HW 8 Solutions
• Extra#2*: A random sample of 9 overweight men tested an experimental
diet. (The actual test also included a control group, but in this problem
we consider only the treatment group.) Their change in weight over
a 2 week period is given below. A negative value means they lost weight.
Would you conclude the diet is successful? Use a 95% confidence
interval to answer.
-2.9, -5.7, 1.3, 2.0, 0.0, 1.6, -9.1, 2.1, -4.2
First, note that the population we're considering here consists of all overweight
men who will someday take our pills. The diet could be considered a success
if the mean weightloss for this population were negative. Of course,
the more negative the better. So our goal here is to estimate the mean
weight loss for this population.
We can't see the population, but we got to see what happens on a small sample
of 9 people form this population. The mean of this sample is -1.656
and the standard deviation is s= 4.021.
A 95% CI is therefore -1.656 +/- t (4.021/sqrt(9))
where t comes from a table of t-statistics with 9 degrees of freedom. For
a 95% CI, we use the column in Appendix A6 that has P(T>t) = .025
which is 2.262.
Our CI is -1.656 +/- 2.262*(4.0121/3)
-1.656 +/- 3.025
or (-4.68, 1.37)
Because this interval contains 0, we can't rule out the possibility that
the true mean weight loss is 0. In fact, we can't rule out the possibility
that the mean weight loss will actually be a weight gain! So we must
conclude the diet doesn't work.
Extra #4*: If the mean speed on a road is faster than the posted speed
limit, people can contest speeding tickets (at least in some cities). Below
are recorded speeds on a street in which the speed limit is 30 mhp. We
will examine whether there is evidence that the mean speed of all cars
is faster than the posted speed limit. The observed speeds were 29,
34, 34, 28, 30, 29, 38, 31, 29, 34, 32, 31, 27, 37, 29, 26, 24, 34, 36,
31, 34, 36, 21
The five number summary is (21.00, 29.00, 31.00, 34.00, 38.00). The average
of the observed speed was 31.04 and the standard deviation was 4.24.
a) Make an appropriate plot of the data and describe the distribution
of speeds.
A histogram would show the distribution. The distribution is slightly
skewed left, but if your histogram had fewer bins you might see it as symmetric.
Skewed left is an unexpected shape for these data (you'd expect skewed
right, I think.) The reason for checking is that t-tests assume the
population is normal, or nearly normal. Without some experience it's
hard to tell, but this is fairly symmetric and so given a sample size in the
20's, we'd have good reason to believe the t-test will be appropriate.
b) State the null and alternative hypotheses to test.
H0: mean speed = 30
Ha: mean speed > 30
c) Compute the appropriate test statistic.
t = (31.04 - 30)/(4.24/sqrt(23))
= 1.177
d) Find the p-value for the observed test statistic.
p-value = P(T > 1.177) where T follows a t distribution with n-1=22
degrees of freedom.
From a computer: p-value = .126
OR....
From Appendix A6: Using the row for 22 degrees of freedom, 1.177
is between 1.061 and 1.321. So the p-value is between .10 and .15.
In other words, it's bigger than .05.
e) Would you conclude tht the true mean speed is faster than 30 miles
per hour?
No. There's no evidence that the true mean differs from 30 mhp using
a 5% significance level (alpha = .05)