Prefinal for AMS310
SOLUTIONS
I1
086
.
0
15
*
2
58
.
2
2
2
/
=
=
=
n
Z
E
α
I2
48
.
15
10
60
*
58
.
2
2
/
=
=
=
n
Z
E
σ
α
I3 (a)
I4 (d)
I5
2
.
4
2
44
.
1
*
841
.
5
2
/
=
=
=
n
s
t
E
α
I6 (b)
I7 (a)
I8 (d)
I9
20
10
*
2
2
=
=
=
X
Y
σ
σ
I10 (a)
I11
25
.
0
25
.
0
50
.
0
)
2
(
)
3
(
)
3
(
=

=

=
=
F
F
X
P
I12
323
.
0
667
.
0
1
)
2
(
1
)
3
(
=

=
≤

=
≥
X
P
X
P
Longer problem:
II1(a)
p
ˆ
=
170/2835
=0.06. So point estimate of 100p =100x 0.06=6
%
II1(b) First estimate SE of
p
ˆ
=
n
n
x
n
x
)
1
(

=
2835
)
2835
170
1
(
2835
170

=0.0045. Thus
)
ˆ
100
(
p
SE
=0.45%
II1(c).
The estimate of this quantity is
p
ˆ
=70/940=0.074 Thus and our estimate of the percentage of
defiectives given an item is produced on the night shift equals 100x0.074=
7.4%. This is more
than the
overall percentage of defectives produced by workers (6%) in this sample.
II1(d)
H
0
: p
1
= p
2=
p
3.
Here p
1,
p
2,
p
3
denote the proportion of
defectives produced on Day shift, Evening shift
and Night shift repectively.
II1(e) ANS
P= P(
χ
2
2
> 7.378)=0.04. Compare Pvalue to
α
, if P
≤α
then reject Ho at
α
level ; if P>
α
then
accept Ho.
Since P= 0.04 < 0.05=
α
, we have P<
α
so
we can reject Ho at the 0.05 level and conclude
that the proportion of defectives is NOT the same for each shift.
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 Fall '03
 Mendell
 Statistics, Statistical hypothesis testing

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