STAT 554: HW #2
due March 3, 2008
As with HW #1, put all of your answers on the answer sheet (posted on the course web site). Express all
confidence intervals in the form (lower confidence bound, upper confidence bound), as opposed to (point
estimate ±δ). Also, for the first problem, round the confidence bounds to the nearest tenth, and for all of
the sample size computations report the exact integer value. For the power values requested in the second
problem, round to the nearest thousandth. Otherwise, round answers to two significant digits, unless specific
instructions are given to do something different. The p oints total to 13.5, but I’ll truncate scores at 13 for
this assignment.
1) Suppose that the percent moisture was measured for batches of a dozen seeds from the wheat heads of
six randomly selected wheat plants (by weighing, drying, and then reweighing each batch of seeds), and that
the following measurements were obtained: 62.7, 66.3, 60.6, 63.0, 62.7, 63.7.
(a) (0.5 point) Assume (near) normality and give a 95% confidence interval for the mean percent moisture
for all wheat plants grown in the same conditions as those in the sample.
(b) (0 points) Now give a 90% confidence interval (and note that it is narrower).
(c) (0.5 point) Now give a 99% confidence interval (and note that it is widest interval of the three).
(d) (1 point) Suppose one is disappointed that the 95% confidence interval of part (a) is as wide as it is, and
wants to supplement the sample with additional observations and produce a 95% confidence interval that
has a half-width no greater than 0.5 (making the total width of the interval less than or equal to 1.0). If
one merely combines new data with the original six observations to produce a single sample and forms
the confidence interval from that sample in the usual way, it’s not clear how many new observations
should be added since the width of the resulting interval depends upon the sample standard deviation,
and that value will not be known until the new observations are taken. However, if Stein’s two-stage
procedure is used, one can predetermine the width of the confidence interval which results from adding
a given number of new observations to the sample. Here’s how Stein’s method works: determine the
sample standard deviation s0from an initial sample of n0observations, then add n1more observations
to produce a combined sample of size n∗=n0+n1and compute the sample mean x∗from this combined
sample. A confidence interval follows from the fact that
√n∗ÃX∗−µX
S0!∼ Tn0−1,
and this result means that once the value s0is obtained from the initial sample, the width of the final
interval that will result from adding n1observations can be determined before those data values are
obtained. I want you to use the initial sample of six observations to determine the fewest number
of additional observations that will result in a 95% confidence interval formed using Stein’s two-stage
procedure having a half-width no greater than 0.5. Also, give the resulting confidence interval if the
sample mean of the n1additional observations is 61.38.
2) X1, X2, . . . , Xnare iid N(µ, 400.0) random variables. Consider using a ztest to do a test about the
(unknown) mean.
(a) (1.5 points) If n= 25, what is the power of a size α= 0.05 ztest of H0:µ= 200 vs. H1:µ6= 200
against the alternative µ= 195 ?
(b) (1.5 points) For a size α= 0.05 ztest of H0:µ= 200 vs. H1:µ6= 200, what sample size is needed in
order for the power of the test against the alternative µ= 195 to be approximately 0.80? (Hint: The
value of the power gets a contribution from each of the two portions of the rejection region. Initially,
assume that the contribution from one of the tails is so small that it’s negligible. Then determine the
sample size that makes the contribution from the other tail equal to 0.80. Once you get a sample size
this way, you can wrap things up by confirming that the contribution that was initially assumed to be
negligible really is negligible.)
3) Suppose that you observe 11 black outcomes in 15 trials with my toy roulette wheel. Let pbe the
probability of obtaining a black outcome, and consider a test of H0:p= 9/19 against H1:p6= 9/19.
