Answer all numeric questions to 3 decimal places unless the number has fewer dig

Question

Answer all numeric questions to 3 decimal places unless the number has fewer digits or you are otherwise instructed.

I just need answer. I do not need R code. I want to make sure if I have correct answers. Answer all. Thank you

Question Detalls My NotesAsk Your Teacher 0 According to the American Heart Association, "Cardiomyopathy refers to diseases of the heart muscle. These diseases have many causes, signs and symptoms, and treatments. The heart muscle becomes enlarged, thick or rigid in cardiomyopathy, and in rare cases the muscle tissue is replaced with scar tissue." The following is the left ventricular ejection fraction for a random sample of 17 patients with acute dilated cardiomyopathy 0.19, 0.24, 0.40, 0.23, 0.30, 0.19, 0.24, 0.32, 0.28, 0.24 0.18, 0.22, 0.14, 0.30, 0.07, 0.12, 0.17 a) Calculate the upper 8% percentile(Same as 92nd percentile) of the t distribution with 16 degrees of freedom. b) Calculate the upper 406 percentile(same as 96th percentile) of the t distribution with 16 degrees of freedom. c) Calculate the upper 4% percentile(same as 96th percentile) of the t distribution with 17 degrees of freedom. d) Calculate the upper 496 percentile(Same as 96th percentile) of the standard normal distribution. e)Calculate the sample variance for this data f) Calculate the sample standard deviation for this data? g) Calculate the sample mean for this data. h) Calculate the maximum likelihood estimate for using this data i) Compute a 92% Confidence Interval for . Compute a 92% Prediction Interval for a single future weight measurement. k) Copy your R script for the above into the text box here.

Explanation / Answer

(a) Upper 8% percentile with degree of freedom (16) = 92% percentile = Pr(0.92,16) = 1.4736

(b) Upper 4% percentile with degree of freedom (16) = 96% percentile = Pr(0.92,16) = 1.8693

(c) Upper 8% percentile with degree of freedom (17) = 92% percentile = Pr(0.92,16) = 1.4694

(d)  Upper 4% percentile of the standard normal distribution = 96% percentile = 1.7507

(e) Sample variance = 0.006526

(f) Sample standard deviation = 0.080787

(g) sample mean x = 0.2253

(h) MAximum likelihood estimate for = 0.2253

(i) 92% confidence interval = x +- tdf,0.08 (s/ sqrt(n)

= 0.2253 +- 1.8619 * (0.080787/sqrt(17)

= (0.1888, 0.2618)

(ii) 92 % prediction interval for a single future measurement = x +- Z92%s

= 0.2253 +- 1.7507 * 0.080787

= (0.0839, 0.2596)

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