Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

The exercise is:

Assume that the population mean is actually 110 grams and that the distribution is normal with standard deviation 4 grams. In a z test of $H_0$: u = 113 against $H_a$: u < 113 with $\alpha = 0.05$, find the probability of rejecting $H_0$ with six observations.

From part (a) in that exercise i found out:

$$\hat x = 112.967 - s = 4.28 - n = 6- t,0.05,5 = -2.015$$

My approach is: Find the type II error and do $$ 1 - \beta(110) $$


$$P\left( \frac{\hat x - 113}{\frac{4.28}{\sqrt{6}}} > -2.015 | u = 110, \sigma=4\right) = \alpha $$ $$1 - P\left(\hat x < -2.015\cdot{\frac{4.28}{\sqrt{6}}} + 113\right) $$ $$1 - P\left(\frac{\hat x - 110}{\frac{4}{\sqrt{6}}} < \frac{-2.015\cdot{\frac{4.28}{\sqrt{6}}} + 3)}{\frac{4}{\sqrt{6}}}\right) $$


$$1-P\left(\frac{\hat x - 110}{\frac{4}{\sqrt{6}}} < -0.32\right)$$

which gives (1-(1-0.3745)) = 0.3745, while the solution is 0.58.

share|cite|improve this question
up vote 1 down vote accepted

The sample mean doesn't enter into this calculation at all. You're answering a question about the probability that (under repeated sampling) you'd reject the null when you're drawing from a population with mean 110 while the hypothesized mean is 113.

Further note that the question specifies a z test, not a t-test - and specifies $\sigma$ - so your critical value is wrong. You need to pay careful attention to the question.

(By comparison, your sample either leads to rejection or it doesn't.)

Here's a calculation (done in the statistical package R) you may find of some value, by considering where the inputs come from, and what they mean:

> pnorm(-1.645-(110-113)/(4/sqrt(6)))
[1] 0.5761748

Note that used with a single argument, pnorm is just the cdf of the standard normal.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.