Mean = $18,459.25
St. Dev = $6,543.70
b) 95% CI: (15396.74, 21521.76)
c) No, the 95% CI for µ is an interval that has a 95% chance of containing the mean winnings of
the entire population, NOT the winnings of 95% of similar winners.
d)
I.
H
o
:
μ
=
μ
o
H
a
:
μ
≠
μ
o
μ
o
= 34064
This is a twosided alternative because the question asks whether or not the population
mean
differs
from Jennings’ average (alternative hypothesis has a “not equal” sign),
which means that
μ
o
can either be larger or smaller than
μ
.
II.
t =
= 10.6647
Pvalue = 2 x Pr (t >tstat) < 0.01.
Pvalue = 2 x < 0.01
Pvalue < 0.02.
At the 5% significance level, 0.02 < alpha, therefore we have significant evidence to prove that
the mean winnings of the population
are
different from Ken Jennings’ mean winnings.
III.
The 95% CI from part b) does support the conclusion of our hypothesis test. The CI we
found was (15396.74, 21521.76). Ken Jennings’ mean winnings are not inside this
interval, which means that we reject the null hypothesis.
e) In this situation, Type I error would mean that Ken Jennings’ average of $34,064 actually
represented the population mean winnings, but this was rejected in the hypothesis test. Type II
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View Full Documenterror would mean that the previous statement was actually false, but we accepted the hypothesis.
The only error that could have happened in part d) was the Type I error.
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 Winter '10
 E.Fowler
 Statistics, Null hypothesis, Statistical hypothesis testing, Statistical significance, Statistical power, Ken Jennings

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