Homework 5 Solutions

# Homework 5 Solutions - CE 398 Homework 5 Solutions QUESTION...

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CE 398 Homework 5 Solutions QUESTION 1 (TEST OF HYPOTHESIS) For each problem, clearly show all steps and detailed calculations and provide a sketch of the graph associated with your final answer. 1. As the site engineer at a large construction site, you have been asked to oversee the concrete production process. You are particularly worried about the slump of the concrete. If slump is too small, it suggests the concrete is too stiff. If slump is too much, then the concrete is too watery. The contract specifications state that the slump for that kind of concrete should be 1 inch, with a 90% level of confidence. So during the concrete production process, you instruct the laboratory technician to take 20 random samples of fresh concrete and measure the slump using slump test equipment. The technician obtained the following test results (in inches): 0.92 1.21 1.03 1.10 1.01 0.99 0.89 0.97 1.01 0.99 1.05 1.11 0.95 1.00 1.00 1.04 0.88 1.02 0.97 1.01 Would you accept that day’s production of concrete at the given level of confidence? Assume that from past slump test results, concrete slumps are known to be normally distributed. SOLUTION Here, the claim could be attributed to the contractor – the concrete produced that day was good because it met the slump requirements statistically. [Note that the claim could also be your claim that the concrete met or did not meet the slump requirements]. We are simply asked to test a hypothesis. The hypothesis in words, is as follows: H 0 : The concrete produced that day was good (that is, the average slump was 1 inch) H 1 : The concrete produced that day was not good (that is, the average slump was 1 inch) As you are interested in the average slump, the statistical parameter of interest is the mean. In math notation, the hypothesis is: H 0 : μ = 1 inch H 1 : μ ≠ 1 inch From the formulated hypothesis, it is clear that the test is two-tailed. Because (i) we are interested in the mean, (ii) the population parameter is normally distributed, the appropriate statistical distribution to use is the Z distribution. The decision rule is to reject the null is the calculated value of the test statistic falls in the rejection region. The level of significance, α, = 1 – Confidence Level = 1 – 0.90 = 0.10. Thus α is 0.1 Because the test is two-tailed, the C value is α/2 = 0.05/2 = 0.05

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The critical values of the test statistic are – Z C and + Z C that is: Z 0.05 and + Z 0.05 From the statistical tables, these are determined as follows: –1.645 and +1.645 The calculated value of the test statistic can be found after calculating the sample mean and standard deviation. The mean of slumps from the sample is: x = (0.92 + 1.05 + 1.21 + 1.11 . . . ) / 20 = 1.0075 inches The standard deviation of slumps from the sample is: σ of (0.92 + 1.05 + 1.21 + 1.11 . . . ) = 0.0755 inches g = G − ± ² = 1.0075 − 1 0.0755 u20 = 0.4445 Z* does not fall in the rejection region, therefore we fail to reject the null hypothesis. In other words, there is no statistical evidence to conclude that the mean slump differs from 1 inch. Thus
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## This note was uploaded on 01/20/2012 for the course CIVIL ENGI 398 taught by Professor Labi during the Spring '11 term at Purdue.

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Homework 5 Solutions - CE 398 Homework 5 Solutions QUESTION...

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