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Question

The mentioned exercise concerns a test about the mean contents of cola bottles. The hypotheses are

H0:μ=300Ha:μ<300\begin{aligned} & H_0: \mu=300 \\ & H_a: \mu<300 \end{aligned}

The sample size is n=6, and the population is assumed to have a normal distribution with σ=3\sigma=3. A 5% significance test rejects H0H_0 if z1.645z \leq-1.645, where the test statistic z is

z=xˉ3003/6z=\frac{\bar{x}-300}{3 / \sqrt{6}}

Power calculations help us see how large a shortfall in the bottle contents the test can be expected to detect.

Find the power of this test against the alternative μ=299\mu=299.

Solution

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Given:

n=Sample size=6μA=Alternative mean=299σ=Population standard deviation=3H0:μ=300Ha:μlt;300\begin{aligned} n&=\text{Sample size}=6 \\ \mu_A&=\text{Alternative mean}=299 \\ \sigma&=\text{Population standard deviation}=3 \\ H_0&:\mu=300 \\ H_a&:\mu&lt;300 \end{aligned}

We need to determine the power of the test.

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