Question

(3 points) The study by Schlaich et al. (1998) investigated lung function in patients with spinal osteoporosis. The volume of forced expiration in one second was recorded on each of 34 patients, this figure being adjusted for associated variables such as age and gender and then recorded as a percentage of some standardised figure. Suppose the mean and standard deviation of the data were 95 and 14.2 respectively, both in percentage points. It is assumed the data are from a Normal distribution with unknown mean μ and standard deviation σ, and the researchers were interested in estimating μ. Provide answers to the following to two decimal places. Part a) Compute a 90% confidence interval for μ, basing your inference on a Student's t-distribution. Provide the width of the interval. Part b) Before Student derived the t-distribution, it was common to use the standard Normal distribution for estimating confidence intervals as above. Recompute the confidence interval you found in (a) but using the standard Normal rather than the t-distribution. Provide the width of the interval. Part c) Taking the normal width as the base, compare the relative widths of the intervals you found in (a) and (b) rounded to two decimal places. By what percentage of the width of the interval in (b) is the width of the interval in (a) larger or smaller? Schlaich, C., Minne, H. W., Bruckner, T., Wagner, G., Gebest, H. J., Grunze, M., Ziegler, R., and Leidig-Bruckner, G. (1998): "Reduced Pulmonary Function in Patients with Spinal Osteoporotic Fractures." Osteoporosis International 8, 261-267.

Answer 1

a)

sample mean 'x̄=	95.000
sample size   n=	34.00
sample std deviation s=	14.200
std error 'sx=s/√n=	2.4353

for 90% CI; and 33 df, value of t=		1.692
margin of error E=t*std error    =		4.12

b)

for 90 % CI value of z=		1.645
margin of error E=z*std error =		4.01

c)

=(4.01-4.12)*100/4.01 =2.74 % smaller (please try 2.67 if this comes wrong and reply)