Answer:
We conclude that there is not enough evidence to support the claim compute technique performs differently than the traditional method.
Step-by-step explanation:
We are given the following in the question:
Population mean, μ = 101 hours
Sample mean, [tex]\bar{x}[/tex] =100 hours
Sample size, n = 140
Alpha, α = 0.01
Population standard deviation, σ = 6 hours
First, we design the null and the alternate hypothesis
[tex]H_{0}: \mu = 101\text{ hours}\\H_A: \mu \neq 101\text{ hours}[/tex]
We use Two-tailed z test to perform this hypothesis.
Formula:
[tex]z_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}} }[/tex]
Putting all the values, we have
[tex]z_{stat} = \displaystyle\frac{100 - 101}{\frac{6}{\sqrt{140}} } = -1.972[/tex]
Now, [tex]z_{critical} \text{ at 0.01 level of significance } = \pm 2.58[/tex]
Since,
The calculated z statistic lies in the acceptance region, we fail to reject the null hypothesis and accept it.
We conclude that there is not enough evidence to support the claim that compute technique performs differently than the traditional method.