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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.