finding the probablity of type 2 error in a normally distributed RV using a Z test

Question

I am getting a $0.913$ answer as opposed to $0.903$ from a text, or are we both wrong on this?

Find the probability of type II error in

$H_0 : X \sim \mathcal{N}( 84, 100)$

$H_1$: $\mu_{X} \neq 84$

at $5\%$ significance, with $n = 4$, and in reality $\mu_{X} = 81$.

Answer 1

We begin with computing the standard deviation of the mean,.

n = 4 =>sample size

$\sigma = 10$ => population standard deviation $ \sigma_{\bar X} = \sigma/\sqrt{n} = \frac{10}{2} = 5$; standard error

We next compute the lower and upper bounds of sample means for which the null hypothesis μ = 84 would not be rejected.

$\alpha= .05 $=> significance level

$\mu0 = 84$ => hypothetical mean

$I = c(\frac{\alpha}{2}, 1-\frac{\alpha}{2}) $

$I =([5\times (-1.96) + 84] , [5\times (1.96)+84]) =>(74.2,93.8)$ , Critical value = 1.96 on either end for 0.025 and 0.975.

Therefore, so long as the sample mean is between 74.2 and 93.8 in a hypothesis test, the null hypothesis will not be rejected. Since we assume that the actual population mean is 81, we can compute the lower tail probabilities of both end points. $ \mu = 81$=> actual mean Probability of Type II error $=P(\frac{(74.2-81)}{5} \lt z \lt \frac{(93.8-81)}{5}) = 0.994766392 - 0.086914962 = 0.90785143 $

It is neither .903 nor 0.913. I get a different value which is 0.908.

Thanks

Satish