Question

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)

Answer 1

Answer:

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, $\bar{x}$ = 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

$H_{0}: \mu = 3400\text{ psi}\\H_A: \mu > 3400\text{ psi}$

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

Formula:

$z_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}} }$

Putting all the values, we have

$z_{stat} = \displaystyle\frac{3450 - 3400}{\frac{75}{\sqrt{12}} } = 2.309$

Now,

$z_{critical} \text{ at 0.05 level of significance } = 2.33$

Since,  

$z_{stat} < z_{critical}$

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