A neurological disease is characterized by gait imbalance. To provide effective treatment, it is important to detect the disease as soon as possible. In an experiment, age at onset of symptoms and age at diagnosis for 18 patients suffering from the disease were recorded for the same patients (in months). The differences in age were observed to be -38.60 months with a sample standard distribution of 23.18. Calculate a lower 95% confidence bound for the population mean of age differences and interpret the answer.
Q: In a test of the effectiveness of garlic for lowering cholesterol, 45 subjects were treated with…
A: The mean is 4.7 and standard deviation is 17.1.
Q: Controls 2.4 0.0 0.9 -0.2 1.0 1.7 2.9 -0.6 1.1 -0.1 0.3 1.2 -1.6 -0.1 -1.5 0.7 -0.4 2.2 -0.4 -0.4…
A: 1) The stem and leaf plot is Control Breast-feeders -8 3 -7 8,0 -6 8,8,5,5,2 -5…
Q: In a test of the effectiveness of garlic for lowering cholesterol, 42 subjects were treated with…
A: Given,sample size(n)=42sample mean(x¯)=4.5standard deviation(s)=17.6degrees of…
Q: In a test of the effectiveness of garlic for lowering cholesterol, 45 subjects were treated with…
A: Obtain the 99% confidence interval estimate of the mean net change in LDL cholesterol after the…
Q: In a test of the effectiveness of garlic for lowering cholesterol, 48 subjects were treated with…
A: From the provided information, Sample size (n) = 48 Sample mean (x̄) = 2.7 Sample standard deviation…
Q: In a test of the effectiveness of garlic for lowering cholesterol, 44 subjects were treated with…
A: Given Information: Sample size n=44 Sample mean x¯d=4.2 Sample standard deviation sd=19.6 Confidence…
Q: In a test of the effectiveness of garlic for lowering cholesterol, 48 subjects were treated with…
A:
Q: In a test of the effectiveness of garlic for lowering cholesterol, 50 subjects were treated with…
A: From the provided information, Sample size (n) = 50 Sample mean (x̅) = 4.5 Sample standard deviation…
Q: A doctor wants to estimate the mean HDL cholesterol of all 20- to 29-year-old females. How many…
A: The formula for the sample size is as follows:
Q: In a test of the effectiveness of garlic for lowering cholesterol, 48 subjects were treated with…
A: According to the provided information,
Q: In a test of the effectiveness of garlic for lowering cholesterol, 42 subjects were treated with…
A: From the provided information, Sample size (n) = 42 Sample mean (x̄) = 3.5 Sample standard deviation…
Q: na test of the effectiveness of garlic for lowering cholesterol 45 subjects were treated with garlic…
A: From the given information, x¯d=4.6sd=15.3n=45 Confidence level is 90% Significance level is 10%.…
Q: In a test of the effectiveness of garlic for lowering cholesterol, 44 subjects were treated with…
A:
Q: In a test of the effectiveness of garlic for lowering cholesterol, 50 subjects were treated with…
A: From the provided information, Sample size (n) = 50 Sample mean (x̅) = 5.1 Sample standard deviation…
Q: In a test of the effectiveness of garlic for lowering cholesterol, 50 subjects were treated with…
A: Given n=50 Mean=5.5 Standard deviations=17.9 Alpha=0.01
Q: In a test of the effectiveness of garlic for lowering cholesterol, 50 subjects were treated with…
A: Sample size, Sample mean, Sample standard deviation, The confidence level is 0.90.
Q: In a test of the effectiveness of garlic for lowering cholesterol, 42 subjects were treated with…
A:
Q: In a test of the effectiveness of garlic for lowering cholesterol, 42 subjects were treated with…
A:
Q: In a test of the effectiveness of garlic for lowering cholesterol, 42 subjects were treated with…
A: From the provided information, Sample size (n) = 42 Sample mean (x̄) = 4.4 Sample standard deviation…
Q: In a test of the effectiveness of garlic for lowering cholesterol, 48 subjects were treated with…
A: We have given that, Sample mean (x̄) = 5.5 , standard deviation (s) = 17.5 and sample size (n)…
Q: population sta ke use of the C periment might
A: Sample sizes are: nA=26nB=26 Sample means are: x¯A=20.1x¯B=24.1 So, x¯A-x¯B=4 Population standard…
Q: confidence interval estimate of the mean net change in LDL cholesterol after the garlic treatment.…
A: From the provided information, Sample mean (x̄ d ) = 4.4 Sample size (n) = 50 standard deviation (s…
Q: n LDL cholesterol after the garlic treatment. What does the confidence interval suggest about the…
A: The mean change (before - after) is 5.8 SD of change is 19.6 Sample size n = 50 then the degrees of…
Q: In a test of the effectiveness of garlic for lowering cholesterol, 50 subjects were treated with…
A:
Q: for lowering cholesterol,45 subjects were treated with garlic in a processed tablet form.…
A: Given Data : Sample Size, n = 45 Sample Mean, x̄ = 5.1 sample standard…
Q: What does the confidence interval suggest about the effectiveness of garlic in reducing LDL…
A:
Q: In a test of the effectiveness of garlic for lowering cholesterol, 44 subjects were treated with…
A: We have given that Mean = 3.3, sample size n = 44 standard deviation s = 18.4 significance level =…
Q: In a test of the effectiveness of garlic for lowering cholesterol, 45 subjects were treated with…
A: Given that, x¯=2.9,s=19.1,n=45 The degree of freedom is, df =n-1 =45-1 =44 Critical value:…
Q: In a test of the effectiveness of garlic for lowering cholesterol, 44 subjects were treated with…
A: Given that: Sample size, n=44 Sample mean, x¯=4.4 Standard deviation, s=19.4 Confidence level is…
Q: In a test of the effectiveness of garlic for lowering cholesterol, 50 subjects were treated with…
A: It is given that sample mean is 4.4 and the standard deviation is 19.3.
