he drying time of a certain type of paint under specified test conditions is kno
ID: 3323117 • Letter: H
Question
he drying time of a certain type of paint under specified test conditions is known to be normally distributed with mean value 75 min and standard deviation 9 min. Chemists have proposed a new additive designed to decrease average drying time. It is believed that drying times with this additive will remain normally distributed with =9. Because of the expense associated with the additive, evidence should strongly suggest an improvement in average drying time before such a conclusion is adopted. Let denote the true average drying time when the additive is used. The hypotheses H0:=75 versus HA:<75 are to be tested using a random sample of n=27 observations.
(a) If x¯=72.7, what is the conclusion using =0.04?
A. Reject H0H0.
B. Do Not Reject H0H0.
(b) What is for the test procedure that rejects H0 when z<2.88?
(c) For the test procedure of part (b), what is (70)?
(d) If the test procedure of part (b) is used, what n is necessary to ensure that (70)=0.05?
Explanation / Answer
a.
Given that,
population mean(u)=75
standard deviation, =9
sample mean, x =72.7
number (n)=27
null, Ho: =75
alternate, H1: <75
level of significance, = 0.04
from standard normal table,left tailed z /2 =1.751
since our test is left-tailed
reject Ho, if zo < -1.751
we use test statistic (z) = x-u/(s.d/sqrt(n))
zo = 72.7-75/(9/sqrt(27)
zo = -1.328
| zo | = 1.328
critical value
the value of |z | at los 4% is 1.751
we got |zo| =1.328 & | z | = 1.751
make decision
hence value of |zo | < | z | and here we do not reject Ho
p-value : left tail - ha : ( p < -1.328 ) = 0.092
hence value of p0.04 < 0.092, here we do not reject Ho
ANSWERS
---------------
null, Ho: =75
alternate, H1: <75
test statistic: -1.328
critical value: -1.751
decision: do not reject Ho
p-value: 0.092
option :B
b.
for the test procedure that rejects H0 when z<2.88
= 0.002
c.
Given that,
Standard deviation, =9
Sample Mean, X =72.7
Null, H0: =75
Alternate, H1: >75
Level of significance, = 0.04
From Standard normal table, Z /2 =1.7507
Since our test is right-tailed
Reject Ho, if Zo < -1.7507 OR if Zo > 1.7507
Reject Ho if (x-75)/9/(n) < -1.7507 OR if (x-75)/9/(n) > 1.7507
Reject Ho if x < 75-15.756/(n) OR if x > 75-15.756/(n)
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Suppose the size of the sample is n = 27 then the critical region
becomes,
Reject Ho if x < 75-15.756/(27) OR if x > 75+15.756/(27)
Reject Ho if x < 71.968 OR if x > 78.032
Implies, don't reject Ho if 71.968 x 78.032
Suppose the true mean is 70
Probability of Type II error,
P(Type II error) = P(Don't Reject Ho | H1 is true )
= P(71.968 x 78.032 | 1 = 70)
= P(71.968-70/9/(27) x - / /n 78.032-70/9/(27)
= P(1.136 Z 4.637 )
= P( Z 4.637) - P( Z 1.136)
= 1 - 0.872 [ Using Z Table ]
= 0.128
For n =27 the probability of Type II error is 0.128
d.
Suppose the size of the sample is n = ? then the critical region
becomes,
Reject Ho if x < 75-14.804/(n) OR if x > 75+14.804/(n)
Reject Ho if x < 72.151 OR if x > 77.849
Implies, don't reject Ho if 72.151 x 77.849
Suppose the true mean is 70
Probability of Type II error,
P(Type II error) = P(Don't Reject Ho | H1 is true )
= P(72.151 x 77.849 | 1 = 70)
= P(72.151-70/9/(n) x - / /n 77.849-70/9/(n)
type2 error = 0.496 from part:C
0.496 = (72.151-70/9/(n)
covert into Z value
-1.135= (72.151-70/9/(n)
-0.135 = 0.239*(n)
n = 22.55 = 23
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