Homework SourseMean tensile strength for a steel product is 24100 psi The s
Mean tensile strength for a steel product is 24100 psi. The standard deviation is 160 psi. 16 random observatioms were used. What percentage of the time will a test detect a possible increase of 180psi in mean strength at significance level 5%?
Mean tensile strength for a steel product is 24100 psi. The standard deviation is 160 psi. 16 random observatioms were used. What percentage of the time will a test detect a possible increase of 180psi in mean strength at significance level 5%?
Solution
Given that
mean tensile strength (mu) = 24100 psi
sd = 160 psi
n = number of observations = 16
Xbar= 24100 + 180 = 24280
Here we have to test the hypothesis that
H0 : mu = 24100 Vs H1 : mu > 24100
The test statistic is,
z = (Xbar - mu) / sd/sqrt(n)
z = (24280 - 24100) / 160 / sqrt(4) = 180 / 80 = 2.25
We can find P-value for decision by using EXCEL.
syntax :
=normsdist(z)
where z is test statistic value.
P-value = 0.012
alpha = 0.05
P-value < alpha
Reject H0 at 5% level of significance.
Conclusion : Mean tensile strength is greator than 24100.
