A study was done to determine whether there is a relationship between snoring and the risk of heart disease. Among 1,105 snorers in the study, 88 had heart disease, while only 22 of1,379 nonsnorers had heart disease.(a) Determine a 95% confidence interval that estimates p, – p = difference in proportions of snorers and nonsnorers who have heart disease. (Let p, be the proportion of snorers andp, be the proportion of nonsnorers. Round your answers to three decimal places.)to(b) On the basis of the confidence interval determined in part (a), can we infer that the population proportions with heart disease differ for nonsnorers and snorers? Explain.Yes, because the interval does not cover 0.No, because the interval covers 0.Yes, because the interval covers 0.No, because the interval does not cover 0.You may need to use the appropriate table in the Appendix of Tables to answer this question.

Confidence interval:The (1-α)100 % confidence interval for proportion p1 – p2 is given by,

Answered: A study was done to determine whether…

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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A union of restaurant and foodservice workers would like to estimate the mean hourly wage, u, of foodservice workers in the U.S. this year The mean hourly wage last year was $8.08, and there is good reason to believe that this year's value is different from last year's. The union decides to do a statistical test to see if the value has indeed changed. The union chooses a random sample of this year's wages, computes the mean of the sample to be $7.78, and computes the standard deviation of the sample to be $1.20. Based on this information, complete the parts below. (a) What are the null hypothesis H. and the alternative hypothesis H₁ that should be used for the test? H₂ : O H₁ :0 (b) Suppose that the union decides not to reject the null hypothesis. What sort of error might it be making? (Choose one) ▼ (c) Suppose the true mean hourly wage for foodservice workers in the U.S. this year is $8.08. Fill in the blanks to describe a Type I error. A Type I error would be (Choose one) when, in…
4. Researchers are interested in the average heart rate of male college students. Assume a random sample of 130 male college students were taken from 200 students for the study. Find a 95% confidence interval for the actual proportion of students.
A study was performed to determine the percentage of people who wear life vests while out on the water. A researcher believed that the percentage was different for those who rode jet skis compared to those who were in boats. Out of 400 randomly selected people who rode a jet ski, 88% wore life vests. Out of 250 randomly selected boaters, 82.4% wore life vests. Using a 0.01 level of significance, test the claim that the proportion of people who wear life vests while riding a jet ski is not the same as the proportion of people who wear life vests while riding in a boat. Let jet skiers be Population 1 and let boaters be Population 2. Step 1 of 3: State the null and alternative hypotheses for the test. Fill in the blank below. H0: p1=p2 Ha: p1⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯p2 Step 2 of 3: Compute the value of the test statistic. Round your answer to two decimal places. Step 3 of 3: Draw a conclusion and interpret the decision.
Assume a researcher wants to compare the mean Alanine Aminotransferase (ALT) levels in two populations, individuals who drink alcohol and individuals who do not drink alcohol. The mean ALT levels for the individuals who do not drink alcohol is 32 with a standard deviation of 14, and 37 individuals were in the sample. The mean ALT levels for individuals who drink alcohol is 69 with a standard deviation of 19, and 38 individuals were in the sample. Construct and interpret a 95% confidence interval demonstrating the difference in means for those individuals who drink alcohol when compared to those who do not drink alcohol.
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