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.)
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 |
Get Answers For Free
Most questions answered within 1 hours.