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= 876 and the test statistic is t=54.443. Use technology to find the P-value. Based on the result, what is the final conclusion? Use a significance level of 0.10 State the null and alternative hypotheses H_0: \mu H_1: \mu (Type integers or decimals. Do not round.) The test statistic is (Round to two decimal places as needed) The P-value is (Round to three decimal places as needed) Based on the P-value, there sufficient evidence at a significance level of 0.10 to the claim that smokers have a mean cotinine level greater than the level of 2.84 ng/mL found for nonsmokers
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The null hypothesis (H0) states that there is no difference in the mean nicotine level between smokers and non-smokers. The alternative hypothesis (Ha) states that the mean nicotine level for smokers is greater than the mean nicotine level for non-smokers. H0: Show more…
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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 = 805, and the test statistic is t = 55.744. Use technology to find the P-value. Based on the result, what is the final conclusion? Use a significance level of 0.10. State the null and alternative hypotheses. Ho: The mean cotinine level for smokers is less than or equal to 2.84 ng/mL. Ha: The mean cotinine level for smokers is greater than 2.84 ng/mL. The test statistic is t = 55.744. (Round to two decimal places as needed.) The P-value is 0.000. (Round to three decimal places as needed.) Based on the P-value, there is sufficient evidence at a significance level of 0.10 to reject the null hypothesis and support the claim that smokers have a mean cotinine level greater than the level of 2.84 ng/mL.
Adi S.
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 = 932 and the test statistic is t = 53.088. Use technology to find the P-value. Based on the result, what is the final conclusion? Use a significance level of 0.01. H0: μ H1: μ (Type integers or decimals. Do not round.) The test statistic is (Round to two decimal places as needed.) The P-value is (Round to three decimal places as needed.) Based on the P-value, there sufficient evidence at a significance level of 0.01 to 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=755 and the test statistic is t=54.004. Use technology to find the P-value. Based on the result, what is the final conclusion? Use a significance level of 0.01 a. State the null and alternative hypotheses? b. The Test Statistic? c. The P-Value?
Ana Carolina D.
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