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= 892 and the test statistic is t = 52.657. 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. Họ: H (Type integers or decimals. Do not round.) The test statistic is. (Round to two decimal places as needed.) The P-value isN. (Round to three decimal places as needed.) Based on the P-value, there V 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.

According to a survey conducted by the Association for Dressings and Sauces, 20% of American adults eat salad once a week. A nutritionist suspects that this percentage is not accurate. She conducts a survey of 579 American adults and finds that 104 of them eat salad once a week. Use a 0.10 significance level to test the claim that the proportion of American adults who eat salad once a week is equal to 20%. Claim: p = 0.2 v which corresponds to H0: p = 0.2 Opposite: p 0.2 v which corresponds to H1 p + 0.2 The test is: right-tailed The test statistic is: z= (to 2 decimals) The Critical Value is: z= Based on this we: Fail to reject the null hypothesis v Conclusion: There does not proportion of American adults who eat salad once a week is equal to 20%. v appear to be enough evidence to support the claim that the Submit Question thing 近

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 = 911 and the test statistic is t= 56.944. Use technology to find the P-value. Based on the result, what is the final conclusion? Use a significance level of 0.01. State the null and alternative hypotheses. Hot = 2.84 H₁: > 2.84 (Type integers or decimals. Do not round.) The test statistic is 56.94. (Round to two decimal places as needed.) The P-value is. (Round to three decimal places as needed.) C

Use the data below to test the claim that the weight loss was greater with the low fat diet compared to the low carbohydrate diet.  Use a 10% level of significance. Low Fat Low Carbohydrate x¯1=3.7lbs 1=5.2lb n1=57 x¯1=2.7 lbs1=4.9lb n1=62   Identify the tail of the test.       What is the P-value?        Will the null hypothesis be rejected?        Is the initial claim supported?