180 Solutions to Selected Exercises- Explain how the percentile method bootstrap confi dence interval for a param-
eter is obtained.
To obtain a percentile method bootstrap confi dence interval with confi dence
level 100(1 − α )%, generate many (e.g., 1000) bootstrap samples. Calculate the
estimate of interest for each bootstrap sample. Order the estimates from lowest to
highest. Find the fi rst integer greater than or equal to 100 α /2. Let ’ s call that m.
Then look for the m th bootstrap estimate, call it E()m. Then fi nd the fi rst integer
greater than 100(1 − α /2). Call that integer ml. Find the ml th bootstrap estimate in
the ordered list and call that E()ml. Then the interval [ E()m, E()ml] is a two - sided
100(1 − α ) % bootstrap percentile confi dence interval for the parameter being
estimated. - The mean weight of 100 men in a particular heart study is 61 kg, with a
standard deviation of 7.9 kg. Construct a 95% confi dence interval for the
mean.
We assume that the 100 men constitute a random sample of size 100, and that the
central limit theorem will apply to the average weight. So fi rst we compute
the Z - statistic for the lower and upper bounds of the 95% confi dence interval. Call
the confi dence interval [ L , U ], then for the Z - scores the interval is [( L − 61)/7.9,
( U − 61)/7.9]. To make this a symmetric two - sided interval, we want ( L − 61)/7.9
to be the 2.5 percentile of a standard normal random variable and ( U − 61)/7.9 to
be the 97.5 percentile. From our table for the standard normal, we look for the
point X with area from 0 to X equal to 0.975/2 = 0.4875. We see that this gives a
value of X = 2.24. So ( U − 61)/7.9 = 2.24. By symmetry, ( L − 61)/7.9 = − 2.24. We
can now solve for U and L. U = 2.24(7.9) + 6 1 = 17.696 + 6 1 = 78.696 and L =
61 − 2.24(7.9) = 6 1 − 17.696 = 43.304.
Chapter 6- How are equivalence tests different from standard hypothesis tests?
The standard Neyman – Pearson method for hypothesis testing make the hypothesis
of no signifi cant difference the null hypothesis with the power of the test controlled
for the alternative by the necessary sample size. However, in equivalence testing,
we want no signifi cant difference to be the alternative. Some people call this
“ proving the null hypothesis. ” In the Neyman – Pearson approach, we cannot “ prove
the null hypothesis ” without formally making it the alternative. That is because the
type I error is only the probability that we reject the null hypothesis when the null
hypothesis is “ true. ” It does not control the probability of accepting the null
hypothesis when the null hypothesis is “ true. ” If the sample size is small, it is hard
to reject the null hypothesis regardless of whether or not it is “ true. ” So to control
that probability, we make the null hypothesis the alternative and then controlling
the power of the test controls the probability of accepting the hypothesis when it
is “ true, ” since that is precisely the defi nition of power. By no signifi cant difference
we mean that the difference between the two groups is in absolute value less than
a defi ned margin of equivalence δ.
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