Answer the following true or false. If false, explain why. a. We can reduce α...

naivlingr

Answered question

2021-01-19

Answer the following true or false. If false, explain why.
a. We can reduce $\alpha $ (alpha) by pushing the critical regions further into the tails of the distribution.
b. A decrease in the probability of the type II error always results in an increase in the probability of the type I error, provided that the sample size n does not change.
c. The p-value of a left-sided hypothesis test is always half of a two-sided hypothesis test.

Answer & Explanation

SkladanH

Skilled2021-01-20Added 80 answers

a.
The quantity, $\alpha $ is also known as the probability of Type-I error. It is the probability of rejecting the null hypothesis when it is true.
In a hypothesis testing problem, the probability assigned to the critical region is alpha, which is proportional to the area of the critical region. This critical region is the region, which lies in the extremities of the sampling distribution of the statistic, and is the region that marks the extreme or unlikely values of the distribution.
If one pushes the critical regions further into the tails, the probability assigned to it will become smaller, thus reducing alpha
Hence, the statement is True.
b.
The probability of Type-II error is the probability of failing to reject the null hypothesis when it is false.
In reality, if the data and all the conditions attached to the hypothesis test are kept unchanged, there is only one way to decrease the probability of Type-II error, which is, by increasing the probability of Type-I error. It is not possible to simultaneously reduce both errors keeping all other factors in the study exactly identical.
In a study, the only way to decrease both types of errors simultaneously, is by increasing the sample size.
Hence, the statement is True.
c.
The p-value of a left-sided hypothesis test will be equal to half of that of a two-sided test, only in case of a symmetric sampling distribution of the test statistic (example: t, Normal distributions).
For a symmetric distribution, the probability to the right of a certain value is equal to the probability to the left of the value which is at the mirror-image location of that value.
However, this is not true for any distribution that is not symmetric (example: $F,{\chi}^{2}$ distributions).
Hence, the statement is False.