Question

QUESTION 4

1. The developer of a new welding rod claims that spot welds using his product will have greater strength than conventional welds. For 45 welds using the new rod, the average tensile strength is 23,500 pounds per square inch, with a standard deviation of 600 pounds. For 40 conventional welds on the same materials, the average tensile strength is 23,140 pounds per square inch, with a standard deviation of 750 pounds. Using the 0.01 level, what is the calculated value of t statistic? (Specify your answer to the 3rd decimal.)

QUESTION 5

1. The developer of a new welding rod claims that spot welds using his product will have greater strength than conventional welds. For 45 welds using the new rod, the average tensile strength is 23,500 pounds per square inch, with a standard deviation of 600 pounds. For 40 conventional welds on the same materials, the average tensile strength is 23,140 pounds per square inch, with a standard deviation of 750 pounds. Using the 0.01 level, what is the calculated value of degree of freedom? (Specify your answer to the 2nd decimal.)

QUESTION 6

1. The developer of a new welding rod claims that spot welds using his product will have greater strength than conventional welds. For 45 welds using the new rod, the average tensile strength is 23,500 pounds per square inch, with a standard deviation of 600 pounds. For 40 conventional welds on the same materials, the average tensile strength is 23,140 pounds per square inch, with a standard deviation of 750 pounds. Using the 0.01 level, what is your conclusion based on calculated value of test statistic and critical value?

   		Reject H0
   		Do not reject H0

Answer 1

Assume equal population variances.
Standard Deviation=
pooled standard deviation=sqrt(((45-1)*600^2+(40-1)*750^2)/(45+40-2))
=674.65
standard error for difference=674.65*sqrt((1/45)+(1/40))=146.61
A)Test statistic:
t=(23500-23140)/146.61
t=2.456

B)df=45+40-2=83

C)p-value=tdist(2.456,83,1)=0.0081 {Excel Function To find P value)
As,p-value<0.01,we reject the null hypothesis.