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, [tex]\bar{x}[/tex] = 62
Sample size, n = 50
Alpha, α = 0.051
Sample standard deviation, s = 8
First, we design the null and the alternate hypothesis
[tex]H_{0}: \mu = 64\\H_A: \mu < 64[/tex]
We use one-tailed(left) z test to perform this hypothesis.
Formula:
[tex]z_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}} }[/tex]
Putting all the values, we have
[tex]z_{stat} = \displaystyle\frac{62 - 64}{\frac{8}{\sqrt{50}} } = -1.767[/tex]
Now, [tex]z_{critical} \text{ at 0.05 level of significance } = -2.33[/tex]
Since,
[tex]z_{stat} > z_{critical}[/tex]
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.
