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4. 4'8 polnts | Previous Answers Devorestat9 8.E.012 My Notes Ask Your Teacher A mixture of pulverized fuel ash and Portland cement to be used for grouting should have a compressive strength of more than 1300 KN/m2. The mixture will not be used unless experimental evidence indicates conclusively that the strength specification has been met. Suppose compressive strength for specimens of this mixture is normally distributed with = 67, Let denote the true average compressive strength (a) What are the appropriate null and altenative hypotheses? Ho: = 1300 Ha: 1300 Ho: > 1300 Ha: = 1300 Ho:

Explanation / Answer

4.

Given that,
population mean(u)=1300
standard deviation, =67
sample mean, x =1340
number (n)=15
null, Ho: =1300
alternate, H1: >1300
level of significance, = 0.01
from standard normal table,right tailed z /2 =2.326
since our test is right-tailed
reject Ho, if zo > 2.326
we use test statistic (z) = x-u/(s.d/sqrt(n))
zo = 1340-1300/(67/sqrt(15)
zo = 2.3122
| zo | = 2.3122
critical value
the value of |z | at los 1% is 2.326
we got |zo| =2.3122 & | z | = 2.326
make decision
hence value of |zo | < | z | and here we do not reject Ho
p-value : right tail - ha : ( p > 2.3122 ) = 0.0104
hence value of p0.01 < 0.0104, here we do not reject Ho
ANSWERS
---------------
a.
null, Ho: =1300
alternate, H1: >1300
b.
test statistic is normal distributed
test statistic: 2.3122
critical value: 2.326
decision: do not reject Ho
p-value: 0.0104

c.
if u take mean = 1350 then test statistic is normally distributed

standard deviation =67

Given that,
Standard deviation, =67
Sample Mean, X =1340
Null, H0: =1350
Alternate, H1: !=1350
Level of significance, = 0.01
From Standard normal table, Z /2 =2.5758
Since our test is two-tailed
Reject Ho, if Zo < -2.5758 OR if Zo > 2.5758
Reject Ho if (x-1350)/67/(n) < -2.5758 OR if (x-1350)/67/(n) > 2.5758
Reject Ho if x < 1350-172.5786/(n) OR if x > 1350-172.5786/(n)
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Suppose the size of the sample is n = 15 then the critical region
becomes,
Reject Ho if x < 1350-172.5786/(15) OR if x > 1350+172.5786/(15)
Reject Ho if x < 1305.4404 OR if x > 1394.5596
Implies, don't reject Ho if 1305.4404 x 1394.5596
Suppose the true mean is 1350
Probability of Type II error,
P(Type II error) = P(Don't Reject Ho | H1 is true )
= P(1305.4404 x 1394.5596 | 1 = 1350)
= P(1305.4404-1350/67/(15) x - / /n 1394.5596-1350/67/(15)
= P(-2.5758 Z 2.5758 )
= P( Z 2.5758) - P( Z -2.5758)
= 0.995 - 0.005 [ Using Z Table ]
= 0.99
For n =15 the probability of Type II error is 0.99

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