Midterm Exam Solution
1.(i) 24:,24:
0
≠=
µ
µ
a
HH
186.1
90/32.
2496.23
/−=
−
=
−
=ns
X
t
µ
, so we can not reject the null hypothesis at significance level 0.05.
987.1||
89,025.
=< tt
(ii) The 99% CI for
µ
is
)0487.24,8713.23(089.96.2390/32.*63.296.23/
89,005.
=±=±=± nstX
(iii)
))
1
1(,0(~
2
σ
n
NXX +−
, an estimate of the proportion P(X < 23.76) is
2671.)6216.(
)
90/1132.
96.2376.23
()
/11
96.2376.23
()96.2376.23()76.23(
ˆ
=−<=
+
−
<=
+
−
<=−<−=<
TP
TP
ns
TPXXPXP
2. (a)
FMaFM
HH
µ
µ
µ
µ
≠= :,:
0
Since we assume equal variance, we need to calculate the pooled standard error.
64.83
279145
9294.82*7802.84*144
2
)1()1(
22
21
2
2
2
1
=
−+
+
=
−+
−+−
=nn
SnSn
s
YX
p
97.1
.9968.3
64.83
0253.5657724.611
222,025.
79
1
145
111
21
=>
=
+
−
=
+
−
=
tt
s
YX
t
nn
p
So is rejected at significance level 0.05.
0
H
(b) 10:,10:
0
>
−
=−
FMaFM
HH
µ
µ
µ
µ
34.2
.1418.3
64.83
100253.5657724.61110
222,01.
79
1
145
111
21
=>
=
+
−−
=
+
−−
=
tt
s
YX
t
nn
So is rejected at significance level 0.01.
0
H
