Hypothesis Testing

  • Thread starter Klungo
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  • #1
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Statistics doesn't come to me as naturally as math.

I'm curious as to how to make a hypothesis test under the assumptions that the population standard deviation is unknown and using tables only.

Here is my understanding.

Given
Suppose:
[itex]H_0: \mu = \mu_0[/itex].
Suppose also that:
[itex]\bar{x}, s[/itex] is the mean and standard deviation of a sample of size [itex]n[/itex].
Suppose a significance level of [itex]\alpha[/itex].

Question: If [itex]\bar{x} < \mu_0[/itex], do we use [itex]H_A: \mu < \mu_0[/itex]?
When do we use [itex]<,>,\neq[/itex]?

Test Statistic
Since the population standard deviation is unknown, we "standardize" to a random variable [itex]T[/itex] with a [itex]t[/itex] distribution with [itex]n-1[/itex] degrees of freedom.

[itex]T_{test} = \displaystyle\frac{\bar{x} - \mu_0}{s/\sqrt{n}}[/itex].

Determining P-value
Assuming [itex]H_A: \mu < \mu_0[/itex], we have a left-tail test
So, [itex]p-value = P(T < T_{test}) = P(-T > -T_{test})[/itex] by symmetry. (My tables only show the right tail.)

So, now we look at the table.
Question I'm not sure if I know how to read and apply the values of the table. Is my work below correct?

Suppose [itex]df = 10, T_{test} = 2 [/itex].
Here is a t-table: http://3.bp.blogspot.com/_5u1UHojRiJk/TEdJJc6of2I/AAAAAAAAAIE/Ai0MW5VgIhg/s1600/t-table.jpg.

We have [itex]T_{0.05}=1.812 < T_{test}= 2 < T_{0.025} = 2.228 [/itex]
Thus, [itex]0.025 < p-value < 0.05[/itex].

Decision
Finally, if [itex]p-value < \alpha[/itex], then we reject [itex]H_0[/itex]. Otherwise, we do not reject [itex]H_0[/itex].

The not equal case
If [itex]H_A : \mu \neq \mu_0[/itex], then we have
[itex]p-value = 2 P(|T| > T_{test}) = 2 P(-T < -T_{test}) + 2 P(T > T_{test})[/itex].

Question What now? And is the equality above correct?


Thanks.
 

Answers and Replies

  • #2
Stephen Tashi
Science Advisor
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Question: If [itex]\bar{x} < \mu_0[/itex], do we use [itex]H_A: \mu < \mu_0[/itex]?
When do we use [itex]<,>,\neq[/itex]?
The alternative hypothesis is determined by the context of the particular problem. For example if [itex] \mu_0 [/itex] were the mean weight of a type of candy bar and you are a consumer then your concern might be that you are not "short changed" by getting a batch of candy bars whose mean weight is less than [itex] \mu_0 [/itex]. if you are a candy bar producer then your concern might be that the factory is not making candy bars that too heavy or too light, so your alternative hypothesis would be that the mean is not equal to [itex] \mu_0 [/itex].

The effect of alternative hypothesis on the mechanics of the problem is that it determines what kind of "acceptance region" you use in the hypothesis test. The typical decision about acceptance regions is whether to do a "one tailed" or "two tailed" test.

You must realize that hypothesis testing is simply a procedure for making a decision. It is not a proof that the decision is correct. It doesn't even tell you the probability that the decision is correct. There is no mathematical proof that it is an optimal procedure. So if you are seeking a mathematical understanding of why hypothesis testing should be done a certain way, you won't find any mathematical justifications based on those goals. If you want a procedure that aims for those goals, you have to study Bayesian statistics.
 
  • #3
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You must realize that hypothesis testing is simply a procedure for making a decision. It is not a proof that the decision is correct. It doesn't even tell you the probability that the decision is correct. There is no mathematical proof that it is an optimal procedure. So if you are seeking a mathematical understanding of why hypothesis testing should be done a certain way, you won't find any mathematical justifications based on those goals.
I just want to know how it's done as it is in a stats 101 course. I know that it is not a proof, but rather an "educated" guess that could be wrong.

[Edit:] That is, the general procedure for solving problems given those assumptions.
 
  • #4
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I declare this thread over. I now understand the procedure.
 

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