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Using $n=16$ observations from normally distributed population $H_0: \mu = 30$ is tested against $H_A : \mu> 30$. If power of the test, $1-β = 0.85$ when $\mu_A = 34$, what is the probability of making Type 1 Error? Assume that $σ = 9$.

We have $Power = P(\text{Rejecting } H_0 \text{ when } \mu = 34) = 0.85$

Let $a$ be the point that we reject $H_0$ if $\bar{x} \gt a$. Then,

$P(\text{Rejecting } H_0 \text{ when } \mu = 34) = P(\bar{x} \gt a |\mu=34)= P(Z \gt \frac{a-\mu}{\sigma/\sqrt{n}}) = P(Z \gt \frac{a-34}{9/4}) = 0.85 => P(Z \lt \frac{a-34}{9/4}) = 0.15 $

Then, from z-table I find $\frac{a-34}{9/4} \approx -2.17 => a \approx 29.11 $.

What makes me wonder about what I did until here is the value of $a$. According to the way I defined $a$, a sample with mean $29.5$ would cause me to reject $H_0:\mu=30$ and accept $H_A:\mu \gt 30$.

I feel like I made some mistakes, can I get some help please?

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Your only problem is in saying the $15\%$ quantile of the normal distribution is $-2.17.$ Not sure what table you looked it up in but it's very wrong. It should be $-1.04,$ and then we get a threshold of $31.67$.

Edit

I see the error you made... $-2.17$ is the $1.5\%$ quantile, so you just accidentally typed $.015$ instead of $.15$ somewhere. Note that if we actually demanded this very high $98.5\%$ power, we would get a threshold for rejection below $30$ as you have calculated. This is fine... if we want to reject the null hypothesis a lot, we need to set a threshold where we reject it a lot and there's no hard-fast rule that says this needs to be above our null hypothesis. However, notice there's a trade-off here and our type one error will be very bad here... well over 50%. (Not that it was great even at $85\%$ power, as you will calculate.)

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  • $\begingroup$ It is sad to make suck a stupid mistake. I even checked the z-table several times to make sure. Also thanks for the further explanation about $98.5%$ power. $\endgroup$
    – user666150
    Dec 28, 2019 at 21:28
  • $\begingroup$ By the way, I was looking up that value from an actual table with rows and columns, trying to find the value 0.150 with my eyes. Maybe I should use a software or a calculator. $\endgroup$
    – user666150
    Dec 28, 2019 at 21:30
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    $\begingroup$ @zzlawlzz It's better than making non-stupid mistakes. I usually use octave. Frustratingly, I can't easily find a web-app that does this... I can find some stuff that will compute tail probabilities, but not the inverse problem of finding where a given quantile is. $\endgroup$ Dec 28, 2019 at 21:58

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