random sample of 10 cigarettes of a certain brand has an average nicotine content of 3.4 mg and a standard deviation of 0.8 mg. Construct a 99% confidence interval for the true average of nicotine content of this particular brand of cigarettes, assuming a normal distribution of population.
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- Assume researchers are trying to find out the mean concentration levels of a specific drug in a population of individuals’ blood during a clinical trail. The researchers find that the sample population n= 100 has a mean concentration of 225 ng/mL with a standard deviation of 47 ng/mL. Calculate a 95% confidence interval for the mean concentration level of the medication for the population’s blood and interpret it.In a random sample of 24 people, the mean commute time to work was 31.9 minutes and the standard deviation was 7.2 minutes. Assume the population is normally distributed and use a t-distribution to construct a 90% confidence interval for the population mean u. What is the margin of error of u? Interpret the results.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 4.4 and a standard deviation of 19.3. 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 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 normaldistribution table. What is the confidence interval estimate of the population mean u? mg/dLA 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.Assume researchers are trying to find out the mean concentration levels of a specific drug in a population of individuals’ blood during a clinical trail. The researchers find that the sample population of 30 has a mean concentration of 6.7 mcg/mL with a standard deviation of 1.3 mcg/ mL. Calculate a 95% confidence interval for the mean concentration level of the medication for the population’s blood and interpret it.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 4.1 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/dLA college admissions director wishes to estimate the mean age of all students currently enrolled. In a random sample of 30 students the mean age is found to be 21.5 years. From past studies the population standard deviation at the college is known to be 1.2 years and the population is normally distributed. Construct a 95% confidence interval of the population median age.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 < µSuppose a sample of 30 students are given an IQ test. The sample has a standard deviation of 12.23 points. Assuming IQ test score follow a normal distribution, suppose the researcher wishes to obtain a 90% confidence interval for the population standard deviation.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 womenSEE MORE QUESTIONSRecommended textbooks for youA First Course in Probability (10th Edition)ProbabilityISBN:9780134753119Author:Sheldon RossPublisher:PEARSONA First Course in Probability (10th Edition)ProbabilityISBN:9780134753119Author:Sheldon RossPublisher:PEARSON