Answer the question ### Hypothesis Testing on Carbon Monoxide Levels in U.S. Cities**Study 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. **Data Collected**:- Mean amount of carbon monoxide for a random sample of 66 cities: 2.41 parts per million.- Standard deviation: 2.09 parts per million.**Significance Level**:- \(\alpha = 0.10\)**Assumption**:- Population is normally distributed.**Procedure**:Complete parts (c) through (e) below.---#### (c) Finding the Standardized Test Statistic, \( t \)The standardized test statistic \( t \) is calculated as follows:\[ t = 0.389 \](Rounded to two decimal places as needed.)---#### (d) Decision on the Null HypothesisDecide whether to reject or fail to reject the null hypothesis \( H_0 \).\[ \text{Fail to reject } H_0 \text{ because the standardized test statistic } \boxed{\ }\text{ in the rejection region.} \]---#### (e) Interpretation in Context of the Original ClaimInterpret the decision within the context of the original claim.\[ \text{There is } \boxed{\text{not}} \text{ enough evidence at the } \boxed{\ } \text{ \% level of significance to } \boxed{\ } \text{ the claim that the mean amount of carbon monoxide in the air in U.S. cities is } \boxed{\text{less than}} \boxed{2.31} \text{ parts per million.} \]---This dataset and its analysis help in concluding whether the determined sample mean justifies rejecting the hypothesis that the mean amount of carbon monoxide in U.S. cities is less than specified levels based on the given data and statistical calculations. **Title: Testing a Claim about Mean Carbon Monoxide Levels in U.S. Cities****Objective:** 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.**Given Data:**- Mean amount of carbon monoxide for a random sample of 66 cities: **2.41 parts per million**- Standard deviation: **2.09 parts per million**- Significance level (\(\alpha\)): **0.10****Assumption:** The population is normally distributed.## Steps for Hypothesis Testing:### (a) Null and Alternative Hypotheses:- **Null Hypothesis (\(H_0\))**: The mean carbon monoxide level is 2.31 parts per million (μ = 2.31).- **Alternative Hypothesis (\(H_a\))**: The mean carbon monoxide level is less than 2.31 parts per million (μ < 2.31).### (b) Test Statistic:A t-test statistic is used for this scenario, given by the formula: \[ t = \frac{\bar{x} - \mu}{s / \sqrt{n}} \]where:- \(\bar{x}\) is the sample mean (2.41 parts per million),- \(\mu\) is the hypothesized population mean (2.31 parts per million),- \(s\) is the sample standard deviation (2.09 parts per million),- \(n\) is the sample size (66).### (c) Find the Standardized Test Statistic:The standardized test statistic \(t\) is calculated as follows:\[ t = \frac{2.41 - 2.31}{2.09 / \sqrt{66}} \approx 0.389 \](Rounded to two decimal places.)### (d) Decision Rule:Based on the calculated t-value, determine if we should reject the null hypothesis:Given that \(\alpha = 0.10\), we look up the critical t-value for a one-tailed test with 65 degrees of freedom (n - 1). - **Decision:** We **fail to reject** \(H_0\) because the test statistic \(t = 0.389\) is not in the rejection region.### (e) Interpretation:Since we fail to reject \(H_0\),

The sample size is 66, the sample mean is 2.41 and the sample standard deviation is 2.09.

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. (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 in the rejection region. level of significance to less than the claim that the mean amount of 2.31 parts per million.

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. (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 in the rejection region. level of significance to less than the claim that the mean amount of 2.31 parts per million.

Glencoe Algebra 1, Student Edition, 9780079039897, 0079039898, 2018

18th Edition

ISBN:9780079039897

Author:Carter

Publisher:Carter

Chapter10: Statistics

Section10.4: Distributions Of Data

Problem 8PPS

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