Question

A concrete mix must withstand an average compressive stress of more than 1,300 KN / m2. The mixture will not be used unless we have sufficient evidence that the specification is being met. Assume that effort is a normally distributed random variable with a standard deviation of 60. A sample of 20 specimens is taken:

1350.343	1284.595	1248.794	1233.162
1314.349	1392.296	1420.502	1411.886
1330.357	1338.123	1353.439	1210.946
1319.671	1375.284	1311.355	1299.507
1231.943	1295.214	1355.426	1530.636

1. 1. Clearly state the null and alternative hypothesis.
2. 2. With the evidence provided, would you use the mixture? Justify. Calculate the P-Value of the test.
3. 3. Calculate by hand, the probability of making a type II error if the mean compressive stress was 1325 KN / m2. Explain in your words what it means.

Answer 1

a.
Given that,
population mean(u)=1300
standard deviation, σ =60
sample mean, x =1330.3914
number (n)=20
null, Ho: μ=1300
alternate, H1: μ>1300
level of significance, α = 0.05
from standard normal table,right tailed z α/2 =1.645
since our test is right-tailed
reject Ho, if zo > 1.645
we use test statistic (z) = x-u/(s.d/sqrt(n))
zo = 1330.3914-1300/(60/sqrt(20)
zo = 2.265
| zo | = 2.265
critical value
the value of |z α| at los 5% is 1.645
we got |zo| =2.265 & | z α | = 1.645
make decision
hence value of | zo | > | z α| and here we reject Ho
p-value : right tail - ha : ( p > 2.265 ) = 0.012
hence value of p0.05 > 0.012, here we reject Ho
ANSWERS
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a.
null, Ho: μ=1300
alternate, H1: μ>1300
b.
test statistic: 2.265
critical value: 1.645
decision: reject Ho
p-value: 0.012
we have enough evidence to support the claim that A concrete mix must withstand an average compressive stress of more than 1,300 KN / m2.
c.
Given that,
Standard deviation, σ =60
Sample Mean, X =1330.3914
Null, H0: μ=1300
Alternate, H1: μ>1300
Level of significance, α = 0.05
From Standard normal table, Z α/2 =1.6449
Since our test is right-tailed
Reject Ho, if Zo < -1.6449 OR if Zo > 1.6449
Reject Ho if (x-1300)/60/√(n) < -1.6449 OR if (x-1300)/60/√(n) > 1.6449
Reject Ho if x < 1300-98.694/√(n) OR if x > 1300-98.694/√(n)
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Suppose the size of the sample is n = 20 then the critical region
becomes,
Reject Ho if x < 1300-98.694/√(20) OR if x > 1300+98.694/√(20)
Reject Ho if x < 1277.9314 OR if x > 1322.0686
Implies, don't reject Ho if 1277.9314≤ x ≤ 1322.0686
Suppose the true mean is 1325
Probability of Type II error,
P(Type II error) = P(Don't Reject Ho | H1 is true )
= P(1277.9314 ≤ x ≤ 1322.0686 | μ1 = 1325)
= P(1277.9314-1325/60/√(20) ≤ x - μ / σ/√n ≤ 1322.0686-1325/60/√(20)
= P(-3.5083 ≤ Z ≤-0.2185 )
= P( Z ≤-0.2185) - P( Z ≤-3.5083)
= 0.4135 - 0.0002 [ Using Z Table ]
= 0.4133
For n =20 the probability of Type II error is 0.4133