Consider the following observations on a receptor binding measure (adjusted dist

Question

Consider the following observations on a receptor binding measure (adjusted distribution volume) for a sample of 13 healthy individuals: 23, 39, 40, 42, 43, 48, 52, 57, 64, 66, 67, 69, 71. (a) Is it plausible that the population distribution from which this sample was selected is normal? ,it plausible that the population distribution is normal. (b) Calculate an interval for which you can be 95% confident that at least 95% of all healthy individuals in the population have adjusted distribution volumes lying between the limits of the interval. (Round your answers to three decimal places.) (c) Predict the adjusted distribution volume of a single healthy individual by calculating a 95% prediction interval. (Round your answers to three decimal places.) How does this interval's width compare to the width of the interval calculated in part (b)? This interval's width is less than the width of the interval calculated in part (b). You may need to use the appropriate table in the Appendix of Tables to answer this question. Need Help? Read It T Talk to a Tutor

Explanation / Answer

Solutionb:

In R

Recbindm <- c(23,39,40,42,43,48,52,57,64,66,67,69,71)
t.test(Recbindm)

ootput:

data: Recbindm
t = 12.859, df = 12, p-value = 2.231e-08
alternative hypothesis: true mean is not equal to 0

data: Recbindm
t = 12.859, df = 12, p-value = 2.231e-08
alternative hypothesis: true mean is not equal to 0
95 percent confidence interval:
43.50835 61.26088
sample estimates:
mean of x
52.38462
sample estimates:
mean of x
52.38462

95% confidnce interval for mean is

43.508,61.261

Solutionc:

t crit=2.179 for 12 df and 0.05 level of significance

sample mean =x bar= 52.38462

sample sd=s= 14.68865

95% PI is

xbar-t*s*sqrt(1+1/n),xbar+t*s*sqrt(1+1/n)

=52.38462-2.179 *14.688658*sqrt(1+1/13),52.38462+2.179 *14.688658*sqrt(1+1/13)

=19.17,85.599

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