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Before the launch of a commercial product, a company makes a market survey to know the price that buyers are willing to pay. It is assumed that this price is normally distributed with a desviation of \$10. The marketing department informs that the public considers appropriate the price of \$30. To test this hypothesis against a price of \$40, a sample of 25 people is selected and adopted the following decision rule: if the sample mean is less than \$35, are considered it is appropriate to set a price of \$30.

  1. Find the probability of committing an error of type 1.
  2. Find the probability of committing an error of type 2.
  3. Find de power.

Let be $\alpha$ the probabilityof committing an error of type 1 and $\beta$ the probability of committing an error of type 2: $\alpha=P($accept $H_0/H_0$ is false$)$ and $\beta=P($refuse $H_0/H_0$ is false$)$

The problem is thatI can't find the hypothesis. My criterion is to define it as follows:

$H_0: \mu = \$30$ -> null hypothesis

$H_1: \mu \not= \$30$ -> alternative hypothesis

This is ok?, but what can I do with the price of $\$40$?

And, whatis the sample mean?, I have to find it to calculate the $\alpha$. Or I have to suppose an $\alpha=0,01$?

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  • $\begingroup$ 35\$ is your threshold. and NO you wont assume $\alpha=0.01$, you will determine it using the given threshold. Check your notes, you should have already a formula to calculate the false alarm and miss detection probabilities. $\endgroup$ – Seyhmus Güngören Feb 22 '16 at 23:23
  • $\begingroup$ I've made a lot of exercices of hypothesis testing, but nothing like that. I don't have any formula to calculate this $\endgroup$ – Valentina Ramirez Feb 23 '16 at 0:25
  • $\begingroup$ Okay, then we can find it together. Just let me remind you. This is probably the simplest example. Let me ask you first how do you calculate $\alpha=P(\bar{X}>35|X\sim \mathcal{N}(30,10^2))$? $\endgroup$ – Seyhmus Güngören Feb 23 '16 at 0:44
  • $\begingroup$ $\alpha = 1-P(X<35)=P\left(z<{{X-\mu}\over{\sigma\over{\sqrt n}}}\right)$, right? $\endgroup$ – Valentina Ramirez Feb 23 '16 at 4:32
  • $\begingroup$ If $X>35$ then I have to accept $H_0$ hence $\mu=30$, but, alpha is the probability of accept $H_0$ when $H_0$ is false, so the correct hypothesis is $H_1$. The rule says "if $X<35$, then $\mu = 30$", but I have to calculate what happen when $H_0$ is false, i.e. the correct is $H_1$, so I have that: $\alpha = P(X>35) = 1 - P(X<35)$, right? $\endgroup$ – Valentina Ramirez Feb 23 '16 at 4:43
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let $X$ be a random variable that represents the price.

$H_0: X \sim \mathcal{N}(30, 100) $

$H_1: X \sim \mathcal{N}(40, 100) $

Sample mean is $\bar{X} = \frac{1}{25}\sum\limits_{i=1}^{25}X_i$, where $X_i$ are i.i.d. having the same distribution as $X$.

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