Test a claim that the mean amount of carbon monoxide in the air in U.S. cities is less than 2.34 parts per million. It was found that the mean amount of carbon monoxide in the air for the random sample of 62 cities is 2.41 parts per million and the standard deviation is 2.09 parts per million. At a = 0.01, can the claim be supported? Complete parts (a) through (e) below. Assume the population is normally distributed. C ▼ (a) Identify the claim and state Ho and H₂. Which of the following correctly states Ho and H₂? Но Ha ▼ ▼ (Type integers or decimals. Do not round.) The claim is the hypothesis. (b) Use technology to find the critical value(s) and identify the rejection region(s). The critical value(s) is/are to = - (Use a comma to separate answers as needed. Round to two decimal places as needed.) Choose the graph which shows the rejection region. OP. M
Test a claim that the mean amount of carbon monoxide in the air in U.S. cities is less than 2.34 parts per million. It was found that the mean amount of carbon monoxide in the air for the random sample of 62 cities is 2.41 parts per million and the standard deviation is 2.09 parts per million. At a = 0.01, can the claim be supported? Complete parts (a) through (e) below. Assume the population is normally distributed. C ▼ (a) Identify the claim and state Ho and H₂. Which of the following correctly states Ho and H₂? Но Ha ▼ ▼ (Type integers or decimals. Do not round.) The claim is the hypothesis. (b) Use technology to find the critical value(s) and identify the rejection region(s). The critical value(s) is/are to = - (Use a comma to separate answers as needed. Round to two decimal places as needed.) Choose the graph which shows the rejection region. OP. M
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
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![### Hypothesis Testing: Mean 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 \(2.34\) parts per million (ppm). It was found that the mean amount of carbon monoxide in the air for the random sample of \(62\) cities is \(2.41\) ppm and the standard deviation is \(2.09\) ppm. At \(\alpha = 0.01\), 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 \(H_a\)?
- \(H_0\) : [Dropdown selection]
- \(H_a\) : [Dropdown selection]
(Type integers or decimals. Do not round.)
*The claim is the [Dropdown selection] hypothesis.*
#### (b) Use Technology to Find the Critical Value(s) and Identify the Rejection Region(s)
The critical value(s) is/are \( t_0 = \) [Textbox for input].
(Use a comma to separate answers as needed. Round to two decimal places as needed.)
*Choose the graph which shows the rejection region.*
[Radio button selection]
- \( \text{Graph A} \)
- \( \text{Graph B} \)
- \( \text{Graph C} \)
- \( \text{Graph D} \)
### Explanation of Graphs (if Provided)
If there are graphs included, they typically represent the critical value and the rejection regions. For example:
- **Graph A:** May show a two-tailed test with two rejection regions on either end of the distribution.
- **Graph B:** Might depict a one-tailed test with a rejection region on the left end of the distribution.
- **Graph C:** Could illustrate a one-tailed test with a rejection region on the right end of the distribution.
- **Graph D:** Would potentially show no rejection region (for illustration purposes if none apply).
Please refer to the drop-down selections and input areas for specific hypothesis definitions and critical value entries. Use statistical software or tables to find precise critical values for the stated \(\alpha\) level.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F15e40907-5866-4802-8bee-6781a38984f9%2F58ba5eb4-06ce-479e-bc96-c28f1fffe6d8%2Fue6b52e_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Hypothesis Testing: Mean 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 \(2.34\) parts per million (ppm). It was found that the mean amount of carbon monoxide in the air for the random sample of \(62\) cities is \(2.41\) ppm and the standard deviation is \(2.09\) ppm. At \(\alpha = 0.01\), 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 \(H_a\)?
- \(H_0\) : [Dropdown selection]
- \(H_a\) : [Dropdown selection]
(Type integers or decimals. Do not round.)
*The claim is the [Dropdown selection] hypothesis.*
#### (b) Use Technology to Find the Critical Value(s) and Identify the Rejection Region(s)
The critical value(s) is/are \( t_0 = \) [Textbox for input].
(Use a comma to separate answers as needed. Round to two decimal places as needed.)
*Choose the graph which shows the rejection region.*
[Radio button selection]
- \( \text{Graph A} \)
- \( \text{Graph B} \)
- \( \text{Graph C} \)
- \( \text{Graph D} \)
### Explanation of Graphs (if Provided)
If there are graphs included, they typically represent the critical value and the rejection regions. For example:
- **Graph A:** May show a two-tailed test with two rejection regions on either end of the distribution.
