Klungo
- 135
- 1
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:
H_0: \mu = \mu_0.
Suppose also that:
\bar{x}, s is the mean and standard deviation of a sample of size n.
Suppose a significance level of \alpha.
Question: If \bar{x} < \mu_0, do we use H_A: \mu < \mu_0?
When do we use <,>,\neq?
Test Statistic
Since the population standard deviation is unknown, we "standardize" to a random variable T with a t distribution with n-1 degrees of freedom.
T_{test} = \displaystyle\frac{\bar{x} - \mu_0}{s/\sqrt{n}}.
Determining P-value
Assuming H_A: \mu < \mu_0, we have a left-tail test
So, p-value = P(T < T_{test}) = P(-T > -T_{test}) 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 df = 10, T_{test} = 2.
Here is a t-table: http://3.bp.blogspot.com/_5u1UHojRiJk/TEdJJc6of2I/AAAAAAAAAIE/Ai0MW5VgIhg/s1600/t-table.jpg.
We have T_{0.05}=1.812 < T_{test}= 2 < T_{0.025} = 2.228
Thus, 0.025 < p-value < 0.05.
Decision
Finally, if p-value < \alpha, then we reject H_0. Otherwise, we do not reject H_0.
The not equal case
If H_A : \mu \neq \mu_0, then we have
p-value = 2 P(|T| > T_{test}) = 2 P(-T < -T_{test}) + 2 P(T > T_{test}).
Question What now? And is the equality above correct?Thanks.
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:
H_0: \mu = \mu_0.
Suppose also that:
\bar{x}, s is the mean and standard deviation of a sample of size n.
Suppose a significance level of \alpha.
Question: If \bar{x} < \mu_0, do we use H_A: \mu < \mu_0?
When do we use <,>,\neq?
Test Statistic
Since the population standard deviation is unknown, we "standardize" to a random variable T with a t distribution with n-1 degrees of freedom.
T_{test} = \displaystyle\frac{\bar{x} - \mu_0}{s/\sqrt{n}}.
Determining P-value
Assuming H_A: \mu < \mu_0, we have a left-tail test
So, p-value = P(T < T_{test}) = P(-T > -T_{test}) 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 df = 10, T_{test} = 2.
Here is a t-table: http://3.bp.blogspot.com/_5u1UHojRiJk/TEdJJc6of2I/AAAAAAAAAIE/Ai0MW5VgIhg/s1600/t-table.jpg.
We have T_{0.05}=1.812 < T_{test}= 2 < T_{0.025} = 2.228
Thus, 0.025 < p-value < 0.05.
Decision
Finally, if p-value < \alpha, then we reject H_0. Otherwise, we do not reject H_0.
The not equal case
If H_A : \mu \neq \mu_0, then we have
p-value = 2 P(|T| > T_{test}) = 2 P(-T < -T_{test}) + 2 P(T > T_{test}).
Question What now? And is the equality above correct?Thanks.