A mixture of pulverized fuel ash and Portland cement to be u

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 = 65. Let denote the true average compressive strength.

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

MEAN = 1350 kn/M^3 and Standard deviation= 18.028KN/m^2

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.)

THE ANSWER IS 0.3274 BUT I NEED TO KNOW HOW TO GET THAT ANSWER

Solution

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. now compressive strength for specimens of this mixture is normally distributed with = 65.

denote the true average compressive strength.

here we want to test whether is greater than 1300 or not

so null hypothesis is H0: =1300 and the alternative hypothesis is H1: >1300

to test this we have the test statistic

T=(Xbar-)*sqrt(n)/ where Xbar is the average of the sample at hand. and n is the number of sample observations considered

. now Xbar~N(,²/n)

now =65 and =1300 under H0

so T=(Xbar-1300)*sqrt(n)/65 would follow a N(0,1) distribution under H0

the mixture would be considered satisfactory iff t>tao_alpha where t is the observed value of T and tao_alpha is the upper alpha point of a N(0,1) distribution.

given that alpha=0.01

so tao_0.01=2.32635 [using minitab]

so the mixture would be considered satisfactory iff t>2.32635

hence the mixture would be considered unsatisfactory iff t<2.32635

now when = 1350 Xbar~N(1350,65²/n)

so Z=(Xbar-1350)*sqrt(n)/65 ~N(0,1)

so the probability that the mixture will be judged unsatisfactory when = 1350 is

P[(Xbar-1300)*sqrt(n)/65<2.32635]

=P[(Xbar-1350+1350-1300)*sqrt(n)/65<2.32635]

=P[(Xbar-1350)*sqrt(n)/65+50*sqrt(n)/65<2.32635]

P[Z+50*sqrt(n)/65<2.32635]    where Z~N(0,1)

nwo the standard deviation of Xbar is 65/sqrt(n)

it is given that 65/sqrt(n)=18.028

so P[Z+50*sqrt(n)/65<2.32635]=P[Z+50*(1/18.028)<2.32635]

=P[Z+2.77346<2.32635]

=P[Z<-0.477]    where Z~N(0,1)

=0.3274 [using minitab]

so the probability that the mixture will be judged unsatisfactory when = 1350 is 0.3274 [answer]