These questions cover content that will be on the final.
A. It is claimed that the mean heihgt of men in a particular population
exceeds 69 inches. Write the null and alternative hypothesis for
a test of this claim.
B. Suppose that heights are normally distributed in the population
mentioned in A. Then we collect a random sample of 100 people and
measure their heights. The average
of this sample, Xbar, is a random number and is also normally distributed.
What is the mean and SD of Xbar?
C. We want to test the effectiveness of the drug Laevo, which is a
sleeping pill. The table on p. 614 (Table 12.1) gives the differences in
hours of sleep for 10 patients on two successive nights. On one night,
the patients took the standard medication, Dextro. On the other night,
they took Laevo. For example, patient A got .7 more additional hours
of sleep with Dextro, and 1.9 additional hours with Laevo. The difference
is 1.2, which means that this patient got 1.2 additional hours of sleep
using Laevo than using Dextro. The makers of Laevo claim that their
drug has a higher mean number of additional hours of sleep then Dextro.
Dextro would claim that the observed difference is due to chance.
a) State the null and alternative hypotheses.
b) Our test statistic is based on the average. If we think of
the average as being based on a random sample of 10 people, and that the
differences in the table should be normally distributed in the population,
then Xbar, the average of these 10 random people , is normally distributed.
According to the null hypothesis, what is the mean for Xbar?
c) What is the observed value for the average?
d) What is the standard deviation of the observed differences?
Note that this is the standard deviation of the sample, but we will
use it as an estimate of the standard deviation of the population.
(Hint, use s. Think way back to the second week of the course.)
e) If the SD of the population were known, then the SD of Xbar would
be SD/sqrt(n). And then, Z = (Xbar - mu)/(SD/sqrt(n)) would
be a standard normal random variable. However, SD is not known for
the population, but we can substitute s/sqrt(n) as an estimate. In
that case, T = (Xbar - mu)/(s/sqrt(n)) is not a standard normal random
variable. But instead, its pdf is something called a t-distribution
with n-1 degrees of freedom. In this problem, n=10, so this is a t-distribution
with 9 degrees of freedom. Using the table in Appendix F, find the
probability P(T > 1.38) and P(T > 2.82).
f) What's the observed value of T for the observed data?
g) If we use a significance level of 5%, then we would reject the null
hypothesis if p-value < 0.05. Here, the p-value is P(T > observed
value of T). Using Appendix F, what is the smallest value of the
observed value of T for which we would reject the null hypothesis?
h) Do you reject the null hypothesis for your data?
p. 616, #3. (Note, pvalue = P(T < observed T). Why?)
#4, #5