Surgical complications: A medical researcher wants to construct a 90% confidence interval for the proportion of knee replacement surgeries that result in complications. (a) An article in a medical journal suggested that approximately 10% of such operations result in complications. Using this estimate, what sample size is needed so that the confidence interval will have a margin of error of 0.04? (b) Estimate the sample size needed if no estimate of p is available.
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A: Given: Sample proportion p^=0.28 Confidence interval =0.95 So, α=1-0.95=0.05 Margin of error (E)…
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A: Given, Favorable Cases XX = 140140 Sample Size NN = 420019420019 To find, construct a…
Q: Surgical complications: A medical researcher wants to construct a 90% confidence interval for the…
A: From the provided information,Confidence level = 90%Proportion (p) = 0.10Margin of error (E) = 0.04
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A: Given Data Sample Size, n = 420087 Number of successes, x = 135.0 confidence…
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A: Given: n=420,092x=130 The sample proportion is, p^=xn=130420,092=0.0003 From the Standard Normal…
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A: Given that Sample size n = 420078 Number of cell phone user who developed cancer, X = 130 The…
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A: We have given that, Margin of error (E) = 3% = 0.03, confidence interval = 95% Then, We will find…
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A: Given confidence level, C = 95% = 0.95 Level of significance, α = 1-0.95 = 0.05 Using formula,
Q: A study of 420,018 cell phone users found that 139 of them developed cancer of the brain or nervous…
A: https://drive.google.com/file/d/1S_b_En4yZMZPhsaGfBA2eE_M0IJFOVeW/view?usp=sharing
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A: Answer Given X1 =43 n1 =91 x2 =51 n2 =81 Confidence interval =99%
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A: a) From the provided information, Sample size (n) = 420,015 139 of them developed cancer of the…
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A: here for sample1 and sample2 data are respectively n1 = 31+47 = 78 n2 = 92 x1 = 31 x2 = 45…
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A: From the given information, it is obtained that: σ (population standard deviation)=12Confidence…
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A: Given, n = 420007 X = 136 p^ = Xn = 136420007 = 0.000324 Confidence level = 0.90 To find, a. 90%…
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A: Introduction: A histogram for the apartment sizes in a sample of 48 apartments is given.
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A: Given that, n=420048 x=137 p¯=xn=137420048=0.000326 90% C.I. for p P(-z<p<z)=0.90 z=1.645 […
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A: From the provided information, Sample size (n) = 34006 Proportion of people who responded “yes” to…
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A: Given: n=30
Q: A study of 420,082 cell phone users found that 136 of them developed cancer of the brain or nervous…
A: Given, p^=136420082=0.00032 Rate of cancer for the people who did not use cell phones = 0.0415%
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A: The confidence interval for proportion can be calculated as follows:
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A: Sample size n1=70 Favorable cases x1=55 Sample size n2=130 Favorable cases x2=120
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- A computer store near campus is stocking its shelves and wants to know the relative proportions of students who use PCs vs Macs. A random sample of 60 students is surveyed, and each student is asked whether they use a PC or Mac more often. 55% select ”PC.” (a) What is the population and what is the sample in this study? (b) Calculate a 95% confidence interval for the proportion of UCI students that are PC users. (c) Provide an interpretation of this confidence interval in the context of this problem. (d) The confidence interval is pretty wide and leaves a lot of uncertainty over the proportion of UCIundergarduates who use PCs. With the goal to estimate a narrower 95% confidence interval, what is a simple change to this study that you could suggest for the next time that a similar survey is conducted?A medical researcher wants to construct a 95% confidence interval for the proportion of knee replacement surgeries that result in complications. A journal article suggested that approximately 6% of such operations result in complications. Using this estimate, what sample size is needed so that the confidence interval will have a margin of error of 0.05?Suppose we know that a confidence interval for a population proportion is (0.105,0.355), with a sample proportion of p̂=0.23. What is the margin of error?
- 2. You are going to construct a 90% confidence interval for the proportion ofadults that have a post-secondary education. You want the margin of errorto be .03. Your initial guess is that 35% of the population has a postsecondary education. What should the sample size be to achieve thismargin of error?Suppose that you are interested in the number of people in each household in the U.S. Your sample includes 10 households, with a mean 2.5 and a standard deviation 1.2. Construct a 95% confidence interval (CI) of the mean number of household members. (A) 1.2-3.8 (B) 1.4-3.6 (C)1.6-3.4 (D) 1.7-3.3 (E) 1.8-3.2A study of 420,031 cell phone users found that 136 of them developed cancer of the brain or nervous system. Prior to this study of cell phone use, the rate of such cancer was found to be 0.0316% for those not using cell phones. Complete parts (a) and (b). a. Use the sample data to construct a 90% confidence interval estimate of the percentage of cell phone users who develop cancer of the brain or nervous system. %Early-childhood-development studies indicate that the more often a child is read to from birth, the earlier the child begins to read. A local parents’ group wants to test this theory and samples families with young children. They find the following results. Construct a 99% confidence interval to estimate the true difference between the proportions of children who read at an early age when they are read to frequently compared to those who were read to less often, as described in the table of results. Let Population 1 be the children who were read to frequently and Population 2 be the children who were read to less often. Round the endpoints of the interval to three decimal places, if necessary. Ages when Children Begin to Read Read to at Least Three Times per Week Read to Fewer than Three Times per Week Started Reading by age 4 60 35 Started Reading after age 4 36 51 Lower end point: Upper end point:A study of 420,039 cell phone users found that 131 of them developed cancer of the brain or nervous system. Prior to this study of cell phone use, the rate of such cancer was found to be 0.0443% for those not using cell phones. Complete parts (a) and (b). ..... a. Use the sample data to construct a 95% confidence interval estimate of the percentage of cell phone users who develop cancer of the brain or nervous system. • D%In a study of speed dating, male subjects were asked to rate the attractiveness of their female dates, and a sample of the results is listed below (1=not attractive; 10=extremely attractive). Construct a confidence interval using a 90% confidence level. What do the results tell about the mean attractiveness ratings of the population of all adult females? 5,7,2,9,4,4,6,9,7,8,6,8In the Star and Tribune newspaper, there is an article entitled "1 in 8 school buses fail tests." The article says that the overall statewide proportion of school buses that fail safety tests is 9%. However, one bus company has a higher rate of over 12%. Assume that we want a new estimate for the statewide proportion of school buses that fail the safety inspection. How large a sample would you have to take in order to construct a 95% confidence interval estimate for the true proportion of buses that fail the inspection and achieve a margin of error of plus or minus 2%? (your answer should be an integer)A study of 420,097 cell phone users found that 137 of them developed cancer of the brain or nervous system. Prior to this study of cell phone use, the rate of such cancer was found to be 0.0311% for those not using cell phones. Complete parts (a) and (b). .... a. Use the sample data to construct a 90% confidence interval estimate of the percentage of cell phone users who develop cancer of the brain or nervous system. %Recommended textbooks for youMATLAB: An Introduction with ApplicationsStatisticsISBN:9781119256830Author:Amos GilatPublisher:John Wiley & Sons Inc
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