Solutions to Homework #11
(W4150 ‘Intro to Probability and Statistics’, S04)
Sec. 10.12. (1)
H
0
: p = .4
H
1
: p > .4
α
= .05
If we denote the number of those who choose lasagne by X then under H
0
X ~ binomial(20;.4) and so
Pvalue is given by P[X >
9  p = .4] = 1 – P[X <
8  p = .4] = .4044 > .05
Therefore we cannot reject nullhypothesis.
Sec.10.12. (2)
H
0
: p = .4
H
1
: p > .4
α
= .05
Let’s denote the number of those who favored capital punishment by X. Then under H
0
X ~
binomial(15;.4) and so Pvalue is given by P[X >
8  p = .4] = 1 – P[X <
7  p = .4] =
= .2131 > .05
Therefore we cannot reject nullhypothesis.
Sec. 10.12. (6)
H
0
: p = .25
H
1
: p > .25
α
= .05
Since n = 90 > 30 we’d like to use normal approximation for binomial distribution here.
μ
= np = 22.5;
σ
=
√
npq = 4.107 and critical region is defined as X >
28 where X is the number of
students using bicycles. Therefore critical value of Z is z = (28.5 – 22.5)/(4.107) = 1.43 and P[X >
28] = P[Z > 1.43] = 1  .9236 = .0724 > .05
formally we cannot reject H
0
; however, relatively
low Pvalue clearly indicates that there is some evidence in favor of alternative hypothesis.
Sec. 10.12 (9)
H
0
: p
1
= p
2
H
1
: p
1
≠
p
2
, where p
1
is proportion of urban residents and p
2
– proportion of suburban residents
who favor the nuclear plant.
Following Example 10.12 we compute
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 Spring '04
 GALLEGO
 Null hypothesis, Probability theory, Statistical hypothesis testing, critical region

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