•..Section II: Free Response Show all your work. Indicate clearlythe correctness of your methods as well as on the accuracy andAP2.22 A health worker is interested in determining ifomega-3 fish oil can help reduce cholesterol inadults. She obtains permission to examine thehealth records of 200 people in a large medicalclinic and classifies them according to whetheror not they take omega-3 fish oil. She also obtainstheir latest cholesterol readings and finds that themean cholesterol reading for those who are takingomega-3 fish oil is 18 points lower than the meanfor the group not taking omega-3 fish oil.AF(a) Is this an observational study or an experiment?Justify your answer.(b) Explain the concept of confounding in the contextof this study and give one example of a variablethat could be confounded with whether or notpeople take omega-3 fish oil.(c) Researchers find that the 18-point difference inthe mean cholesterol readings of the two groupsis statistically significant. Can they conclude thatomega-3 fish oil is the cause? Why or why not?

Introduction: The study involves 200 people in a large clinic, and consider whether these people…

Answered: AP2.22 A health worker is interested in…

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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The Economist collects data each year on the price of a Big Mac in various countries around the world. The price of a Big Mac for a sample of McDonalds restaurants in Europe in May 2009 resulted in the following Big Mac prices (after conversion to U.S. dollars): 3.80 5.89 4.92 3.88 2.65 5.57 6.39 3.24The mean price of a Big Mac in the U.S. in May 2009 was $3.57. For purposes of this exercise, assume it is reasonable to regard the sample as representative of European McDonalds restaurants. Does the sample provide convincing evidence that the mean May 2009 price of a Big Mac in Europe is greater than the reported U.S. price? Test the relevant hypotheses using a = .05. Assume a normal distribution of the data. x = S= Ho: μ = Ha: Select an answer (Put in the correct symbol and value) α= P-value: = Based on the above we choose t✓ Select an answer Question Help: Post to forur Reject the null hypothesis B
South Dakota prices vary $350,000: $400,000:$400,000:$425,000:$429,000: $440,000: $560,000: $585,000,: $585,000: $785,000 : $1,300,000. We found the following Mean = 569000 SD = 272703.1353 T score = .7113 P Value = .4932 CI: 385795.5409,752204.4591 If you had a margin of error of $13,000 how large of sample would you need with a 90% confidence interval using the given tscore.
On occasion, medical studies need to model the proportion of the population that has a disease and compare that to observed frequencies of the disease actually occurring. Suppose the end-stage renal failure in south-west Wales was collected for different age groups. Do the data in the table show that the observed frequencies are in agreement with proportion of people in each age group? Test at the 1% level. Renal Failure Frequencies Age group 16-29 30-44 45-59 60-74 75+ Total Observed Frequencies 125 152 134 124 48 583 Expected Proportion 0.23 0.25 0.22 0.21 0.09 State the hypotheses.H0: The observed frequencies in agreement with proportion of people in each age group.Ha: The observed frequencies in agreement with proportion of people in each age group. Calculate the χ2 test statistic. Round expected values to two decimal places. Round χ2 to 3 decimal places.χ2 = Calculate the p-value. Round to 3 decimal places.p-value = State your decision.Since p-value α,…
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According to data from the Tobacco Institute Testing Laboratory, a certain brand of cigarette contains an average of 1.4 milligrams of nicotine. An advocacy group questions this figure, and commissions an independent test to see if the the mean nicotine content is higher than the industry laboratory claims. The test involved randomly selecting n = 15 cigarettes, measuring the nicotine content (in milligrams) of each cigarette. The data is given below: 1.8, 1.8, 1.5, 1.1, 1.4, 1.3, 1.3, 1.4, 1.1, 1.3, 1.6, 1.2, 1.8, 1.4, 1.6; Download .csv file Not sure how to perform some of the tasks with R-Studio? Click to learn more (a) Do the data follow an approximately Normal distribution? Use alpha = 0.05. yes (b) Determine the P-value for this Normality test, to at least three decimal places. P (c) Choose the correct statistical hypotheses. O A. Ho: > 1.4, O B. Ho = 1.4, O C. Ho X = 1.4, O D. Ho: X = 1.4, ⒸE. Ho - 1.4 HA: > 1.4 OF. Ho: X > 1.4, HA: X < 1.4 HA: < 1.4 HA: 1.4 HA: X 1.4 HA: X <…
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Bottles for room spray are supposed to weigh exactly 16 oz. Inspectors want to develop process control charts. They take six boxes of five bottles each and weigh them. They obtain the following data. The mean weights of each of the 6 samples are 15.43, 14.47, 18.07, 17.34, 12.74, and 15.58 respectively. Likewise, the range values for the each of 6 samples are 1.63,0.66, 0.47,0.40, 0.20, and 0.90 respectively. Find out Grand Mean (X double bar), and mean range (R bar) O a. 16.538, 0.86 O b. 15.61,0.71 16.53,0.71 O d. 15.82, -0.65 O e. 0,0
A marketing firm is doing research for an Internet-based company. It wants to appeal to the age group of people who spend the most money online. The company wants to know if there is a difference in the mean amount of money people spend per month on Internet purchases depending on their age bracket. The marketing firm looked at two age groups, 1818-2424 years and 2525-3030 years, and collected the data shown in the following table. Let Population 1 be the amount of money spent per month on Internet purchases by people in the 1818-2424 age bracket and Population 2 be the amount of money spent per month on Internet purchases by people in the 2525-3030 age bracket. Assume that the population variances are not the same. Internet Spending per Month 1818-2424 Years 2525-3030 Years Mean Amount Spent 62.4962.49 62.8462.84 Standard Deviation 20.0420.04 15.6415.64 Sample Size 1616 2828 Step 1 of 2 : Construct a 95%95% confidence interval for the true difference…
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