238 ANSWERS TO SELECTED PROBLEMS
6.7 Ps5 = 148.2 pounds.
6.8 With a sample size of 16 and weights perhaps being a somewhat skewed dis-
tribution, we would expect the sample means to be approximately normally
distributed.6.9 Try a log or a square-root transformation on the data since it is skewed to the
right.6.10 z = 00, 1.65, 1.96, and 2.83.
6.11 z = 1.96.
Chapter 7
7.1 The 95% confidence limit is 1.04 < p < 2.09. Since 1.407 is almost in
the middle of the confidence limit, it appears that the pressure may not have
changed the mean bleeding time.7.2 No answer.7.3 No answer.7.4 The 95% confidence limit is 1.358 < p1 - p2 < 5.559.
7.5 The 95% confidence limit is ,364 < p1- p2 < .972. Yes, since the confidence
limit does not cover 0.7.6 (a) No, the 95% confidence limit is -.11 to .99 kg. (b) If the sample size is 50,
the 95% confidence limit is .13 to .75, so the answer is yes.7.7 (a) The mean difference of -.433 degrees Celsius. (b) The 95% confidence
interval is -.7455 to -.1211. In 95% of such experiments the computed
intervals cover the true difference. Since the confidence limit does not include
zero, the conclusion is that the rectal temperatures are higher than the oral
temperatures. (c) The 95% confidence limit for the mean oral temperature is
37.3-37.9. (d) The 95% confidence limit for the mean rectal temperature is
37.6-3 8.5.7.8 (a) sp = ,0762. (b) 18 d.f. (c) The 99% confidence interval is -.410 to -.214.
(d) Yes, since the confidence limit does not include 0.7.9 The 99% confidence limit is -50.12 to 2.12 with 11 d.f. A difference may exist
but the data have failed to establish one since the interval covers 0.7.10 The sample size in each group should be 217.