Concept explainers
Cushing’s disease is characterized by muscular weakness due to adrenal or pituitary dysfunction. To provide effective treatment, it is important to detect childhood Cushing’s disease as early as possible. Age at onset of symptoms and age at diagnosis (months) for 15 children suffering from the disease were given in the article “Treatment of Cushing’s Disease in Childhood and Adolescence by Transphenoidal Microadenomectomy” (New Engl. J. of Med., 1984: 889). Here are the values of the differences between age at onset of symptoms and age at diagnosis
-24 | -12 | -55 | -15 | -30 | -60 | -14 -21 |
-48 | -12 | -25 | -53 | -61 | -69 | -80 |
a. Does the accompanying normal probability plot cast strong doubt on the approximate normality of the population distribution of differences?
b. Calculate a lower 95% confidence bound for the population mean difference, and interpret the resulting bound.
c. Suppose the (age at diagnosis) – (age at onset) differences had been calculated. What would be a 95% upper confidence bound for the corresponding population mean difference?
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Chapter 9 Solutions
Probability and Statistics for Engineering and the Sciences
- please help me out. show full working out for better understanding Hereditary hemochromatosis (HH) is a recessively inherited genetic disorder resulting from mutations in the HFE gene. There are several possible mutations of different degrees of importance but the disease expresses itself only when at least 2 arepresent in any one individual. A blood analysis was performed on 3,000 blood samples from newborn babies of Caucasian descent in the state of Michigan, USA. The results showed that 163 of those sampled carried two mutations in the HFE gene. a. What is the proportion in the sample that carry the two mutations of the HFE gene? b. Calculate the 95% confidence interval for the proportion of this population that carries two mutations in the HFE Check the data for the necessary conditions and show full working. c. Interpret your interval in context d. If the researcher wanted to halve (x ½) the margin of error in the confidence interval found, what sample size would be required…arrow_forward“Passive and Active Smoke” in Appendix B includes cotinine levels measured in a group of nonsmokers exposed to tobacco smoke (n = 40, Mean = 60.58 ng>mL, s = 138.08 ng>mL) and a group of nonsmokers not exposed to tobacco smoke (n = 40, Mean = 16.35 ng>mL, s = 62.53 ng>mL). Cotinine is a metabolite of nicotine, meaning that when nicotine is absorbed by the body, cotinine is produced. Use a 0.05 significance level to test the claim that nonsmokers exposed to tobacco smoke have a higher mean cotinine level than nonsmokers not exposed to tobacco smoke. 1. Construct a confidence interval estimate of the difference betwen the mean continen levels fo the two groups of nonsmokers. What confidence level would be appropriate? 2. Find the margin of error E using the formula. 3. What is the confidence interval? Explain the meaning of the confidence interal and what the limit represents.arrow_forward“Passive and Active Smoke” in Appendix B includes cotinine levels measured in a group of nonsmokers exposed to tobacco smoke (n = 40, Mean = 60.58 ng>mL, s = 138.08 ng>mL) and a group of nonsmokers not exposed to tobacco smoke (n = 40, Mean = 16.35 ng>mL, s = 62.53 ng>mL). Cotinine is a metabolite of nicotine, meaning that when nicotine is absorbed by the body, cotinine is produced. Use a 0.05 significance level to test the claim that nonsmokers exposed to tobacco smoke have a higher mean cotinine level than nonsmokers not exposed to tobacco smoke. Based on your hypothesis tests and confidence intervals, what do you conclude about the effects of second hand smoke? Why?arrow_forward
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- The data in the attached image represents blood pressure based on a new trial medication used in an experiment involving a total of 50 male and female subjects. There were two levels of treatment: 0 = placebo and 1 = treatment group. Family History Blood Pressure = Fam-Hist-BP; from the paternal side = PS; from the maternal side = MS; from both sides = BS; none on both sides or not known = N.; Blood pressure before the experiment = BP-Before-Exp; Blood pressure after the experiment = BP-After-Exp. Task: Generate at least two different cross tabulations.arrow_forwardThe data in the attached image represents blood pressure based on a new trial medication used in an experiment involving a total of 50 male and female subjects. There were two levels of treatment: 0 = placebo and 1 = treatment group. Family History Blood Pressure = Fam-Hist-BP; from the paternal side = PS; from the maternal side = MS; from both sides = BS; none on both sides or not known = N.; Blood pressure before the experiment = BP-Before-Exp; Blood pressure after the experiment = BP-After-Exp. Task: Construct a side-by-side boxplot to compare the variables BP-Before-Exp and BP-After-Exp.arrow_forwardThe data in the attached image represents blood pressure based on a new trial medication used in an experiment involving a total of 50 male and female subjects. There were two levels of treatment: 0 = placebo and 1 = treatment group. Family History Blood Pressure = Fam-Hist-BP; from the paternal side = PS; from the maternal side = MS; from both sides = BS; none on both sides or not known = N.; Blood pressure before the experiment = BP-Before-Exp; Blood pressure after the experiment = BP-After-Exp. Task: Construct a histogram to represent the family BP history, the subjects' BP before treatment and the subjects' BP after treatment.arrow_forward
- The data in the attached image represents blood pressure based on a new trial medication used in an experiment involving a total of 50 male and female subjects. There were two levels of treatment: 0 = placebo and 1 = treatment group. Family History Blood Pressure = Fam-Hist-BP; from the paternal side = PS; from the maternal side = MS; from both sides = BS; none on both sides or not known = N.; Blood pressure before the experiment = BP-Before-Exp; Blood pressure after the experiment = BP-After-Exp. Task: Find the mean and standard deviation for the family history group, the before treatment group, and the after treatment group.arrow_forwardShow all work/steps! You must use JMP to analyze data.arrow_forwardCoronary stenosis (narrowing of the artery supply to the heart muscle) is a direct cause of heart disease. A theory suggests that chronic cytomegalovirus (CMV) infection narrows coronary vessels and leads to coronary heart disease. To test this theory, 75 patients undergoing angioplasty were followed for 6 months following their procedure. The 49 patients who were seropositive for CMV experienced an average luminal diameter reduction of 1.24mm (s1 = 0.83 mm). In contrast, the 26 patients who were seronegative for CMV experienced and average luminal reduction of 0.68 (s2 = 0.69).k Test whether this mean difference in luminal reduction is significant. Show all hypothesis-testing steps. Do the data support the hypothesis that chronic CMV virus infection plays a role in coronary luminal reduction? What is the 95% confidence interval?arrow_forward
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