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 = 62. Let denote the true average compressive strength.

(a) What are the appropriate null and alternative hypotheses?

(b) Let

X

denote the sample average compressive strength for n = 11 randomly selected specimens. Consider the test procedure with test statistic

X

itself (not standardized). What is the probability distribution of the test statistic when H0 is true?

The test statistic has a normal distribution.The test statistic has an exponential distribution.    The test statistic has a binomial distribution.The test statistic has a gamma distribution.

c)  If

X = 1340,

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

d) State the mean and standard deviation of the test statistic. (Round your standard deviation to three decimal places.)

e)  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

Solution:-

a) State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: < 3700

Alternative hypothesis: > 3700

Note that these hypotheses constitute a one-tailed test. The null hypothesis will be rejected if the sample mean is too small.

Formulate an analysis plan. For this analysis, the significance level is 0.01. The test method is a one-sample z-test.

Analyze sample data. Using sample data, we compute the standard error (SE), z statistic test statistic (z).

b) The test statistic has a normal distribution.

SE = s / sqrt(n)

S.E = 18.69

D.F = 10

c) z = (x - ) / SE

z = 2.14

where s is the standard deviation of the sample, x is the sample mean, is the hypothesized population mean, and n is the sample size.

The observed sample mean produced a z statistic test statistic of 2.14.

Thus the P-value in this analysis is 0.0162

Interpret results. Since the P-value (0.0162) is greater than the significance level (0.01), we cannot reject the null hypothesis.

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