Test a claim that the mean amount of carbon monoxide in the air in U.S. cities is less than 2.31 parts per million. It was found that the mean amount of carbon monoxide in the air for the random sample of 66 cities is 2.41 parts per million and the standard deviation is 2.09 parts per million. At a=0.10, can the claim be supported? Complete parts (a) through (e) below. Assume the population is normally distributed.
Test a claim that the mean amount of carbon monoxide in the air in U.S. cities is less than 2.31 parts per million. It was found that the mean amount of carbon monoxide in the air for the random sample of 66 cities is 2.41 parts per million and the standard deviation is 2.09 parts per million. At a=0.10, can the claim be supported? Complete parts (a) through (e) below. Assume the population is normally distributed.
MATLAB: An Introduction with Applications
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Author:Amos Gilat
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![### Hypothesis Testing for Mean Amount of Carbon Monoxide in U.S. Cities
**Problem Statement:**
Test a claim that the mean amount of carbon monoxide in the air in U.S. cities is less than 2.31 parts per million. It was found that the mean amount of carbon monoxide in the air for the random sample of 66 cities is 2.41 parts per million and the standard deviation is 2.09 parts per million. At α = 0.10, can the claim be supported? Complete parts (a) through (e) below. Assume the population is normally distributed.
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#### (a) Identify the claim and state \(H_0\) and \(H_a\). Which of the following correctly states \(H_0\) and \(H_a\)?
\[ H_0: \mu \geq 2.31 \]
\[ H_a: \mu < 2.31 \]
(Type integers or decimals. Do not round.)
The claim is the \( H_a \) hypothesis.
---
#### (b) Use technology to find the critical value(s) and identify the rejection region(s).
The critical value(s) is/are \( t_0 = \_\_\_\_ \)
(Use a comma to separate answers as needed. Round to two decimal places as needed.)
---](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F952bcd3e-4a7c-474a-8dd6-ddd08471460a%2Fb6ed644b-e87d-4841-97f9-aba2f48936d2%2Fhowcm0s_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Hypothesis Testing for Mean Amount of Carbon Monoxide in U.S. Cities
**Problem Statement:**
Test a claim that the mean amount of carbon monoxide in the air in U.S. cities is less than 2.31 parts per million. It was found that the mean amount of carbon monoxide in the air for the random sample of 66 cities is 2.41 parts per million and the standard deviation is 2.09 parts per million. At α = 0.10, can the claim be supported? Complete parts (a) through (e) below. Assume the population is normally distributed.
---
#### (a) Identify the claim and state \(H_0\) and \(H_a\). Which of the following correctly states \(H_0\) and \(H_a\)?
\[ H_0: \mu \geq 2.31 \]
\[ H_a: \mu < 2.31 \]
(Type integers or decimals. Do not round.)
The claim is the \( H_a \) hypothesis.
---
#### (b) Use technology to find the critical value(s) and identify the rejection region(s).
The critical value(s) is/are \( t_0 = \_\_\_\_ \)
(Use a comma to separate answers as needed. Round to two decimal places as needed.)
---
![### Hypothesis Testing of Carbon Monoxide Levels in U.S. Cities
**Problem Statement:**
Test a claim that the mean amount of carbon monoxide in the air in U.S. cities is less than 2.31 parts per million. It was found that the mean amount of carbon monoxide in the air for the random sample of 66 cities is 2.41 parts per million and the standard deviation is 2.09 parts per million. At an alpha level of 0.10, can the claim be supported? Complete parts (a) through (e) below. Assume the population is normally distributed.
**Part (a): Choosing the Graph for the Rejection Region**
Choose the graph which shows the rejection region.
- **Option A:**
Depicts a one-tailed test with the rejection region on the right side (t > t₀).
- **Option B:**
Depicts a two-tailed test with the rejection regions on both sides (−t₀ < t < t₀).
- **Option C:**
Depicts a two-tailed test with the rejection regions on both sides (t < −t₀ and t > t₀).
- **Option D:**
Depicts a one-tailed test with the rejection region on the left side (t < t₀).
**Graphical Explanation:**
- In **Option A**, the shaded region shows where the test statistic falls to the right of the critical value, indicating a right-tailed test.
- In **Option B**, the shaded regions show where the test statistic falls within the critical values on both sides, indicating a two-tailed test.
- In **Option C**, the shaded regions show where the test statistic falls outside the critical values on both sides, indicating another form of two-tailed test.
- In **Option D**, the shaded region shows where the test statistic falls to the left of the critical value, indicating a left-tailed test.
**Find the Standardized Test Statistic**
Part (c) asks to find the standardized test statistic, t. This involves calculating the t-value using the sample data and the known parameters.
To proceed with the calculation of the t-value, you use the formula:
\[ t = \frac{\bar{X} - \mu}{\frac{s}{\sqrt{n}}} \]
Where:
- \(\bar{X}\) = Sample mean (2.41 parts per](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F952bcd3e-4a7c-474a-8dd6-ddd08471460a%2Fb6ed644b-e87d-4841-97f9-aba2f48936d2%2Fxy4xc44_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Hypothesis Testing of Carbon Monoxide Levels in U.S. Cities
**Problem Statement:**
Test a claim that the mean amount of carbon monoxide in the air in U.S. cities is less than 2.31 parts per million. It was found that the mean amount of carbon monoxide in the air for the random sample of 66 cities is 2.41 parts per million and the standard deviation is 2.09 parts per million. At an alpha level of 0.10, can the claim be supported? Complete parts (a) through (e) below. Assume the population is normally distributed.
**Part (a): Choosing the Graph for the Rejection Region**
Choose the graph which shows the rejection region.
- **Option A:**
Depicts a one-tailed test with the rejection region on the right side (t > t₀).
- **Option B:**
Depicts a two-tailed test with the rejection regions on both sides (−t₀ < t < t₀).
- **Option C:**
Depicts a two-tailed test with the rejection regions on both sides (t < −t₀ and t > t₀).
- **Option D:**
Depicts a one-tailed test with the rejection region on the left side (t < t₀).
**Graphical Explanation:**
- In **Option A**, the shaded region shows where the test statistic falls to the right of the critical value, indicating a right-tailed test.
- In **Option B**, the shaded regions show where the test statistic falls within the critical values on both sides, indicating a two-tailed test.
- In **Option C**, the shaded regions show where the test statistic falls outside the critical values on both sides, indicating another form of two-tailed test.
- In **Option D**, the shaded region shows where the test statistic falls to the left of the critical value, indicating a left-tailed test.
**Find the Standardized Test Statistic**
Part (c) asks to find the standardized test statistic, t. This involves calculating the t-value using the sample data and the known parameters.
To proceed with the calculation of the t-value, you use the formula:
\[ t = \frac{\bar{X} - \mu}{\frac{s}{\sqrt{n}}} \]
Where:
- \(\bar{X}\) = Sample mean (2.41 parts per
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