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

We would face a rather close decision and switch to α = .01 for a more powerful test.

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}{(\sigma)/(√(n)) }`

Putting all the values, we have

`t_(stat) = \displaystyle(37.8 - 40)/((5.4)/(√(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.