An engineer who is studying the tensile strength of a steel alloy intended for use in golf club shafts knows that tensile strength is approximately normally distributed with σ=75 psi. a random sample of 12 specimens has a sample mean tensile strength of 3450 psi. is there evidence to support the claim that mean strength is greater than 3400 psi. state and test the hypothesis at significance level α =0.01. (2 points)

We conclude that the mean strength is equal to 3400 psi.

Step-by-step explanation:

We are given the following in the question:

Population mean, μ = 3400 psi

Sample mean, = 3450 psi

Sample size, n = 12

Alpha, α = 0.051

Population standard deviation, σ =75 psi

First, we design the null and the alternate hypothesis

We use one-tailed(right) z test to perform this hypothesis.

Formula:

Putting all the values, we have

Now,

Since,  

We fail to reject the null hypothesis and accept the null hypothesis. Thus, we conclude that the mean strength is equal to 3400 psi.

dc=6 square root of 3 and ac is 12 square root 3

c and d

step-by-step explanation:

square root of 3/2 = 18/c

cross multiply to get 18*2=√3*c

36/√3 * √3/√3 = c

c= 36*√3/3

c=12√3

c

step-by-step explanation: easy math