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Solution to Practice Midterm, Oct. 4, 2003,
October 3, 2003
a) ˆp = 20029 = .145. The 95% confidence interval for p is:
pˆ + zα/ 2 2 n ±^ zα/^2
pˆqˆ n +^
z α/^22 4 n^2 1 + (zα/ 2 )^2 /n
b) Approximately, n =
4 ˆpˆqz α/^22 d^2 , to halve the interval, they need 4(200) people in total. So they need to sample 600 more people. c) ˆp 1 = .145, ˆp 2 = .32, m = 200, n = 300, a α = 0.01, so 99% confidence interval for the difference is:
pˆ 1 − pˆ 2 ± zα/ 2
pˆ 1 qˆ 1 m
pˆ 2 qˆ 2 n
b α = .05, ˆp = .25, to test H 0 : p 1 − p 2 ≥ 0 vs. Ha : p 1 − p 2 < 0.
z =
pˆ 1 − pˆ 2 √ pˆqˆ(1/m + 1/n)
so there is significant evidence that the TV advertising campaign is effective. We carried out one-sided test. c Since z = − 4 .43, so p-value= P (Z < − 4 .43) < .0001.
2 i) ¯xD = 362. 8 − 392 .9 = − 30 .1, nD = 20, α = .01; Test H 0 : μD = 0 vs. Ha : μD 6 = 0.
t = |
x¯d − μD sD /
nD | = 2. 890 > t 19 ,. 005 = 2. 861
Conclusion: There is significant difference in the nutritive value of these two varieties of corn. ii) t = 2.890, p-value= P (|T 19 | > 2 .89) ≈ .01.
a) α = .01, ¯x = 611.8, ¯y = 565.0, s 1 = 84.0, s 2 = 82.9, m = 145 and n = 79; By assuming σ 1 = σ 2 = σ, pooled sample variance:
s^2 p = (
m − 1 m + n − 2
)s^21 + (
n − 1 m + n − 2
)s^22 = (83.6)^2
Test: H 0 : μ 1 − μ 2 = 0 vs. Ha : μ 1 − μ 2 6 = 0.
z = |
x¯ − ¯y sp
1 m +^
1 n
| = 4. 00 > z. 005 = 2. 58
Conclusion: There is significant difference between mean SAT score of male and female students. b) Test: H 0 : μ 1 − μ 2 ≤ 10 vs. Ha : μ 1 − μ 2 > 10.
z =
¯x − ¯y − ∆ sp
1 m +^
1 n
= 3. 15 > z. 01 = 2. 33
Conclusion: There is significant evidence that mean SAT score of male students is at least 10 points higher than female students. c) σ 1 = 84, σ 2 = 83 Test: H 0 : μ 2 ≤. 89 μ 1 vs. H 0 : μ 2 >. 89 μ 1. Since X¯ ∼ N (μ 1 , √^84145 = 6.98), Y¯ ∼ N (μ 2 , √^8379 = 9.34), so Y¯ −. 89 X¯ ∼
N (μ 2 −. 89 μ 1 ,
9. 342 + (6. 98 ∗ .89)^2 = 11.2).
z =
x¯ − .89¯y
- 2
= 1. 894 > z. 05 = 1. 65
Conclusion: There is significant evidence that the mean SAT score of female students is at least 89% of the mean score of male students.
4 a) H 0 : p = .6 vs. Ha : p > .6; α = .05 with rejection region X > K.
P (X > K) = P (
X − np √ npq
K − np √ npq
) = P (Z >
K − np √ npq
K = 1. 65
b) Type I error:
α = P (X > 63 |p = .6) = P (Z >
) = P (Z > .612) =. 271
Type II error:
β = P (X < 63 |p = .7) = P (Z <
) = P (Z < − 1 .53) =. 063
