Type II error probability - is my solution correct?

Question

When conducting a hypothesis test for a normal sample, with sample size = 5, with known standard deviation 0.02, where $H_0: μ = 1.12$, $H_1 : μ ≠ 1.12$

What is the probability of a type II error if the true mean is 1.14?

The critical values are $\pm 1.645$ so we would not reject $H_0$ if $\lvert \frac{\overline{x} - 1.12}{\frac{0.02}{\sqrt{5}}} \rvert \le 1.645$, so the $\overline{x}$ is not rejected in the range (1.1052, 1.1347).

$\overline{x}$ is distributed $N(1.14, \frac{0.02^2}{5})$ so

$P(1.105 \lt \overline{x} \lt 1.347) = P\left(\frac{1.105-1.14}{\frac{0.02}{\sqrt{5}}} \lt Z \lt \frac{1.1347 - 1.14}{\frac{0.02}{\sqrt{5}}} \right) = P(-3.891 \lt Z \lt -0.593) = 0.27659 - 0.00005 = 0.27654$.

Would greatly appreciate if someone can check if my methodology and solution is correct, as my answer is slightly different from the given solution. Just wondering if this is due to rounding, or if I've made a mistake somewhere.

Thanks.

Answer 1

In R statistical software. In R statistical software, it is possible to compute probabilities for the distribution $Norm(\mu=1.14, \sigma=.02/\sqrt{5})$ directly, without standardizing and without rounding (or interpolating) to use printed standard normal tables. So the probability of 'failing to reject' when $\mu_a = 1.14$ is $0.2767$ to four places. As you can see from the code, this assumes that the rejection region is the complement of $(1.1052, 1.1347),$ that $\sigma=.02,$ and that $n = 5.$ It seems that you are doing a 2-sided test at level $\alpha = 0.1.$ (Notice that in R code, the third parameter of the normal CDF pnorm is the standard deviation.)

 diff(pnorm(c(1.1052,1.1347), 1.14, .02/sqrt(5)))
 ## 0.2766885

Your method seems correct. I used the values you quoted without checking. I don't know the discrepancy between your answer and the given solution, so I will not speculate whether rounding error accounts for it. (I do wonder if the endpoints of your interval are carried to enough places.)

In Minitab statistical software. Minitab statistical software gives the following output for the power in this situation. [Power is $1 - P(\text{Type II Err})$.]

Power and Sample Size 

1-Sample Z Test

Testing mean = null (versus not = null)
Calculating power for mean = null + difference
Alpha = 0.1  Assumed standard deviation = 0.02


            Sample
Difference    Size     Power
      0.02       5  0.722864