Question

A mixture of pulverized fuel ash and Portland cement to be used for grouting should have...

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 σ = 67. Let μ denote the true average compressive strength.

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

H0: μ > 1300
Ha: μ = 1300 H0: μ < 1300
Ha: μ = 1300     H0: μ = 1300
Ha: μ ≠ 1300 H0: μ = 1300
Ha: μ > 1300 H0: μ = 1300
Ha: μ < 1300


(b) Let

X

denote the sample average compressive strength for n = 15 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 binomial distribution. The test statistic has a gamma distribution.     The test statistic has an exponential distribution. The test statistic has a normal distribution.


If

X = 1340,

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

Should H0 be rejected using a significance level of 0.01?

reject H0 do not reject H0    


(c) What is the probability distribution of the test statistic when μ = 1350?

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

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

mean      KN/m2
standard deviation      KN/m2


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

Homework Answers

Answer #1

a)

Ho: μ = 1300
Ha: μ > 1300

b)

test statistic =(x̅-μ)*√n/σ = 2.31
p value = 0.0104

he test statistic has a normal distribution.

do not reject H0    

c(

The test statistic has a normal distribution.

mean= 1350
standard deviation=σ/√n= 17.299
rejection region : x̅ >=μ+z*σ/√n = 1340.307
P(type II error)=P(x̅ <1340.3074)=P(Z<(1340.3074-1350)√15/67)=P(Z<-0.56)= 0.2877
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