## Related questions with answers

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

$\begin{aligned} & H_0: \mu=300 \\ & H_a: \mu<300 \end{aligned}$

The sample size is n=6, and the population is assumed to have a normal distribution with $\sigma=3$. A 5% significance test rejects $H_0$ if $z \leq-1.645$, where the test statistic z is

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

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 $\mu=299$.

Solution

VerifiedGiven:

$\begin{aligned} n&=\text{Sample size}=6 \\ \mu_A&=\text{Alternative mean}=299 \\ \sigma&=\text{Population standard deviation}=3 \\ H_0&:\mu=300 \\ H_a&:\mu<300 \end{aligned}$

We need to determine the power of the test.

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