Question

The Rockwell hardness index for steel is determined by pressing a diamond point into the steel and measuring the depth of penetration. For 50 specimens of a certain type of steel, the Rockwell hardness index averaged 62 with a standard deviation of 8. The manufacturer claims that this steel has an average hardness index of at least 64. Test this claim at the 1% significance level?

Answer 1

Answer:

We conclude that the steel has an average hardness index of at least 64.

Step-by-step explanation:

We are given the following in the question:

Population mean, μ = 64

Sample mean, $\bar{x}$ = 62

Sample size, n = 50

Alpha, α = 0.051

Sample standard deviation, s = 8

First, we design the null and the alternate hypothesis

$H_{0}: \mu = 64\\H_A: \mu < 64$

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

Formula:

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

Putting all the values, we have

$z_{stat} = \displaystyle\frac{62 - 64}{\frac{8}{\sqrt{50}} } = -1.767$

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 steel has an average hardness index of at least 64.