Test2 Revision2_sol.pdf - Test2 Revision2 1 SOLUTIONS[7...

Test2 Revision2 SOLUTIONS
1. [7 marks]
(a) [1]
[1] [total = 2]
(b) If P ( Z > z ) = .03, then z
≈
1.88. [1]
Hence X = μ + zσ = 179.1 + 1.88 × 39.7 = 253.74 mg/dL [1] [total = 2]
(c)
Since this probability is very small (< 1/600), and 152.7 is below the population mean, such a
sample mean would be unusually small. [1, total = 3]
2. [8 marks]
(a) 997 4 993
~
= + = n , 54965 . 997 ) 2 546 (
~
= + = p
90% CI
or .524 to .576 [total = 4]
(b) Couples using non-invasive sex selection methods to increase the chance of having a girl. [1]
(c) [total = 3]
3. [3 marks]
X = number of female babies in sample. X ~ Bin( 12, .9).
P ( X ≥ 11) = 1 – P ( X ≤ 10) = 1 - .3410 = .6590 [total = 3]
4. [9 marks]
(a) H
0
: p = .019 H
1
: p > .019
where p is the proportion of patients who report flu-like symptoms following treatment
with 10 mg of atorvastatin. [1, total = 2]
(b) 02781 . 863 24
ˆ
= = p
89 . 1
00465 .
00881 .
863
981 . 019 .
019 . 02781 .
) 1 (
ˆ
0 0
0
= =
×
-
=
-
-
=
n
p p
p p