In: Statistics and Probability
Group of engineers in a steel company should decide whether or not a shipment of steel meets the required yield strength of 3235 psi. A random sample of 100 specimens is selected with a computed mean of 3210 psi which is following normal distribution. The variance of the population is 25600 psi. In this case, the shipment would be rejected only if the mean strength in the sample is significantly less than 3235 psi with the significance level of α = 0.01. Develop a hypothesis assumption and test and then mention whether they need to accept or reject the shipment
Solution :
= 3235
=3210
2 =25600
=160
n = 100
This is the left tailed test .
The null and alternative hypothesis is ,
H0 : = 3235
Ha : < 3235
Test statistic = z
= ( - ) / / n
= (3210 - 3235) / 160 / 100
= -1.56
Test statistic = z = -1.56
P(z < -1.56 ) = 0.0594
P-value =0.0594
= 0.01
P-value >
0.0594 > 0.01
Fail to reject the null hypothesis .
There is insufficient evidence to suggest that