Benzene is a toxic chemical used in the manufacturing of medicinal chemicals dye

Question

Benzene is a toxic chemical used in the manufacturing of medicinal chemicals dyes, artificial leather, and linoleum. A manufacturer claims that its exit water meets the federal regulation with a mean of less than 7980 ppm of benzene. To assess the benzene content of the exit water, 10 independent water samples were collected and found to have an average of 7906 ppm of benzene. Assume a known standard deviation of 80 ppm and use a significance level of 0.01 (Note: Preliminary data analyses indicate that applying the z-test is reasonable.) (a) Test the manufacturer's claim. Use both critical value and p-value (b) If the true mean is 7920, find the probability of type II error.

Explanation / Answer

a)
Given that,
population mean(u)=7980
standard deviation, =80
sample mean, x =7906
number (n)=10
null, Ho: =7980
alternate, H1: <7980
level of significance, = 0.01
from standard normal table,left tailed z /2 =2.326
since our test is left-tailed
reject Ho, if zo < -2.326
we use test statistic (z) = x-u/(s.d/sqrt(n))
zo = 7906-7980/(80/sqrt(10)
zo = -2.92511
| zo | = 2.92511
critical value
the value of |z | at los 1% is 2.326
we got |zo| =2.92511 & | z | = 2.326
make decision
hence value of | zo | > | z | and here we reject Ho
p-value : left tail - ha : ( p < -2.92511 ) = 0.00172
hence value of p0.01 > 0.00172, here we reject Ho
ANSWERS
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null, Ho: =7980
alternate, H1: <7980
test statistic: -2.92511
critical value: -2.326
decision: reject Ho
p-value: 0.00172
we have evidence that its exit water meets the federal regulation with mean ofl less than 7980 ppm of benze

b)
Given that,
Standard deviation, =80
Level of significance, = 0.01
From Standard normal table, Z /2 =2.33
Since our test is left-tailed
Reject Ho, if Zo < -2.33 OR if Zo > 2.33
Reject Ho if (x-7980)/80/(n) < -2.33 OR if (x-7980)/80/(n) > 2.33
Reject Ho if x < 7980-186.4/(n) OR if x > 7980-186.4/(n)
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Suppose the size of the sample is n = 10 then the critical region
becomes,
Reject Ho if x < 7980-186.4/(10) OR if x > 7980+186.4/(10)
Reject Ho if x < 7921.0551 OR if x > 8038.9449
Implies, don't reject Ho if 7921.0551 x 8038.9449
Suppose the true mean is 7920
Probability of Type II error,
P(Type II error) = P(Don't Reject Ho | H1 is true )
= P(7921.0551 x 8038.9449 | 1 = 7920)
= P(7921.0551-7920/80/(10) x - / /n 8038.9449-7920/80/(10)
= P(0.0417 Z 4.7017 )
= P( Z 4.7017) - P( Z 0.0417)
= 1 - 0.5166   [ Using Z Table ]
= 0.4834
For n =10 the probability of Type II error is 0.4834

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