4) Question: A sample of 40 patients shows that their average cholesterol level is 200 mg/dL with a standard deviation of 20 mg/dL. Construct a 99% confidence interval for the population mean cholesterol level.
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- I need help with this question #44) Question: A sample of 40 patients shows that their average cholesterol level is 200 mg/dL with a standard deviation of 20 mg/dI.. Construct a 99% confidence interval for the population mean cholesterol level.According to a study of political prisoners, the mean duration of imprisonment for 31 prisoners with chronic post-traumatic stress disorder (PTSD) was 30.8 months. Assuming that o = 38 months, determine a 95% confidence interval for the mean duration of imprisonment, μ, of all political prisoners with chronic PTSD. Interpret your answer in words. Click here to view Page 1 of the table of areas under the standard normal curve. Click here to view Page 2 of the table of areas under the standard normal curve. A 95% confidence interval for the population mean is from months to (Round to one decimal place as needed.) months. C... Interpret the confidence interval. Select the correct choice below and fill in the answer boxes to complete your choice. (Round to one decimal place as needed.) O A. There is a 95% chance the mean duration of imprisonment, μ, of all political prisoners with chronic PTSD will equal the mean of the interval from months to OB. We can be 95% confident that the mean…
- What does it mean when you have standard deviations of 27 35 and 32 and its effect on the validitity of the data you are accessing to interpret for statistical anaylsis?In a test of the effectiveness of garlic for lowering cholesterol, 49 subjects were treated with raw garlic. Cholesterol levels were measured before and after the treatment. The changes (before minus after) in their levels of LDL cholesterol (in mg/dL) have a mean of 0.9 and a standard deviation of 2.33 . Use a 0.10 significance level to test the claim that with garlic treatment, the mean change in LDL cholesterol is greater than 0 . What do the results suggest about the effectiveness of the garlic treatment? Assume that a simple random sample has been selected. Identify the null and alternative hypotheses, test statistic, P-value, and state the final conclusion that addresses the original claim. What are the null and alternative hypotheses? A. Upper H 0 : mu greater than0 mg/dL Upper H 1 : mu less than0 mg/dL B. Upper H 0 : mu equals0 mg/dL Upper H 1 : mu less than0 mg/dL C. Upper H 0 : mu…According to a study of political prisoners, the mean duration of imprisonment for 37 prisoners with chronic post-traumatic stress disorder (PTSD) was 34.8 months. Assuming that o = 45 months, determine a 95% confidence interval for the mean duration of imprisonment, μ, of all political prisoners with chronic PTSD. Interpret your answer in words. Click here to view Page 1 of the table of areas under the standard normal curve. Click here to view Page 2 of the table of areas under the standard normal curve A 95% confidence interval for the population mean is from (Round to one decimal place as needed.) months to months. Interpret the confidence interval. Select the correct choice below and fill in the answer boxes to complete your choice. (Round to one decimal place as needed.) O A. There is a 95% chance the mean duration of imprisonment, μ, of all political prisoners with chronic PTSD will equal the mean of the interval from months. months to O B. We can be 95% confident that the mean…
- In a test of the effectiveness of garlic for lowering cholesterol, 49 subjects were treated with raw garlic. Cholesterol levels were measured before and after the treatment. The changes (before minus after) in their levels of LDL cholesterol (in mg/dL) have a mean of 0.8 and a standard deviation of 2.04. Use a 0.10 significance level to test the claim that with garlic treatment, the mean change in LDL cholesterol is greater than 0. What do the results suggest about the effectiveness of the garlic treatment? Assume that a simple random sample has been selected. Identify the null and alternative hypotheses, test statistic, P-value, and state the final conclusion that addresses the original claim.The weights of students in a large statistics class vary according to a normal distribution with mean 145 pounds and standard deviation 18 pounds what percent of students weigh less than 100 pounds ?An investigation of the relationship between traffic flow (measured in cars per day) and lead content (measured in micrograms per gram of dry weight) in the bark of trees near the highway yielded the following summary statistics: Sample Mean Sample Standard Deviation Lead Content (micrograms per gram dry weight) y¯=y¯= 680 sysy = 240 Traffic Flow (cars / day) x¯=x¯= 1750 sxsx = 800 The correlation between lead content and traffic flow was found to be r = 0.6 and a scatterplot showed the form to be linear. Since Traffic flow is the X-variable, its mean and standard deviation have been labelled x¯ and sxx¯ and sx. Since Lead content is the Y-variable, its mean and standard deviation have been labelled y¯ and syy¯ and sy . Determine slope and y-intercept of the least squares regression line for using X to predict Y. Group of answer choices Slope y-intercept
- Researchers from a certain country were interested in how characteristics of the spleen of residents in their tropical environment compare to those found elsewhere in the world. The researchers randomly sampled 91 males and 109 females in their country. The mean and standard deviation of the spleen lengths for the males were 10.9 cm and 0.9 cm, respectively, and those for the females were 9.8 cm and 1.1 cm, respectively. Determine a 95% confidence interval for the difference between mean spleen lengths of males and females in this country. Click here to view page 1 of the table of critical values of t. Click here to view page 2 of the table of critical values of t. Let population 1 be all males in the country and let population 2 be all females in the country. The 95% confidence interval for µ1 - 4½ is ( | D (Round to three decimal places as needed. Use ascending order.)Low-density lipoprotein, or LDL, is the main source of cholesterol buildup and blockage in the arteries. This is why LDL is known as "bad cholesterol." LDL is measured in milligrams per deciliter of blood, or mg/dL. In a population of adults at risk for cardiovascular problems, the distribution of LDL levels is normal, with a mean of 123 mg/dL and a standard deviation of 41 mg/dL. If an individual's LDL is at least 1 standard deviation or more above the mean, he or she will be monitored carefully by a doctor. What percentage of individuals from this population will have LDL levels 1 or more standard deviations above the mean? Use the 68–95–99.7 rule. (Enter your exact answer as a whole number.) percentage %Almost all medical schools in the United States require applicants to take the Medical College Admission Test (MCAT). On one exam, the scores of all applicants on the biological sciences part of the MCAT were approximately Normal with mean 9.4 and standard deviation 2.6. For applicants who actually entered medical school, the mean score was 10.6 and the standard deviation was 1.7. (a) What percent of all applicants had scores higher than 11? (b) What percent of those who entered medical school had scores between 8 and 12?