Answer :
Complete question:
new concrete mix is being designed to provide adequate compressive strength for concrete blocks. The specification for a particular application calls for the blocks to have a mean compressive strength µ greater than 1350 kPa. A sample of 100 blocks is produced and tested. Their mean compressive strength is 1356 kPa and their standard deviation is 70 kP
a) Find the p value.
b) Do you believe it is plausible that the blocks do not meet the specification, or are you convinced that they do? Expain your reasoning.
Answer:
a) p-value= 0.1949
b) It is possible the blocks do not meet the specifications
Explanation:
Given:
n = 100
Sample mean, X' = 1356
Standard deviation, s.d = 70
a) To find p- value.
Null hypothesis:
H0: u ≤ 1350
Alternative hypothesis:
H1 : u > 1350
The test statistic wll be:
[tex] z= \frac{X'-u_o}{s.d/ \sqrt{n}}[/tex]
[tex] = \frac{1356-1350}{70/\sqrt{100}}[/tex]
=0.86
The p value will be:
= P(z>0.86)
= 1-P(z≤0.86)
Using the normal distribution table, we now have:
1 - 0.8051
= 0.1949
P value = 0.1949
b) Since our p-value is 0.1949, we do not reject the null hypothesis, because the p-value, 0.1949, is not small. This means that it is possible the blocks do not meet the specifications.
The probability that the compressive strength is greater than 1350 kPa is 80.51%
Z score is used to determine by how many standard deviations the raw score is above or below the mean. The z score is given by:
[tex]z=\frac{x-\mu}{\sigma/\sqrt{n} } \\\\Where\ x=raw\ score,\mu=mean, \sigma=standard\ deviation, n=sample\ size[/tex]
Given that n = 100, μ = 1356, σ = 70. For x > 1350 kPa:
[tex]z=\frac{1350-1356}{70/\sqrt{100} } =-0.86[/tex]
From the normal distribution table, P(x > 1350) = P(z > -0.86) = 1 - P(z < -0.86) = 1 - 0.1949 = 80.51%
Hence the probability that the compressive strength is greater than 1350 kPa is 80.51%
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