this information to determine if the improvements in the program have significantly increased the success rate above 80%.
### Conclusion:
Determine if there is enough statistical evidence, based on the performed test, to support the marriage counselor's new claim that her improved program can prevent divorce in more than 80% of married couples.
For more detailed steps on calculating and interpreting the results, you can refer to the provided links on statistical concepts.
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