The average daily intake of dietary fiber for a random sample of 53 young male adults (ages 19-49) is 9.05 grams/day with a standard deviation of 0.04 grams/day. Construct a 95% confidence interval for the average daily intake of dietary fiber for male young adults and compute its width.
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, 43 subjects were treated with…
A: The random variable LDL cholesterol level follows normal distribution. We have to construct 99%…
Q: n a test of the effectiveness of garlic for lowering cholesterol, 43 subjects were treated with…
A: In a test of the effectiveness of garlic for lowering cholesterol, 43 subjects were treated with…
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, 49 subjects were treated with…
A: Here we need to find a 95% confidence interval for 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:
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, 47 subjects were treated with…
A: Obtain the 90% 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, 49 subjects were treated with…
A:
Q: and after the treatment. The changes (before-after) in their levels of LDL cholesterol (in mg/dL)…
A: Given n=sample size=49, Sd=17.1, mean difference x̄d=2.8 Level of significance ɑ=0.05
Q: In a test of the effectiveness of garlic for lowering cholesterol, 49 subjects were treated with…
A: Given : The changes (before−after) in their levels of LDL cholesterol (in mg/dL) have a mean of…
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: 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, 43 subjects were treated with…
A: Given Information: Sample size n=43 Sample mean x¯d=3.2 Sample standard deviation sd=17.9 Confidence…
Q: In a test of the effectiveness of garlic for lowering cholesterol, 47 subjects were treated with…
A:
Q: In a test of the effectiveness of garlic for lowering cholesterol, 47 subjects were treated with…
A: The random variable net change in LDL cholesterol follows normal distribution. The sample size is…
Q: In a test of the effectiveness of garlic for lowering cholesterol, 47 subjects were treated with…
A: Let d denotes the changes (before-after) in the levels of LDL cholesterol (in mg/dL) of the…
Q: In a test of the effectiveness of garlic for lowering cholesterol, 49 subjects were treated with…
A: Given,sample size(n)=49sample mean(x¯)=5.6Sample standard deviation(s)=16.9degrees of…
Q: en analyses of a chemical in soil gave a mean of 20.92 mg/kg and a standard deviation of 0.45 mg/kg.…
A: n=10x¯=20.92S=0.45CI=0.95α=1-0.95=0.05we have to find out the 95 % CI for the given data
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Q: nd a standard deviation of 18.9. Construct a 90% confidence interval estimate of the mean net change…
A:
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 data sample size (n) = 42sample mean ( xd¯ ) = 5.5sample standard deviation (sd) =17.490% ci…
Q: Construct a 99% confidence interval estimate of the mean net change in LDL cholesterol after the…
A: In a test of the effectiveness of garlic for lowering cholesterol, 47 subjects were treated with…
Q: In a test of the effectiveness of garlic for lowering cholesterol, 49 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, 48 subjects were treated with…
A:
Q: In a test of the effectiveness of garlic for lowvering cholesterol, 45 subjects were treated with…
A:
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: In a test of the effectiveness of garlic for lowering cholesterol, 44 subjects were treated with…
A: From the provided information, Sample size (n) = 44 Sample mean (x̄) = 4.4 Sample standard deviation…
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: In a test of the effectiveness of garlic for lowering cholesterol, 44 subjects were treated with…
A: The degrees of freedom is, df=n-1=44-1=43 The critical-value is obtained by using Excel function…
Q: In a test of the effectiveness of garlic for lowering cholesterol,43 subjects were treated with…
A: Since population standard deviation is unknown, Use t-distribution to find t-critical value. Find…
Q: Construct a 95% confidence interval estimate of the mean net change in LDL cholesterol after the…
A:
Q: In a test of the effectiveness of garlic for lowering cholesterol, 48 subjects were treated with…
A: Given Information : In a test of the effectiveness of garlic for lowering cholesterol, 48 subjects…
Q: In a test of the effectiveness of garlic for lowering cholesterol, 43 subjects were treated with…
A: The sample size is 43, the mean is 3.4 and the standard deviation is 18.7.
Q: In a test of the effectiveness of garlic for lowering cholesterol, 43 subjects were treated with…
A:
Q: In a test of the effectiveness of garlic for lowering cholesterol, 49 subjects were treated with…
A: Since population standard deviation is unknown, Use t-distribution to find t-critical value. Find…
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, 49 subjects were treated with…
A: sample size(n)=49Mean()=3.6standard deviation(s)=17.7confidence level=90%
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Q: An automobile manufacturer needs to estimate the average highway fuel economy (miles per gallon) for…
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Q: In a test of the effectiveness of garlic for lowering cholesterol, 49 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, 48 subjects were treated with…
A: Sample size (n) = 48Sample mean (x̄) = 2.8 mg/dLStandard deviation (s) = 17.7 mg/dLConfidence…
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A: We have given that Sample size n= 1300 , sample mean = 12.3 mu=13 , standard deviation s=0.5
Q: In a test of the effectiveness of garlic for lowering cholesterol, 44 subjects were treated with…
A: From the provided information, Sample size (n) = 44 Sample mean (x̄) = 5.7 mg/dL Standard deviation…
Q: a test of the effectiveness of garlic for lowering cholesterol, 47 subjects were treated with…
A:
The average daily intake of dietary fiber for a random sample of 53 young male adults (ages 19-49) is 9.05 grams/day with a standard deviation of 0.04 grams/day. Construct a 95% confidence interval for the average daily intake of dietary fiber for male young adults and compute its width.
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- 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 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 3.5 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?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 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, 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, 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 4.1 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? Click here to view a t distribution table. LOADING...Bone mineral density (BMD) is a measure of bone strength. Studies show that BMD declines after age 45. The impact of exercise may increase BMD. A random sample of 59 women between the ages of 41 and 45 with no major health problems were studied. The women were classified into one of two groups based upon their level of exercise activity: walking women and sedentary women. The 39 women who walked regularly had a mean BMD of 5.96 with a standard deviation of 1.22. The 20 women who are sedentary had a mean BMD of 4.41 with a standard deviation of 1.02. Which of the following inference procedures could be used to estimate the difference in the mean BMD for these two types of womenIn 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 5.3 and a standard deviation of 17.4. 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? . What is the confidence interval estimate of the population mean μ? enter your response here mg/dL<μ<enter your response here mg/dL (Round to two decimal places as needed.)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(beforeminus after) in their levels of LDL cholesterol (in mg/dL) have a mean of 5.6 and a standard deviation of 17.5. 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? What is the confidence interval estimate of the population mean mu? mg/dLless than muless than mg/dL (Round to two decimal places as needed.)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 3.2 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? What is the confidence interval estimate of the population mean μ? ___mg/dL<μ<____mg/dL (Round to two decimal places as neededThe mean and standard deviation of system blood pressure of 200 employees in a factory were 127mmhg and 13 mmHg respectively. Calculate and interprete the 95% confidence interval of systolic blood pressure of the employeesIn a test of the effectiveness of garlic for lowering cholesterol, 47 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 3.7 and a standard deviation of 18.2. 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 p? Question Viewer mg/dL < p< 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…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 3.1 and a standard deviation of 15.8. 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 µ? mg/dL mg/dL<µ< (Round to two decimal places as needed.) What does the confidence interval suggest about the effectiveness of the treatment? O A. 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