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 ist = 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.Ho: H(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, theresufficient evidence at a significance level of 0.10 tothe claim that smokers have a mean cotinine level greater than the level of 2.84 ng / mL found for nonsmokers.

Claim : The smokers have a mean cotinine level greater than the level of 2.84 ng/ml found for…

Answered: The claim is that smokers have a mean…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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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?
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 5% level of significance. Sample Data Low Fat Low Carbohydrate T1 = 4.7lb s1 = 7.2lb n1 = 77 T1 = 2.6lb 81 = 5.9lb nị = 79 1. Identify the tail of the test. [ Select ] 2. What is the P-value? [ Select ] 3. Will the null hypothesis be rejected? [Select ] 4. Is the initial claim supported? [ Select ]
Assume that the significance level is α=0.01. Use the given information to find the P-value and the critical value(s). With H1: p≠ 2 3, the test statistic is z=−1.41. Click here to view page 1 of the Normal table. LOADING... Click here to view page 2 of the Normal table. LOADING... Question content area bottom Part 1 P-value=enter your response here (Round to four decimal places as needed.) Part 2 The critical value(s) is/are enter your response here. (Round to two decimal places as needed. Use a comma to separate answers as needed.)
In the population of senior students at Shermer High School, the proportion who plan to attend college is 0.64. In a random sample of 100 students from this population, the proportion who plan to attend college is 0.75. We would call the value 0.64 a P-value. z-score. parameter statistic significance level.
Suppose a statistician named Big Daddy thinks that the average monthly electric bill for Houston in August is more than $150. Using sample data from a normal population, Big Daddy obtains a test statistic of z = 1.06. Determine the P-value and state the conclusion at the 5% level of significance.
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 810 and the test statistic is t=55.337. 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. Ho H H₁H (Type integers or decimals. Do not round.)
Assume that a simple random sample has been selected from a normally distributed population. State the final conclusion.Test the claim that the mean lifetime of car engines of a particular type is greater than 220,000 miles. Sample data are summarized as n = 23, and s = 11,500 miles. Use a significance level of α = 0.01.H0: μ = 220,000 HA: μ > 220,000State your conclusion about H0. Choose the best answer below: A. Reject HA B. Do not reject H0 C. Do not reject HA D. Reject H0 E. Cannot draw a conclusion for information given
The US Department of Energy reported that 48% of homes were heated by natural gas. A random sample of 331 homes in Oregon found that 132 were heated by natural gas. Test the claim that proportion of homes in Oregon that were heated by natural gas is different than what was reported. Use a 1% significance level. Give answer to at least 4 decimal places. What are the correct hypotheses? (Select the correct symbols and use decimal values not percentages.) H0: H1: Based on the hypotheses, compute the following:Test Statistic = p-value = Based on the above we choose to The correct summary would be: that the proportion of homes in Oregon that were heated by natural gas is different than what the DOE reported value of 48%.
During the 1990s, the fatality rate for lung cancer was 80 per 100,000 people. After the turn of the century and the establishment of newer treatments and adjustment in public health advertising, a random sample of 10,000 people exhibits only six deaths due to lung cancer. Test at the 0.05 significance level whether that data are proof of a reduced fatality rate for lung cancer. b. Make the decision rule. (Negative amount should be indicated by a minus sign. Round your answer to 2 decimal places.) c. Evaluate the test statistic. (Negative amount should be indicated by a minus sign. Round your answer to 3 decimal places.)
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