ENGG2430A-Homework 8
Due on April 9, 2014 via the assignment box.
1. Ex 10.55 from textbook A marketing expert for a pasta-making company believes that 40% of pasta
lovers prefer lasagna. If 9 out of 20 pasta lovers choose lasagna over other pastas, what can be concluded
about the expert’s claim? Use a 0.05 level of significance.
Solution: Let p be the proportion of pasta lovers who prefer lasagna. The null and alternative
hypotheses are:
H0 :p = p0 := 0.4
H1 :p 6= p0
The sample size is n = 20 and the test statistics is Y denoting the number of people in a random
sample who choose lasagna. The observed value is y = 9 and np0 = 20(0.4) = 8. For the two-sided
test, since y > np0 , the P-value is
P = 2 Pr( Y ≥ y| p = p0 )
n X
n i
=2
p (1 − p0 )n−i
i 0
i=1
20 X
20
=2
0.4i (1 − 0.4)20−i
i
i=1
≈ 2(0.4044) ≈ 0.81
which is larger than the significance level α = 0.05. Therefore, we accept H0 . The P-value can be
obtained by the octave command 2*(1-binocdf(8,20,0.4)) or from Table A.1 in the textbook.
2. Ex 10.57 from textbook A new radar device is being considered for a certain missile defense system.
The system is checked by experimenting with aircraft in which a kill or a no kill is simulated. If, in 300
trials, 250 kills occur, accept or reject, at the 0.04 level of significance, the claim that the probability of
a kill with the new system does not exceed the 0.8 probability of the existing device.
Solution: Let p be the proportion of kills. The null and alternative hypotheses are:
H0 :p = p0 := 0.8
H1 :p > p0
The sample size is n = 300. Let Y be the number of kills simulated. Given p = p0 , we have Y ∼
Binom(n, p0 ), with mean np0 = 300 × 0.8 = 240 and variance np0 (1 − p0 ) = 300(0.8)(1 − 0.8) = 48.
Since n > 30, we have by the central limit theorem that the standardized random variable
Y − np0
Z := p
np0 (1 − p0 )
is approximately standard normal distributed. We can therefore apply the Z-statistics for the
hypothesis test. Under α = 0.04 significance level for a one-sided test, the critical region is “z >
zα ≈ 1.75”. The observed number of kills simulated is y = 250, and so the observed z-value is
y − np0
250 − 240
√
=
≈ 1.44
z := p
48
np0 (1 − p0 )
which is not in the critical region. Thus, we accept H0 .
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