A team of eye surgeons has developed a new technique for a risky eye operation to restore the sight of people blinded from a certain disease, Under the oldmethod, it is known that only 34% of the patients who undergo this operation recover their eyesight.Suppose that surgeons in various hospitals have performed a total of 325 operations using the new method and that 124 have been successful (i.e., the patientsfully recovered their sight). Can we justify the claim that the new method is better than the old one? (Use a 2% level of significanice.).What is your decision and why?p-value aReject the Hop-value >aFail to reject the Ho357

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Answered: A tearm of eye surgeons has developed a…

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 study was performed to test the effectiveness of a new treatment for migraines. There were 125 patients in the study. Of these, 60 were randomly assigned to receive the standard treatment and 65 to receive the new treatment. All were instructed to take the medication at the onset of their next migraine and to notice if the pain of their migraine felt reduced after half an hour. Fifty patients receiving the new treatment reported reduced pain, compared with only 40 of the patients receiving the standard treatment. The results seem to favor the new treatment, but could the difference just be due to chance? In this problem, consider the null hypothesis that the percentage of patients with a reduction in pain is the same between the two treatments, versus the alternative hypothesis that the percentages are different. (a) Under the null hypothesis, the difference in the sample percentages is expected to be ___%. The standard error for the difference is estimated to be ___%. (b) Calculate…
There are X-Mart and Y-Mart convenience stores. It is assumed that each customer purchases once a week at one of the two supermarkets (but not both). In a study, a sample of 100 buyers was taken for 10 weeks. From this research it is known that if buyers go to X-mart store one week, 85 people will still buy at X-mart store the following week and if buyers go to Y-mart store one week, 70 people will still buy at Y-mart store. mart the following week. 1. Make a table and matrix of transition probabilities. 2. If John purchases at X-mart in the first week, what is the probability that Bob purchases at X-store in the third and fourth weeks?
3) A medical association claims that less than 20% of adults in the US are regular smokers of tobacco. To test this claim the survey a sample of 125 US adults and find that 19 of them are regular smokers. Test the medical association's claim using α=0.05.
Is the national crime rate really going down? Some sociologists say yes! They say that the reason for the decline in crime rates in the 1980s and 1990s is demographics. It seems that the population is aging, and older people commit fewer crimes. According to the FBI and the Justice Department, 70% of all arrests are of males aged 15 to 34 years†. Suppose you are a sociologist in Rock Springs, Wyoming, and a random sample of police files showed that of 33arrests last month, 25were of males aged 15 to 34 years. Use a 1%level of significance to test the claim that the population proportion of such arrests in Rock Springs is different from 70%. a. What is the level of significance? answer .01 b.State the null and alternate hypotheses. Which one? Its the third. H0: p< 0 .7; H1: p= 0.7 H0: p≠0.7; H1: p= 0.7 H0: p= 0.7; H1: p≠0.7 H0: p= 0 .7; H1: p< 0.7 H0: p= 0.7; H1: p> 0.7 c. What sampling distribution will you use? Its the last one. The Student's t, since np> 5 and…
1. On the basis of her experience with clients, a clinical psychologist thinks that depression may affect sleep. She decides to test this idea. The sleep of nine depressed patients and eight normal controls is monitored for three consecutive nights. The average number of hours slept by each subject during the last two nights is shown in the following table: (Pagano, 2010). If proved significant, compute of the effect size. Hours of Sleep Depressed patients 7.1 Normal controls 8.2 6.8 7.5 6.7 7.3 7.7 7.8 7.5 8.0 6.2 7.4 6.9 7.3 6.5 6.5 7.2
Table of Critical t-Values 12345678910 112151671912092980 Degrees of Freedom 0.25 13 14 23 24 32 33 2 t-Distribution Area in Right Tail 0.20 0.15 0.05 0.025 0.02 1.000 1.376 1.963 3.078 6.314 12.706 15.894 0.816 1.061 1.386 1.886 2.920 4.303 4.849 0.765 0.978 1.250 1.638 2.353 3.182 3.482 0.741 0.941 1.190 1.533 2.132 2.776 2.999 0.727 0.920 1.156 1.476 2.015 2.571 2.757 0.718 0.906 1.134 1.440 1.943 2.447 2.612 3.143 3.707 0.711 0.896 1.119 1.415 1.895 2.365 2.517 2.998 3.499 0.706 0.889 1.108 1.397 1.860 2.306 2.449 2.896 3.355 0.703 0.883 1.100 1.383 1.833 2.398 2.821 1.093 1.372 2.262 3.250 1.812 2.228 2.359 3.169 0.700. 0.879 0.876 2.764 2.328 2.718 0.697 1.796 2.201 0.695 0.873 2.179 2.303 2.681 0.694 0.870 1.088 1.363 1.083 1.356 1.782 1.079 1.350 1.771 2.160 2.282 2.650 1.076 1.345 1.761 2.145 2.264 2.624 1.074 1.753 1.337 1.746 1.333 1.740 2.110 1.330 1.734 2.101 1.328 1.729 1.341 2.131 2.249 1.067 1.066 1.064 1.325 1.725 2.567 2.898 2.552 2.878 2.539 2.861 2.528 2.845 2.093…
Beryllium is an extremely lightweight metal that is used in many industries, such as aerospace and electronics. Long-term exposure to beryllium can cause people to become sensitized. Once an individual is sensitized, continued exposure can result in chronic beryllium disease, which involves scarring of the lungs. In a study of the effects of exposure to beryllium, workers were categorized by their duration of exposure (in years) and by their disease status (diseased, sensitized, or normal). The results were as follows: Duration of Exposure <1 1 to <5 ≥5 Diseased 14 13 18 Sensitized 11 20 15 Normal 78 140 209 Test the hypothesis of independence. Use the a=0.10 level of significance and the P-value method with the TI-84 Plus calculator. What do you conclude? State the null and alternate hypotheses.

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