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 isn 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 HH₁H(Type integers or decimals. Do not round.)

Step 1:Claim: The mean cotinine level is greater than the level of 2.84 ng/mL. From claim hypothesis…

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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Suppose IQ scores were obtained for 20 randomly selected sets of couples. The 20 pairs of measurements yield x = 100.24, y = 102.55, r= 0.848, P-value = 0.000, and %3D y = 12.97 + 0.89x, where x represents the IQ score of the wife. Find the best predicted value of y given that the wife has an IQ of 106? Use a significance level of 0.05. Click the icon to view the critical values of the Pearson correlation coefficient r. The best predicted value of y is (Round to two decimal places as needed.)
Use the attached image to answer the question: Use the data above to construct a bivariate table and explore if there is an association between one’s education and attitudes toward cohabitation. Is there an association? Describe the pattern of association. Assuming that this is a random sample, test a hypothesis (use α = 0.01) that there is an association between one’s education and attitudes toward cohabitation in the population. Show all your work (state the hypotheses, find the critical value, compute the test statistics, state a conclusion) and write one sentence summarizing the results. (Extra-credit: find the p-value). Is this association weak, moderate or strong? Compute and interpret the appropriate statistic.
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.
Interpret the Type I and Type II Errors for the following claim, the mean average drinking age is greater than 18.
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 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=761 and the test statistic is t=55.668. 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. H₁> 2.84 2.84 (Type integers or decimals. Do not round.) The test statistic is (Round to two decimal places as needed.)
The p-value for a t-statistic was .008. What is the best way to report the level of significance? p<.008 p<.05 p<.01 p<.001
If the test requires that the results be statistically significant at a level of α = 0.05, fill in the blanks and complete the sentences that explain the test decision and conclusion. The decision is to reject the null hypothesis. There is enough evidence at a significance level of a = 0.05 that the mean weight for U.S. women of age 40-49 who smoke daily is less than 169.976 pounds.
A teacher would like to determine if quiz scores improve after completion of a worksheet. The students take a pre-quiz before the worksheet and then another quiz after the worksheet. (Pre-quiz score - Post-quiz score) Assume quiz scores are normally distributed. The grades for each quiz are given below. Use a significance level of a = 0.05. H₂: Hd = 0 Ha: Hd <0 pre-quiz 12 16 17 16 11 15 12 7 15 13 post-quiz 13 16 15 16 14 15 13 9 14 15 What is the test statistic? test statistic = (Report answer accurate to 2 decimal places.) What is the p-value for this sample? p-value= (Report answer accurate to 3 decimal places.) The correct decision is to Select an answer Conclusion: Select an answer that the worksheet improved quiz scores.
You want to compare differences in prior arrests across ethnicity in your study, you measured ethnicity as white =0 black =1 hispanic=2, and other =3. Prior arrests was recorded as the number of prior arrests. which test would you perform to see if arrests differ across diffrent ethnic categories? A. ANOVA B. T-test
options for first to blanks : 1- between 0.05 and 0.20 /2- greater than 0.20 / 3- less than 0.2 options for last blank : 1- handrubbing with alcohol / 2- handwashing with soap
Choose the appropriate statistical test. When computing, be sure to round each answer as indicated. A dentist wonders if depression affects ratings of tooth pain. In the general population, using a scale of 1-10 with higher values indicating more pain, the average pain rating for patients with toothaches is 6.8. A sample of 30 patients that show high levels of depression have an average pain rating of 7.1 (variance 0.8). What should the dentist determine? 1. Calculate the estimated standard error. (round to 3 decimals). [st.error] 2. What is thet-obtained? (round to 3 decimals). 3. What is the t-cv? (exact value) 4. What is your conclusion? Only type "Reject" or Retain"

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