Suppose $$ x_{1}, x_{2}, x_{2}, \ldots x_{n} $$ is a random sample from a normal population with mean $$ \mu $$ and variance 16 . Find the test with the best critical region, that is, find the most powerful test, with a sample size of $$ \mathrm{n}=16 $$ and a significance level $$ \propto=0.05 $$ to test the simple null $$ H_{\mathbf{a}}: heta=10 $$ hypothesis against the simple alternative hypothesis $$ H_{1}: heta=15 $$, Also find the power of the test.

Question

Suppose $$ x_{1}, x_{2}, x_{2}, \ldots x_{n} $$ is a random sample from a normal population with mean $$ \mu $$ and variance 16 . Find the test with the best critical region, that is, find the most powerful test, with a sample size of $$ \mathrm{n}=16 $$ and a significance level $$ \propto=0.05 $$ to test the simple null $$ H_{\mathbf{a}}: 	heta=10 $$ hypothesis against the simple alternative hypothesis $$ H_{1}: 	heta=15 $$, Also find the power of the test.

UpStudy AI · Accepted Answer

We are given a sample from N(µ, 16) with n = 16. We wish to test

H₀: µ = 10  versus  H₁: µ = 15

at a significance level of α = 0.05.

Step 1. Identify the Test Statistic

Since the sample comes from a normal population with a known variance, the sample mean X̄ is normally distributed. Under any given µ, we have

X̄ ~ N(µ, σ²/n)  =  N(µ, 16/16) = N(µ, 1).

Step 2. Obtain the Most Powerful Test (Neyman–Pearson Lemma)

For testing two simple hypotheses, the Neyman–Pearson lemma tells us that the most powerful test rejects H₀ for large values of the likelihood ratio. Given the problem and the alternatives, the decision will be based on X̄. Intuitively, since H₁ has a larger mean than H₀, we reject H₀ when X̄ is large.

Step 3. Determine the Critical Value

We want the test to have a size of 0.05. Therefore, under H₀ (µ = 10) we find the value c such that

P(X̄ > c | µ = 10) = 0.05.

Since X̄ ~ N(10, 1) under H₀, we standardize:

P(X̄ > c) = P(Z > c−10) = 0.05,

where Z ~ N(0, 1).

From the standard normal table, the 95th percentile is approximately 1.645. Therefore,

c − 10 = 1.645  ⟹  c = 11.645.

Thus, the most powerful test rejects H₀ if X̄ > 11.645.

Step 4. Compute the Power of the Test

The power of the test is the probability of rejecting H₀ when H₁ is true (i.e., when µ = 15). Under H₁,

X̄ ~ N(15, 1).

Then, the power is

Power = P(X̄ > 11.645 | µ = 15).

Standardizing under H₁:

P(X̄ > 11.645) = P(Z > 11.645 − 15) = P(Z > −3.355),

where Z ~ N(0, 1).

Since P(Z > −3.355) = 1 − P(Z ≤ −3.355) and using symmetry P(Z ≤ −3.355) = P(Z ≥ 3.355), we note that P(Z ≥ 3.355) is very small. Alternatively, because the standard normal is symmetric, we have

P(Z > −3.355) = Φ(3.355).

Using standard normal tables or a calculator, Φ(3.355) is approximately 0.9996.

Thus, the power is roughly 99.96%.

Step 5. Final Answer

The most powerful test at the 0.05 significance level is:

Reject H₀ if X̄ > 11.645.

The power of this test when µ = 15 is approximately 0.9996 (99.96%).
Reject $$ H_0 $$ if the sample mean $$ \bar{X} $$ is greater than 11.645. The power of the test when $$ \mu = 15 $$ is approximately 99.96%.

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