? 2 j 1 2 i 1 2 o ij e ij 2 e ij 40 33333 2 7 2 16667

Χ
2
=
∑
j = 1
2
∑
i = 1
2
(
O
ij
− E
ij
) 2
E
ij
=
(
40 − 33.333
)
2
33.333
+
(
10 − 16.667
)
2
16.667
+
(
1960 − 1966.667
)
2
1966.667
+
(
990 − 983.333
)
2
983.333
= 4.068
On a Χ
1
2
distribution, p = 0.0437, therefore, we have sufficient data to reject the null hypothesis at the
0.05 level of significance and conclude that MI incidence is higher among smokers than among ex-
smokers.
10.74)
^ p =
1273 + 1293
11036 + 11035
= 0.1163
2

n
1
^ p ^ q = ( 11036 ) ( 0.1163 ) ( 0.8837 ) > 5
and
n
1
^ p ^ q = ( 11035 ) ( 0.1163 ) ( 0.8837 ) > 5
, so use normal approximation
for test for two binomial proportions.
10.75) H
0
: p
1
= p
2
H
A
: p
1
≠ p
2
^ p
1
=
1273
11036
= 0.1153 ^ p
2
=
1293
11035
= 0.1172
z =
^ p
1
− ^ p
2
√
^ p ^ q (
1
n
1
+
1
n
2
)
=
0.1153 − 0.1172
√
(
0.1163
) (
0.8837
)
(
1
11036
+
1
11035
)
=− 0.440
On a normal distribution, p = 0.66, therefore, we have insufficient data to reject the null hypothesis at
the 0.05 level of significance and conclude that there is any difference in cancer incidence between
beta-carotene and placebo groups.
13.1) H
0
: p
1
= p
2
H
A
: p
1
≠ p
2
^ p
1
=
89
283
= 0.3145 ^ p
2
=
640
3833
= 0.1670 ^ p =
729
4116
= 0.1771
z =
^ p
1
− ^ p
2
√
^ p ^ q (
1
n
1
+
1
n
2
)
=
0.3145 − 0.1670
√
( 0.1771 ) ( 0.8229 )
(
1
283
+
1
3833
)
= 6.272
On a normal distribution, p < 0.05, therefore, we have sufficient data to reject the null hypothesis at the
0.05 level of significance and conclude that there is an effect of IUD use on fertility rate.
13.2)
Observed Infertile Fertile Total
IUD use 89 194 283
No IUD use 640 3193 3833
Total 729 3387 4116
Expected Infertile Fertile Total
IUD use 50.12 232.88 283
No IUD use 678.82 3154.18 3833
Total 729 3387 4116
∑
i = 1
2
( O
ij
− E
ij
)
2
E
ij
= ¿
( 89 − 50.12 )
2
50.12
+
( 194 − 232.88 )
2
232.88
+
( 640 − 678.82