Q: In a test of the effectiveness of garlic for lowering cholesterol, 42 subjects were treated with…
A: It is given that sample mean is 2.7 and standard deviation is 18.9.
Q: In a test of the effectiveness of garlic for lowering cholesterol, 50 subjects were treated with…
A:
Q: The pH of rain, measured at a weather station in Michigan, was observed for 39 consecutive rain…
A:
Q: wat distribution tal W. Rage 1 of the sta w Rage 2 of the sta fidence interval estim
A: According to the sum, in a test of the effectiveness of garlic for lowering cholesterol, 43 objects…
Q: In a test of the effectiveness of garlic for lowering cholesterol, 50 subjects were treated with…
A:
Q: In a test of the effectiveness of garlic for lowering cholesterol, 42 subjects were treated with…
A: Given : Sample Size, n = 42 Sample Mean, x̄ = 5.7 sample standard…
It is given that difference in age is –38.60 and the sample standard deviation (s) is 23.18. The sample size (n) is 18.
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- 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 4.1 and a standard deviation of 19.1. 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? O mg/dL < µIn a test of the effectiveness of garlic for lowering cholesterol, 50 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 5.8 and a standard deviation of 17.3. Construct a 90% confidence interval estimate of the mean net change in LDL cholesterol after the garlic treatment. WWhat 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 p? mg/dLthe variance in waiting times(in minutes) of 10 attendees at bethel woods, where attendees enter a single waiting line that feeds three checkpoint windows is 0.227 minutes. A 95% confidence interval for the population variance would beIn 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?In a test of the effectiveness of garlic for lowering cholesterol, 45 subjects were treated with garlic in a processed tablet form. Cholesterol levels were measured before and after the treatment. The changes (beforeafter) in their levels of LDL cholesterol (in mg/dL) have a mean of 3.8 and a standard deviation of 19.2. 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?In a test of the effectiveness of garlic for lowering cholesterol, 44 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 4.6 and a standard deviation of 17.8. 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 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? Omg/dLIn 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 4.1 and a standard deviation of 19.1. 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 < µIn a test of the effectiveness of garlic for lowering cholesterol, 45 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 4.6 and a standard deviation of 16.3. 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? What is the confidence interval estimate of the population mean μ? ______ mg/dL<μ<_____ mg/dL (Round to two decimal places as needed.) What does the confidence interval suggest about the effectiveness of the treatment? A. The confidence interval limits contain 0, suggesting that the garlic treatment did not affect the LDL cholesterol levels. B. The confidence interval limits contain 0, suggesting that the garlic treatment…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 in their levels of LDL cholesterol have a mean of 5.7 and a standard deviation of 17.7. 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 and reducing LDL cholesterol?A study of 22 goats reveals that the average length of a goat’s tail is 6.3 inches. If the standarddeviation is known to be 1.2 inches find a 95% confidence interval for the mean tail length.A technician compares repair costs for two types of microwave ovens (type I and type II). He believes that the repair cost for type I ovens is greater than the repair cost for type II ovens. A sample of 35 type I ovens has a mean repair cost of $80.39. The population standard deviation for the repair of type I ovens is known to be $24.63. A sample of 31 type II ovens has a mean repair cost of $73.47. The population standard deviation for the repair of type II ovens is known to be $10.42. Conduct a hypothesis test of the technician's claim at the 0.05 level of significance. Let i be the true mean repair cost for type I ovens and µz be the true mean repair cost for type II ovens. Step 1 of 5: State the null and alternative hypotheses for the test.Recommended 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. 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