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. www (c) Find the standardized test statistic, t. The standardized test statistic is t = 0.389 (Round to two decimal places as needed.) (d) Decide whether to reject or fail to reject the null hypothesis. Fail to reject Ho because the standardized test statistic (e) Interpret the decision in the context of the original claim. There is not enough evidence at the carbon monoxide in the air in U.S. cities is (Type integers or decimals. Do not round.) in the rejection region. level of significance to less than the claim that the mean amount of 2.31 parts per million.

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### Hypothesis Testing Example: Carbon Monoxide Levels

**Claim**: 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 (ppm). 

**Data Summary**:
- Mean amount of carbon monoxide in a sample of 66 U.S. cities: 2.41 ppm
- Standard deviation: 2.09 ppm
- Significance level, α: 0.10

Assumptions: The population is normally distributed.

### Steps to Test the Claim:

#### (c) Find the standardized test statistic, t.

The standardized test statistic is given by:

\[ t = \frac{\bar{x} - \mu}{s / \sqrt{n}} \]

Where:
- \(\bar{x}\) is the sample mean
- \(\mu\) is the population mean
- \(s\) is the sample standard deviation 
- \(n\) is the sample size

Given:
- Sample mean (\(\bar{x}\)): 2.41 ppm
- Population mean (\(\mu\)): 2.31 ppm
- Sample standard deviation (s): 2.09 ppm
- Sample size (n): 66

The calculated t statistic is:
\[ t = 0.389 \]

(Round to two decimal places as needed.)

#### (d) Decide whether to reject or fail to reject the null hypothesis.

**Decision Rule**:
- If the calculated t value falls into the rejection region, reject the null hypothesis \( H_0 \).

**Conclusion**:
- Fail to reject \(H_0\) because the standardized test statistic \(0.389\) is not in the rejection region.

#### (e) Interpret the decision in the context of the original claim.

**Conclusion**:
There is not enough evidence at the 10% (\(0.10\)) level of significance to support the claim that the mean amount of carbon monoxide in the air in U.S. cities is less than 2.31 ppm.

### Summary:
In this hypothesis test, we have determined that based on the sample data, we do not have sufficient evidence to conclude that the mean amount of carbon monoxide in U.S. cities is less than 2.31 ppm. The test statistic did not fall into the critical region, thus we fail to reject the null hypothesis.
Transcribed Image Text:### Hypothesis Testing Example: Carbon Monoxide Levels **Claim**: 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 (ppm). **Data Summary**: - Mean amount of carbon monoxide in a sample of 66 U.S. cities: 2.41 ppm - Standard deviation: 2.09 ppm - Significance level, α: 0.10 Assumptions: The population is normally distributed. ### Steps to Test the Claim: #### (c) Find the standardized test statistic, t. The standardized test statistic is given by: \[ t = \frac{\bar{x} - \mu}{s / \sqrt{n}} \] Where: - \(\bar{x}\) is the sample mean - \(\mu\) is the population mean - \(s\) is the sample standard deviation - \(n\) is the sample size Given: - Sample mean (\(\bar{x}\)): 2.41 ppm - Population mean (\(\mu\)): 2.31 ppm - Sample standard deviation (s): 2.09 ppm - Sample size (n): 66 The calculated t statistic is: \[ t = 0.389 \] (Round to two decimal places as needed.) #### (d) Decide whether to reject or fail to reject the null hypothesis. **Decision Rule**: - If the calculated t value falls into the rejection region, reject the null hypothesis \( H_0 \). **Conclusion**: - Fail to reject \(H_0\) because the standardized test statistic \(0.389\) is not in the rejection region. #### (e) Interpret the decision in the context of the original claim. **Conclusion**: There is not enough evidence at the 10% (\(0.10\)) level of significance to support the claim that the mean amount of carbon monoxide in the air in U.S. cities is less than 2.31 ppm. ### Summary: In this hypothesis test, we have determined that based on the sample data, we do not have sufficient evidence to conclude that the mean amount of carbon monoxide in U.S. cities is less than 2.31 ppm. The test statistic did not fall into the critical region, thus we fail to reject the null hypothesis.
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