Question

Dullco Manufacturing claims that its alkaline batteries last at least 40 hours on average in a certain type of portable CD player. But tests on a random sample of 18 batteries from a day's large production run showed a mean battery life of 37.8 hours with a standard deviation of 5.4 hours. In a left-tailed test at α = .05, which is the most accurate statement?
We would strongly reject the claim.

We would clearly fail to reject the claim.

We would face a rather close decision.

We would switch to α = .01 for a more powerful test.

Answer 1

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, $\bar{x}$ = 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

$H_{0}: \mu \geq 40\text{ hours}\\H_A: \mu < 40\text{ hours}$

We use one-tailed t test to perform this hypothesis.

Formula:

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

Putting all the values, we have

$t_{stat} = \displaystyle\frac{37.8 - 40}{\frac{5.4}{\sqrt{18}} } = -1.728$

Now,

$t_{critical} \text{ at 0.05 level of significance, 14 degree of freedom } = -1.739$

Since,                    

$t_{stat} < t_{critical}$

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.