The claim is that smokers have a mean cotinine level greater than the level of 2.84 ng/mL found for nonsmokers. (Cotinine is used as a biomarker for exposure to nicotine.) The sample size is n = 920 and the test statistic is t= 56.224. Use technology to find the P-value. Based on the result, what is the final conclusion? Use a significance level of 0.05. State the null and alternative hypotheses. H:H= 2.84 7:μ > 2.84 Type integers or decimals. Do not round.)

MATLAB: An Introduction with Applications
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Chapter1: Starting With Matlab
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*Choose one word that matches the sentences below

 

Based on the P-value, there is or is not sufficient evidence at a significance level of 0.05 to support or warrent rejection of the claim that smokers have a mean continine level greater than the level of 2.84 ng/ mL found for nonsmokers. 

Based on the P-value, there [drop-down menu: "is"/"is not"] sufficient evidence at a significance level of 0.05 to [drop-down menu: "support"/"reject"] the claim that smokers have a mean cotinine level greater than the level of 2.84 ng/mL found for nonsmokers.
Transcribed Image Text:Based on the P-value, there [drop-down menu: "is"/"is not"] sufficient evidence at a significance level of 0.05 to [drop-down menu: "support"/"reject"] the claim that smokers have a mean cotinine level greater than the level of 2.84 ng/mL found for nonsmokers.
The claim is that smokers have a mean cotinine level greater than the level of 2.84 ng/mL found for nonsmokers. (Cotinine is used as a biomarker for exposure to nicotine.) The sample size is n = 920 and the test statistic is t = 56.224. Use technology to find the P-value. Based on the result, what is the final conclusion? Use a significance level of 0.05.

---

**State the null and alternative hypotheses:**

- \( H_0: \mu = 2.84 \)
- \( H_1: \mu > 2.84 \)

(Type integers or decimals. Do not round.)

The test statistic is \( \boxed{\ \ } \).

(Round to two decimal places as needed.)
Transcribed Image Text:The claim is that smokers have a mean cotinine level greater than the level of 2.84 ng/mL found for nonsmokers. (Cotinine is used as a biomarker for exposure to nicotine.) The sample size is n = 920 and the test statistic is t = 56.224. Use technology to find the P-value. Based on the result, what is the final conclusion? Use a significance level of 0.05. --- **State the null and alternative hypotheses:** - \( H_0: \mu = 2.84 \) - \( H_1: \mu > 2.84 \) (Type integers or decimals. Do not round.) The test statistic is \( \boxed{\ \ } \). (Round to two decimal places as needed.)
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