Question

Suppose that the rejection region of a test 𝜓 has the form 𝑅={𝑋⎯⎯⎯⎯⎯𝑛:𝑋⎯⎯⎯⎯⎯𝑛>𝑐} . Find the smallest 𝑐 such that 𝜓 has level 𝛼 .

(If applicable, type abs(x) for |𝑥| , Phi(x) for Φ(𝑥)=𝐏(𝑍≤𝑥) where 𝑍∼N(0,1) , and q(alpha) for 𝑞𝛼 , the 1−𝛼 quantile of a standard normal variable.)

What is
𝑐≥ ?

Answer 1

[AI-generated answer — Explain Bot]

To find the smallest value of 𝑐, we need to determine the critical value that corresponds to the given level of 𝛼. The critical value is the threshold value above which we reject the null hypothesis.

Since the rejection region is given as 𝑅 = {𝑋⎯⎯⎯⎯⎯𝑛: 𝑋⎯⎯⎯⎯⎯𝑛 > 𝑐}, it means that we will reject the null hypothesis if the test statistic, 𝑋⎯⎯⎯⎯⎯𝑛, is greater than 𝑐.

To find 𝑐, we need to determine the value that corresponds to the 𝛼 quantile of the test statistic distribution. In this case, we assume the test statistic distribution to be a standard normal distribution, 𝑍 ∼ N(0,1).

The 𝛼 quantile is the value that leaves an area of 𝛼 in the upper tail of the distribution. Since this is a one-tailed test (rejection in one direction), we are interested in the upper tail of the distribution.

Using the standard normal distribution table or a calculator, we can find the 𝛼 quantile, denoted as 𝑧(𝛼). For example, if 𝛼 = 0.05, we want to find the value 𝑧(0.05), which corresponds to the 0.95 percentile (leaving an area of 0.05 in the upper tail).

Once we have 𝑧(𝛼), we can set 𝑐 to be greater than or equal to 𝑧(𝛼). This ensures that the rejection region contains all values of the test statistic that are in the upper 𝛼 percentile of the distribution.

Therefore, 𝑐 ≥ 𝑧(𝛼) is the answer, where 𝑧(𝛼) is the 𝛼 quantile of the standard normal distribution.