In a random sample of 8 people, the mean commute time to work was 36.5 minutes and the standard deviation was 7.4 minutes. A 98% confidence interval using the t-distribution was calculated to be (28.7,44.3). After researching commute times to work, it was found that the population standard deviation is 8.7 minutes. Find the margin of error and construct a 98% confidence interval using the standard normal distribution with the appropriate calculations for a standard deviation that is known. Compare the results. The margin of error of μ is ______nothing.
Q: In a test of the effectiveness of garlic for lowering cholesterol, 43 subjects were treated with…
A: From the provided information: Sample size, n=43. Sample mean, x¯=3.4. Sample standard deviation,…
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.
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Q: In a test of the effectiveness of garlic for lowering cholesterol, 42 subjects were treated with…
A: The changes in the levels of LDL cholesterol have a mean of and a standard deviation of 18.7. The…
Q: In a test of the effectiveness of garlic for lowering cholesterol, 44 subjects were treated with…
A: Given that Sample size =44 subjects Mean difference=3.4 Sample standard deviation (before-after)…
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, 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 (x̄) = 2.8 Standard deviation (s) =…
Q: test of the effectiveness of garlic for lowering cholesterol, 43 subjects were treated with garlic…
A: Given that Sample size n =43 Sample mean =5.7 Standard deviation =16.1
Q: In a test of the effectiveness of garlic for lowering cholesterol, 48 subjects were treated with…
A: We have given that, The given information is -The sample mean change in the LDL cholesterol level…
Q: In a test of the effectiveness of garlic for lowering cholesterol, 44 subjects were treated with…
A: The sample size is 44, the mean is 4.9 and the standard deviation is 15.7.
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, 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, 42 subjects were treated with…
A: Given that,The sample size is 42.The sample mean is 5.3.The sample standard deviation is 18.5.The…
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.
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Q: In a test of the effectiveness of garlic for lowering cholesterol, 50 subjects were treated with…
A: Givensample size(n)=50Mean(x)=3.4standard deviation(s)=15.3confidence level=95%
Q: In a test of the effectiveness of garlic for lowering cholesterol, 48 subjects were treated with…
A: Given that x¯=5.2s=15.6n=48 Confidence level=90% Significance level=1-0.90=0.10 degrees of freedom,…
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, 43 subjects were treated with…
A: Sample mean = x̅ = 4.5 Sample size = n = 43 Sample S.D = s = 16.8 Standard Error = s/√n =…
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…
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Q: In a test of the effectiveness of garlic for lowering cholesterol, 50 subjects were treated with…
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Q: What does the confidence interval suggest about the effectiveness of garlic in reducing LDL…
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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 =…
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Q: In a test of the effectiveness of garlic for lowering cholesterol, 48 subjects were treated with…
A: Suppose defines the true population mean net change in LDL cholesterol after the garlic treatment.
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Q: In a test of the effectiveness of garlic for lowering cholesterol, 42 subjects were treated with…
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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…
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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…
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Q: na test of the effectiveness of garlic for lowering cholesterol, 42 subjects were treated with…
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- A random sample of 550 credit-card holders shows that the mean annual credit-card debt for individual accounts is $8220, with a standard deviation of $1102. Use these statistics to construct a 90% confidence interval for the mean annual credit-card debt for the populations of all accounts.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 3.1 and a standard deviation of 15.4. 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/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 4.7 and a standard deviation of 16.2. 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. What is the confidence interval estimate of the population mean µ? Omg/dL <µ< mg/dL (Round to two decimal places as needed.)In 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 3.2 and a standard deviation of 15.3. 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, 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 5.7 and a standard deviation of 19.6. 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. ..... What is the confidence interval estimate of the population mean u? mg/dL < µ< mg/dL (Round to two decimal places as needed.) What does the confidence interval suggest about the effectiveness of the treatment? O A. The confidence interval limits do not contain 0,…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 < µA local health center noted that in a sample of 400 patients, the average sample age is 30 years with a population standard deviation of 4.4. Calculate the standard deviation of the sample mean of patients in the study. Provide a 95% confidence interval for the average age of all the patients who are referred to the health center by the hospital.In 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/dLIn 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 5.5 and a standard deviation of 19.8. 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? O mg/dL < µIn 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 3.9 and a standard deviation of 15.3. 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 (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, 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. 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