The drying time of a certain type of paint under specified test conditions is kn

Question

The 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 appropriate hypotheses are H0: = 75 versus Ha: 75. Consider the alternative value = 74, which in the context of the problem would presumably not be a practically significant departure from Ho. (a) For a level 0.01 test, compute at this alternative for sample sizes n-100, 961, and 2500, (Round your answers to four decimal places.) 100 961 2500 0.0006 b) If the observed value of is x = 74, what can you say about the resulting P value when n (Round your z to two decimal places. Round your P-value to four decimal places.) 2500? s the data statistica i significant at any of the standard values of Z z =-5.56 P-value = 0.0000

Explanation / Answer

a.
i.
Given that,
Standard deviation, =9
Sample Mean, X =74
Null, H0: =75
Alternate, H1: <75
Level of significance, = 0.01
From Standard normal table, Z /2 =2.3263
Since our test is left-tailed
Reject Ho, if Zo < -2.3263 OR if Zo > 2.3263
Reject Ho if (x-75)/9/(n) < -2.3263 OR if (x-75)/9/(n) > 2.3263
Reject Ho if x < 75-20.9367/(n) OR if x > 75-20.9367/(n)
-----------------------------------------------------------------------------------------------------
Suppose the size of the sample is n = 100 then the critical region
becomes,
Reject Ho if x < 75-20.9367/(100) OR if x > 75+20.9367/(100)
Reject Ho if x < 72.90633 OR if x > 77.09367
Implies, don't reject Ho if 72.90633 x 77.09367
Suppose the true mean is 74
Probability of Type II error,
P(Type II error) = P(Don't Reject Ho | H1 is true )
= P(72.90633 x 77.09367 | 1 = 74)
= P(72.90633-74/9/(100) x - / /n 77.09367-74/9/(100)
= P(-1.215189 Z 3.437411 )
= P( Z 3.437411) - P( Z -1.215189)
= 0.9997 - 0.1121 [ Using Z Table ]
= 0.8876
For n =100 the probability of Type II error is 0.8876

ii.
Given that,
Standard deviation, =9
Sample Mean, X =74
Null, H0: =75
Alternate, H1: <75
Level of significance, = 0.01
From Standard normal table, Z /2 =2.3263
Since our test is left-tailed
Reject Ho, if Zo < -2.3263 OR if Zo > 2.3263
Reject Ho if (x-75)/9/(n) < -2.3263 OR if (x-75)/9/(n) > 2.3263
Reject Ho if x < 75-20.9367/(n) OR if x > 75-20.9367/(n)
-----------------------------------------------------------------------------------------------------
Suppose the size of the sample is n = 961 then the critical region
becomes,
Reject Ho if x < 75-20.9367/(961) OR if x > 75+20.9367/(961)
Reject Ho if x < 74.324623 OR if x > 75.675377
Implies, don't reject Ho if 74.324623 x 75.675377
Suppose the true mean is 74
Probability of Type II error,
P(Type II error) = P(Don't Reject Ho | H1 is true )
= P(74.324623 x 75.675377 | 1 = 74)
= P(74.324623-74/9/(961) x - / /n 75.675377-74/9/(961)
= P(1.118146 Z 5.770743 )
= P( Z 5.770743) - P( Z 1.118146)
= 1 - 0.8682 [ Using Z Table ]
= 0.1318
For n =961 the probability of Type II error is 0.1318

iii.
Given that,
Standard deviation, =9
Sample Mean, X =74
Null, H0: =75
Alternate, H1: <75
Level of significance, = 0.01
From Standard normal table, Z /2 =2.3263
Since our test is left-tailed
Reject Ho, if Zo < -2.3263 OR if Zo > 2.3263
Reject Ho if (x-75)/9/(n) < -2.3263 OR if (x-75)/9/(n) > 2.3263
Reject Ho if x < 75-20.9367/(n) OR if x > 75-20.9367/(n)
-----------------------------------------------------------------------------------------------------
Suppose the size of the sample is n = 2500 then the critical region
becomes,
Reject Ho if x < 75-20.9367/(2500) OR if x > 75+20.9367/(2500)
Reject Ho if x < 74.581266 OR if x > 75.418734
Implies, don't reject Ho if 74.581266 x 75.418734
Suppose the true mean is 74
Probability of Type II error,
P(Type II error) = P(Don't Reject Ho | H1 is true )
= P(74.581266 x 75.418734 | 1 = 74)
= P(74.581266-74/9/(2500) x - / /n 75.418734-74/9/(2500)
= P(3.229256 Z 7.881856 )
= P( Z 7.881856) - P( Z 3.229256)
= 1 - 0.9994 [ Using Z Table ]
= 0.0006
For n =2500 the probability of Type II error is 0.0006

b.
Given that,
population mean(u)=75
standard deviation, =9
sample mean, x =74
number (n)=2500
null, Ho: =75
alternate, H1: <75
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 = 74-75/(9/sqrt(2500)
zo = -5.5556
| zo | = 5.5556
critical value
the value of |z | at los 1% is 2.326
we got |zo| =5.5556 & | z | = 2.326
make decision
hence value of | zo | > | z | and here we reject Ho
p-value : left tail - ha : ( p < -5.5556 ) = 0
hence value of p0.01 > 0, here we reject Ho
ANSWERS
---------------
null, Ho: =75
alternate, H1: <75
test statistic: -5.5556 = -5.56
critical value: -2.326
decision: reject Ho
p-value: 0

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