Answer:
We would face a rather close decision and switch to α = .01 for a more powerful test.
Step-by-step explanation:
We are given the following in the question:
Population mean, μ = 40 hours
Sample mean, [tex]\bar{x}[/tex] = 37.8 hours
Sample size, n = 108
Alpha, α = 0.05
Sample standard deviation, s = 5.4 hours
First, we design the null and the alternate hypothesis
[tex]H_{0}: \mu \geq 40\text{ hours}\\H_A: \mu < 40\text{ hours}[/tex]
We use one-tailed t test to perform this hypothesis.
Formula:
[tex]t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}} }[/tex]
Putting all the values, we have
[tex]t_{stat} = \displaystyle\frac{37.8 - 40}{\frac{5.4}{\sqrt{18}} } = -1.728[/tex]
Now,
[tex]t_{critical} \text{ at 0.05 level of significance, 14 degree of freedom } = -1.739[/tex]
Since,
[tex]t_{stat} < t_{critical}[/tex]
We fail to reject the null hypothesis and accept the null hypothesis. Thus, we conclude that alkaline batteries last at least 40 hours on average in a certain type of portable CD player.
But we faced a close decision. By decreasing alpha we could get stronger results. We would switch to α = .01 for a more powerful test.
