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 σ = 66. Let μ denote the true average compressive strength.

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

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

(b) Let

X

denote the sample average compressive strength for n = 14 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 gamma distribution.The test statistic has a normal distribution.    The test statistic has a binomial distribution.The test statistic has an exponential 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 H₀do not reject H₀   

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

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

Answer 1

a)

H₀: μ = 1300
H_a: μ > 1300

The test statistic has a normal distribution.

population std dev ,    σ =    66.0000                  
Sample Size ,   n =    14                  
Sample Mean,    x̅ =   1340.0000                  
                          
'   '   '                  
                          
Standard Error , SE = σ/√n =   66.0000   / √    14   =   17.6392      
Z-test statistic= (x̅ - µ )/SE = (   1340.000   -   1300   ) /    17.6392   =   2.268
                          
  
p-Value   =   0.0117  
p-value>α, Do not reject null hypothesis   

do not reject H₀  

c)

The test statistic has a normal distribution.

true mean ,    µ =    1350              
                      
hypothesis mean,   µo =    1300              
significance level,   α =    0.01              
sample size,   n =   14              
std dev,   σ =    66.0000              
                      
δ=   µ - µo =    50              
                      
std error of mean,   σx = σ/√n =    66.0000   / √    14   =   17.63924

P(type II error) , ß =   P(Z < Zα - δ/σx)                  
= P(Z <    2.326   - (   50   /   17.6392   ))
=P(Z<   -0.508   ) =   0.3056   [excel fucntion: =normsdist(z)      

P(type II error ) = 0.3056