A mixture of pulverized fuel ash and Portland cement to be used for grouting sho

Question

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 alternative hypotheses?

a-)If

X = 1340,

find the P-value. (Round your answer to four decimal places.)

b-)State standard deviation of the test statistic. (Round your standard deviation to three decimal places.)

c-)For a test with = 0.01, what is the probability that the mixture will be judged unsatisfactory when in fact = 1350 (a type II error)? (Round your answer to four decimal places.)

Explanation / Answer

a.
Given that,
population mean(u)=1300
standard deviation, =67
sample mean, x =1340
number (n)=2
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(2)
zo = 0.84431
| zo | = 0.84431
critical value
the value of |z | at los 1% is 2.326
we got |zo| =0.84431 & | z | = 2.326
make decision
hence value of |zo | < | z | and here we do not reject Ho
p-value : right tail - ha : ( p > 0.84431 ) = 0.19925
hence value of p0.01 < 0.19925, here we do not reject Ho
ANSWERS
---------------
null, Ho: =1300
alternate, H1: >1300
test statistic: 0.84431
critical value: 2.326
decision: do not reject Ho
p-value: 0.19925

b.
Z test statistic = 0.8443
Z = X-u/s.d/sqrt(n)
0.8443 = 1340-1350/s.d/sqrt(2)
s.d(standard deviation) = 16.750

c.Given that,
Standard deviation, =16.75
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)/16.75/(n) < -2.5758 OR if (x-1350)/16.75/(n) > 2.5758
Reject Ho if x < 1350-43.14465/(n) OR if x > 1350-43.14465/(n)
-----------------------------------------------------------------------------------------------------
Suppose the size of the sample is n = 2 then the critical region
becomes,
Reject Ho if x < 1350-43.14465/(2) OR if x > 1350+43.14465/(2)
Reject Ho if x < 1319.49213 OR if x > 1380.50787
Implies, don't reject Ho if 1319.49213 x 1380.50787
Suppose the true mean is 1340
Probability of Type II error,
P(Type II error) = P(Don't Reject Ho | H1 is true )
= P(1319.49213 x 1380.50787 | 1 = 1340)
= P(1319.49213-1340/16.75/(2) x - / /n 1380.50787-1340/16.75/(2)
= P(-1.73149 Z 3.42011 )
= P( Z 3.42011) - P( Z -1.73149)
= 0.9997 - 0.0417 [ Using Z Table ]
= 0.958
For n =2 the probability of Type II error is 0.958

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