Math Problem Statement

Menentukan uji hipotesis terbaik untuk sampel acak dari populasi normal dengan rata-rata μ, varian 16, n = 16, α = 0.05, H0: μ = 10, dan HA: μ = 15.

Solution

The problem involves hypothesis testing for a normal distribution. Here's the breakdown:

Problem Context:

  • X1,X2,,XnX_1, X_2, \dots, X_n are random samples from a normal distribution with mean μ\mu and variance σ2=16\sigma^2 = 16.
  • Sample size n=16n = 16.
  • Significance level α=0.05\alpha = 0.05.
  • Null hypothesis: H0:μ=10H_0: \mu = 10.
  • Alternative hypothesis: HA:μ=15H_A: \mu = 15.
  • We are to find the most powerful test (Neyman-Pearson Lemma).

Steps to Solve:

  1. Statistical Distribution:

    • Under the null hypothesis H0H_0, the test statistic is: XˉN(μ,σ2n)=N(10,1616)=N(10,1).\bar{X} \sim N\left( \mu, \frac{\sigma^2}{n} \right) = N\left(10, \frac{16}{16}\right) = N(10, 1).

    • Under HAH_A, XˉN(15,1)\bar{X} \sim N(15, 1).

  2. Critical Region:

    • The Neyman-Pearson Lemma tells us that the most powerful test is based on the likelihood ratio: f(XˉHA)f(XˉH0)>k,\frac{f(\bar{X} \mid H_A)}{f(\bar{X} \mid H_0)} > k, for some critical value kk.
  3. Finding the Critical Value:

    • We reject H0H_0 if Xˉ\bar{X} falls in the critical region defined such that the probability of Type I error (α\alpha) is: P(Reject H0H0)=α.P(\text{Reject } H_0 \mid H_0) = \alpha.

    • Since XˉN(10,1)\bar{X} \sim N(10, 1) under H0H_0: P(Xˉ>cH0)=0.05.P(\bar{X} > c \mid H_0) = 0.05.

    • From the standard normal distribution, the critical value is: c=μ0+zασ=10+1.6451=11.645.c = \mu_0 + z_\alpha \cdot \sigma = 10 + 1.645 \cdot 1 = 11.645.

  4. Test Procedure:

    • Reject H0H_0 if Xˉ>11.645\bar{X} > 11.645.
  5. Power of the Test:

    • To calculate the power, compute: β=P(Reject H0HA)=P(Xˉ>11.645μ=15).\beta = P(\text{Reject } H_0 \mid H_A) = P(\bar{X} > 11.645 \mid \mu = 15).
    • Under HAH_A, XˉN(15,1)\bar{X} \sim N(15, 1): P(Xˉ>11.645)=P(Z>11.645151)=P(Z>3.355).P(\bar{X} > 11.645) = P\left(Z > \frac{11.645 - 15}{1}\right) = P(Z > -3.355).
    • From standard normal tables: P(Z>3.355)0.9996.P(Z > -3.355) \approx 0.9996.

Summary:

  1. The critical region is Xˉ>11.645\bar{X} > 11.645.
  2. The power of the test is approximately 0.9996.

Would you like a step-by-step explanation of any specific part?


Related Questions:

  1. What is the significance of the Neyman-Pearson Lemma in hypothesis testing?
  2. How do you interpret the power of a hypothesis test?
  3. What happens if we change the sample size nn? How does it affect the critical value and power?
  4. Can you explain the relationship between α\alpha and β\beta in hypothesis testing?
  5. How would the results change if the variance were unknown?

Tip:

When conducting hypothesis tests, always sketch the sampling distributions under H0H_0 and HAH_A to visualize the critical regions and probability of errors.

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Math Problem Analysis

Mathematical Concepts

Statistical Hypothesis Testing
Normal Distribution
Neyman-Pearson Lemma

Formulas

Z = (X̄ - μ) / (σ / √n)
P(Reject H0 | H0) = α
Critical Value = μ0 + zα × σ
Power = P(Reject H0 | HA)

Theorems

Neyman-Pearson Lemma

Suitable Grade Level

University Level (Statistics)