SECTION 6.15
161
(c) Greater than 0.9. The power is greater than 0.9 against a difference of 3, so it will be greater than 0.9 against
any difference greater than 3.
Section 6.14
1. (a) There are six tests, so the Bonferroniadjusted
P
values are found by multiplying the original
P
values by 6.
For the setting whose original
P
value is 0.002, the Bonferroniadjusted
P
value is therefore 0.012. Since this
value is small, we can conclude that this setting reduces the proportion of defective parts.
(b) The Bonferroniadjusted
P
value is 6(0.03) = 0.18. Since this value is not so small, we cannot conclude that
this setting reduces the proportion of defective parts.
3.
The original
P
value must be 0
05
20
0
0025.
5. (a) No. Let
X
represent the number of times in 200 days that
H
0
is rejected. If the mean burnout amperage is
equal to 15 A every day, the probability of rejecting
H
0
is 0.05 each day, so
X
Bin
200
0
05
.
The probability of rejecting
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 Fall '11
 MattGilg
 Statistics, PValues, mean burnout amperage

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