- **Graph B:** Might depict a one-tailed test with a rejection region on the left end of the distribution.
- **Graph C:** Could illustrate a one-tailed test with a rejection region on the right end of the distribution.
- **Graph D:** Would potentially show no rejection region (for illustration purposes if none apply).
Please refer to the drop-down selections and input areas for specific hypothesis definitions and critical value entries. Use statistical software or tables to find precise critical values for the stated \(\alpha\) level.
![### Hypothesis Testing for Mean Carbon Monoxide Levels
---
**Claim:** The mean amount of carbon monoxide (CO) in the air in U.S. cities is less than 2.34 parts per million.
---
#### Step-by-Step Hypothesis Testing
1. **Given Information:**
- Null Hypothesis (\(H_0\)): The mean amount of carbon monoxide in the air in U.S. cities is equal to 2.34 parts per million.
- Alternative Hypothesis (\(H_1\)): The mean amount of carbon monoxide in the air in U.S. cities is less than 2.34 parts per million.
2. **Graphs Overview:**
- The provided graphs are t-distribution curves that depict critical regions for different scenarios in hypothesis testing.
- The first graph should display a left-tailed test, where the rejection region is on the left side of the curve.
- The second graph includes the t-distribution curve centered around zero, demonstrating the comparison between sample t-value and critical t-value.
- The third graph is similar to the first but highlights the specific rejection region on the left-tail.
- The fourth graph emphasizes the rejection region, possibly with a numerical value at \(t_0\) indicating the critical value for rejection.
3. **Standardized Test Statistic:**
- **Find the standardized test statistic, \( t \).**
- **Formula:**
\[
t = \frac{\bar{x} - \mu}{s / \sqrt{n}}
\]
- \(\bar{x}\): Sample mean
- \(\mu\): Population mean
- \(s\): Sample standard deviation
- \(n\): Sample size
- The standardized test statistic is \( t = [ \, ] \) (Round to two decimal places as needed).
4. **Decision Rule:**
- **Decide whether to reject or fail to reject the null hypothesis.**
- Compare the calculated \( t \)-value to the critical value from the t-distribution table.
- **Conclusion:**
\[
\_\_\_\_\_\_ H_0 \text{ because the standardized test statistic } \_\_\_\_ \text{ in the rejection region}.
\]
- Select the appropriate comparison (e.g., \> or <).
5. **Interpretation:**
- **Interpret](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F15e40907-5866-4802-8bee-6781a38984f9%2F58ba5eb4-06ce-479e-bc96-c28f1fffe6d8%2Fhhc4vb_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Hypothesis Testing for Mean Carbon Monoxide Levels
---
**Claim:** The mean amount of carbon monoxide (CO) in the air in U.S. cities is less than 2.34 parts per million.
---
#### Step-by-Step Hypothesis Testing
1. **Given Information:**
- Null Hypothesis (\(H_0\)): The mean amount of carbon monoxide in the air in U.S. cities is equal to 2.34 parts per million.
- Alternative Hypothesis (\(H_1\)): The mean amount of carbon monoxide in the air in U.S. cities is less than 2.34 parts per million.
2. **Graphs Overview:**
- The provided graphs are t-distribution curves that depict critical regions for different scenarios in hypothesis testing.
- The first graph should display a left-tailed test, where the rejection region is on the left side of the curve.
- The second graph includes the t-distribution curve centered around zero, demonstrating the comparison between sample t-value and critical t-value.
- The third graph is similar to the first but highlights the specific rejection region on the left-tail.
- The fourth graph emphasizes the rejection region, possibly with a numerical value at \(t_0\) indicating the critical value for rejection.
3. **Standardized Test Statistic:**
- **Find the standardized test statistic, \( t \).**
- **Formula:**
\[
t = \frac{\bar{x} - \mu}{s / \sqrt{n}}
\]
- \(\bar{x}\): Sample mean
- \(\mu\): Population mean
- \(s\): Sample standard deviation
- \(n\): Sample size
- The standardized test statistic is \( t = [ \, ] \) (Round to two decimal places as needed).
4. **Decision Rule:**
- **Decide whether to reject or fail to reject the null hypothesis.**
- Compare the calculated \( t \)-value to the critical value from the t-distribution table.
- **Conclusion:**
\[
\_\_\_\_\_\_ H_0 \text{ because the standardized test statistic } \_\_\_\_ \text{ in the rejection region}.
\]
- Select the appropriate comparison (e.g., \> or <).
5. **Interpretation:**
- **Interpret
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