# Why is the power for this test so low?

What follows is a theoretical power calculation for a practice exam, but this seems very low. Did I do something wrong? The problem and solution is as follows:

In a drinks factory the mean fill of the cans is set at 300 ml but there is a concern that the population mean fill of the cans may not in fact be 300 ml. Assume that the standard deviation, $\sigma$, of the amount of liquid in a random can is 1.2 ml. A random sample of 100 cans showed a mean fill of $\hat x = 299.64$

a). Is there evidence at the 1% significance level that the population mean fill differs from 300 ml? Carry out a z test to determine this.

b). Calculate the power of the test in a). when $u = 299.93$

Question b) is the part I'm not sure about. I got a power value of $0.0228$ which seems very low for the power of a test. Here is what I have done -

Power = P(Reject $H_0$ | $u = 299.93$)

We reject $H_0$ when $Z_{obs} < -Z_\frac{\alpha}{2}$, or $Z_{obs} > Z_\frac{\alpha}{2}$

which is equivalent to rejecting $H_o$ when

$$\frac{\hat x - u}{\frac{\sigma}{\sqrt{n}}} < -2.576 \ \ \ \ \ \ {\rm or} \ \ \ \ \ \ \frac{\hat x - u}{\frac{\sigma}{\sqrt{n}}} > 2.576$$

$$\hat x < u-2.576\frac{\sigma}{\sqrt{n}} \ \ \ \ \ \ {\rm or} \ \ \ \ \ \ \hat x >u+2.576\frac{\sigma}{\sqrt{n}}$$

$$\hat x < 300-2.576\frac{1.2}{10} \ \ \ \ \ \ {\rm or} \ \ \ \ \ \ \hat x > 300+2.576\frac{1.2}{10}$$

$$\hat x < 300-0.30912 \ \ \ \ \ \ {\rm or} \ \ \ \ \ \ \hat x > 300+0.30912$$

So we reject $H_0$ when $$\hat x < 299.69 \ \ \ \ \ \ {\rm or} \ \ \ \ \ \ \hat x > 300.309$$

Now using this to find the power we have -

Power = $$P(\hat x < 299.69 | u = 299.93) + P(\hat x > 300.309 | u = 299.93)$$

$$= P(Z < \frac{299.69 - 299.93}{\frac{1.2}{10}}) + P(Z > \frac{300.309 - 299.93}{\frac{1.2}{10}})$$

$$= P(Z < -2) + P(Z > 3.158) \approx 0.0228 + 0$$

where $Z$ is a standard normal random variable. So the power of the test is $0.0228$.

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Some intuition: Your effect size is small, your sample is (relatively) small, and your confidence level is high. Why should you expect high power? –  cardinal May 2 '12 at 12:52
(+1) For showing what you tried and your thought process. –  cardinal May 2 '12 at 12:57
Where did you get 2.576 and how is the first part of the question done? –  dkelly Aug 7 '12 at 12:08

Your effect size is extremely small: the true value is only $.07$ away from the null hypothesis, while the population standard deviation is $1.2$ and the sample size is $100$, making your standard error $.12$, so your standard error is almost twice as large as the distance from the null hypothesis, combining for a very small effect size. Intuitively, it makes sense that you would have very little power to detect a very small effect. In addition, you're using a pretty strict significance criteria $(.01)$, so you're less likely, in general, the reject null hypotheses.
@Jim_CS, the reasoning in this answer is correct, I can assure you :) Re: your calculation, you may be careful about saying $P(Z>3.158) = 0$. I'm not sure how particular your grader is, but including that term will actually change your answer by a bit. Also, welcome to the site! Please consider accepting this answer if you find it to be definitive. –  Macro May 2 '12 at 14:13
It looks like there may be some rounding error on both parts of your answer (you rounded before calculating areas under the normal curve). I'm getting something closer to $0.024$ as the power. Again, if your grader isn't that particular, this isn't a big deal, but if he/she is, you may want to be more careful there. –  Macro May 2 '12 at 14:20