The prevalence rate of osteoporosis among women were measured in a survey, p(%) = 25.3 SE (p) = 1.8 calculate the upper bound of 95% confidence interval
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- The pH of rain, measured at a weather station in Michigan, was observed for 39 consecutive rain storms. The sample mean is 4.6982 and the sample variance is 0.39623. Obtain a 99% confidence interval for the mean pH of the population of storms at that location.The water quality in a popular bay was tested for heavy metal contamination. The average heavy metal concentration from a sample of 36 different locations is 3 grams per millilitre with a standard deviation of 0.5. What is the margin of error (in two decimal places) if the confidence level used is 95%? Confidence Level 1−α Significance Level α α/2 zα/2 90% = 0.90 0.10 0.05 1.645 95% = 0.95 0.05 0.025 1.96 98% = 0.98 0.02 0.01 2.33 99% = 0.99 0.01 0.005 2.576 What is the lower limit(two decimal places) of the 95% confidence interval? What is the upper limit(two decimal places) of the 95% confidence interval?In a test of the effectiveness of garlic for lowering cholesterol, 49 subjects were treated with garlic in a processed tablet form. Cholesterol levels were measured before and after the treatment. The changes (before - after) in their levels of LDL cholesterol (in mg/dL) have a mean of 5.5 and a standard deviation of 19.3. Construct a 90% confidence interval estimate of the mean net change in LDL cholesterol after the garlic treatment. What does the confidence interval suggest about the effectiveness of garlic in reducing LDL cholesterol? Click here to view a t distribution table. Click here to view page 1 of the standard normal distribution table. Click here to view page 2 of the standard normal distribution table. ....e. What is the confidence interval estimate of the population mean u? mg/dL < µAn SRS of 450 high school seniors gained an average of x = 20.80 points in their second attempt at the SAT Mathematics exam. Assume that the change in score has a Normal distribution with standard deviation a = 54.81. We want to estimate the mean change in score in the population of all high school seniors. (a) Using the 68-95-99.7 Rule or the z-table (Table A), give a 68% confidence interval (a, b) for u based on this sample. H (Enter your answers rounded to three decimal places. If you are using Crunchlt, adjust the default precision under Preferences as necessary. See the instructional video on how to adjust precision settings.) a: (b) Based on your confidence interval in part (a), how certain are you that the mean change in score u in the population of all high school seniors is greater than 0? The interval does not contain 0, so we are 68% certain that the mean change in score in the population of all high school seniors is greater than 0. We cannot be certain at all. If we want…In a test of the effectiveness of garlic for lowering cholesterol, 43 subjects were treated with garlic in a processed tablet form. Cholesterol levels were measured before and after the treatment. The changes (before - after) in their levels of LDL cholesterol (in mg/dL) have a mean of 5.1 and a standard deviation of 16.8. Construct a 99% confidence interval estimate of the mean net change in LDL cholesterol after the garlic treatment. What does the confidence interval suggest about the effectiveness of garlic in reducing LDL cholesterol? Click here to view at distribution table. Click here to view page 1 of the standard normal distribution table. Click here to view page 2 of the standard normal distribution table. What is the confidence interval estimate of the population mean u? mg/dL <µ8 helpWhat critical value of t* should be used for a 95% confidence interval for the population mean based on a random sample of 21 observations? Find the t-table here. t* = 1.721 t* = 1.725 t* = 2.080 t* = 2.086Find the critical value needed to construct a 99% confidence interval of the population mean with sample standard deviation of 5.94 of a sample of 26 students.In a test of the effectiveness of garlic for lowering cholesterol, 48 subjects were treated with garlic in a processed tablet form. Cholesterol levels were measured before and after the treatment. The changes (before - after) in their levels of LDL cholesterol (in mg/dL) have a mean of 3.6 and a standard deviation of 17.5. Construct a 95% confidence interval estimate of the mean net change in LDL cholesterol after the garlic treatment. What does the confidence interval suggest about the effectiveness of garlic in reducing LDL cholesterol? Click here to view a t distribution table. Click here to view page 1 of the standard normal distribution table. Click here to view page 2 of the standard normal distribution table. What is the confidence interval estimate of the population mean u? mg/dL <µ< mg/dL (Round to two decimal places as needed.)In a test of the effectiveness of garlic for lowering cholesterol, 42 subjects were treated with garlic in a processed tablet form. Cholesterol levels were measured before and after the treatment. The changes (before - after) in their levels of LDL cholesterol (in mg/dL) have a mean of 2.0 and a standard deviation of 16.5. Construct a 99% confidence interval estimate of the mean net change in LDL cholesterol after the garlic treatment. What does the confidence interval suggest about the effectiveness of garlic in reducing LDL cholesterol?A Gallup Poll in December 2019 reported that 46% of the 5120 adults aged 18 and older in the sample said both their mental and their physical health was good or excellent. Gallup announced, "For results based on the combined sample of 5120 adults, aged 18 and older ... the margin of sampling error is 2 percentage points at the 95% confidence level." Confidence intervals for a percentage follow the form estimate±margin of error Based on the information from Gallup, what is the 95% confidence interval for the percentage of all adults aged 18 and older who would say both their mental and physical health was good or excellent? Give your answer as an interval in the form (lower bound, upper bound). Give your numbers as percentages and as whole numbers. confidence interval: $$ What does it mean to have 95% confidence in this interval? A confidence of 95% in the interval means that 95% of all samples of the same size give an interval that contains the actual…The total of individual weights of garbage discarded by 17 households in one week is normally distributed with a mean of 32.9 lbs with a sample standard deviation of 9.7 lbs. Find the 99% confidence interval of the mean.SEE MORE QUESTIONSRecommended textbooks for youMATLAB: An Introduction with ApplicationsStatisticsISBN:9781119256830Author:Amos GilatPublisher:John Wiley & Sons IncProbability and Statistics for Engineering and th…StatisticsISBN:9781305251809Author:Jay L. DevorePublisher:Cengage LearningStatistics for The Behavioral Sciences (MindTap C…StatisticsISBN:9781305504912Author:Frederick J Gravetter, Larry B. WallnauPublisher:Cengage LearningElementary Statistics: Picturing the World (7th E…StatisticsISBN:9780134683416Author:Ron Larson, Betsy FarberPublisher:PEARSONThe Basic Practice of StatisticsStatisticsISBN:9781319042578Author:David S. 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