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hypothesis Hypothesis testing based on 5 samples

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calbear
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I am reading some books about hypothesis testing, but I am not sure if my following reasoning makes sense:

Assume I have a gaussian random variable $X \sim N(\mu, \sigma)$ with $\sigma=1$. Now I obtain 5 iid samples, $x_1, \cdots, x_5$.

I want to check if $\mu<0$. So I set up the null hypothesis and alternative hypothesis to be $H_0: \mu=0$ and $H_1: \mu<0$

Therefore, for each sample $x_i$, I can compute the p-value = $P(X_i\le x_i)$, and denote it by $\alpha$. Therefore, I have $(1-\alpha)$ confidence to reject $H_0$. Also, by using the rule $(X\le x_i)$ to reject $H_0$, I have type-I error equal to $\alpha$.

Now based on 5 samples, I have 5 p-values $\alpha_i$, ($i=1, \cdots, 5$). Therefore, I have $P(X_1\le x_1, \cdots, X_5\le x_5) = \alpha_1 \times \alpha_2 \times \cdots \times \alpha_5$. Therefore, using the decision rule that (five samples are less than or equal to $x_1, \cdots, x_5$ respectively)$(X_1\le x_1, \cdots, X_5\le x_5)$ to reject H0, I have a type-I error equal to $5!\times \alpha_1 \times \cdots \times \alpha_5$$\times \alpha_1 \times \cdots \times \alpha_5$. And therefore, I have $(1-5!\times \alpha_1 \times \cdots \times \alpha_5)$$(1-\times \alpha_1 \times \cdots \times \alpha_5)$ confidence to reject $H_0$.

Basically, I want to use the 5 $x_i$'s for future testing. Next time I obtain 5 samples, I'll compare them to $x_i$. And I want to see how much confidence level this decision rule gives me. It seems to me that textbooks usually compute the p-value for one sample xi. I am basically trying to compute the "p-value" for 5 samples.

Is the above reasoning correct?

I am reading some books about hypothesis testing, but I am not sure if my following reasoning makes sense:

Assume I have a gaussian random variable $X \sim N(\mu, \sigma)$ with $\sigma=1$. Now I obtain 5 iid samples, $x_1, \cdots, x_5$.

I want to check if $\mu<0$. So I set up the null hypothesis and alternative hypothesis to be $H_0: \mu=0$ and $H_1: \mu<0$

Therefore, for each sample $x_i$, I can compute the p-value = $P(X_i\le x_i)$, and denote it by $\alpha$. Therefore, I have $(1-\alpha)$ confidence to reject $H_0$. Also, by using the rule $(X\le x_i)$ to reject $H_0$, I have type-I error equal to $\alpha$.

Now based on 5 samples, I have 5 p-values $\alpha_i$, ($i=1, \cdots, 5$). Therefore, I have $P(X_1\le x_1, \cdots, X_5\le x_5) = \alpha_1 \times \alpha_2 \times \cdots \times \alpha_5$. Therefore, using the decision rule that (five samples are less than or equal to $x_1, \cdots, x_5$ respectively) to reject H0, I have a type-I error equal to $5!\times \alpha_1 \times \cdots \times \alpha_5$. And therefore, I have $(1-5!\times \alpha_1 \times \cdots \times \alpha_5)$ confidence to reject $H_0$.

Is the above reasoning correct?

I am reading some books about hypothesis testing, but I am not sure if my following reasoning makes sense:

Assume I have a gaussian random variable $X \sim N(\mu, \sigma)$ with $\sigma=1$. Now I obtain 5 iid samples, $x_1, \cdots, x_5$.

I want to check if $\mu<0$. So I set up the null hypothesis and alternative hypothesis to be $H_0: \mu=0$ and $H_1: \mu<0$

Therefore, for each sample $x_i$, I can compute the p-value = $P(X_i\le x_i)$, and denote it by $\alpha$. Therefore, I have $(1-\alpha)$ confidence to reject $H_0$. Also, by using the rule $(X\le x_i)$ to reject $H_0$, I have type-I error equal to $\alpha$.

Now based on 5 samples, I have 5 p-values $\alpha_i$, ($i=1, \cdots, 5$). Therefore, I have $P(X_1\le x_1, \cdots, X_5\le x_5) = \alpha_1 \times \alpha_2 \times \cdots \times \alpha_5$. Therefore, using the decision rule that $(X_1\le x_1, \cdots, X_5\le x_5)$ to reject H0, I have a type-I error equal to $\times \alpha_1 \times \cdots \times \alpha_5$. And therefore, I have $(1-\times \alpha_1 \times \cdots \times \alpha_5)$ confidence to reject $H_0$.

Basically, I want to use the 5 $x_i$'s for future testing. Next time I obtain 5 samples, I'll compare them to $x_i$. And I want to see how much confidence level this decision rule gives me. It seems to me that textbooks usually compute the p-value for one sample xi. I am basically trying to compute the "p-value" for 5 samples.

Is the above reasoning correct?

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calbear
  • 165
  • 6

hypothesis testing based on 5 samples

I am reading some books about hypothesis testing, but I am not sure if my following reasoning makes sense:

Assume I have a gaussian random variable $X \sim N(\mu, \sigma)$ with $\sigma=1$. Now I obtain 5 iid samples, $x_1, \cdots, x_5$.

I want to check if $\mu<0$. So I set up the null hypothesis and alternative hypothesis to be $H_0: \mu=0$ and $H_1: \mu<0$

Therefore, for each sample $x_i$, I can compute the p-value = $P(X_i\le x_i)$, and denote it by $\alpha$. Therefore, I have $(1-\alpha)$ confidence to reject $H_0$. Also, by using the rule $(X\le x_i)$ to reject $H_0$, I have type-I error equal to $\alpha$.

Now based on 5 samples, I have 5 p-values $\alpha_i$, ($i=1, \cdots, 5$). Therefore, I have $P(X_1\le x_1, \cdots, X_5\le x_5) = \alpha_1 \times \alpha_2 \times \cdots \times \alpha_5$. Therefore, using the decision rule that (five samples are less than or equal to $x_1, \cdots, x_5$ respectively) to reject H0, I have a type-I error equal to $5!\times \alpha_1 \times \cdots \times \alpha_5$. And therefore, I have $(1-5!\times \alpha_1 \times \cdots \times \alpha_5)$ confidence to reject $H_0$.

Is the above reasoning correct?