Question:
Let a random sample of size 100 from a normal population with unknown mean \( \mu \) and variance 9 give the sample mean 5.608. Let \( \Phi(\cdot) \) denote the distribution function of the standard normal random variable. If \( \Phi(1.96) = 0.975 \), \( \Phi(1.64) = 0.95 \), and the uniformly most powerful unbiased test based on sample mean is used to test \[ H_0: \mu = 5.02 \quad \text{against} \quad H_1: \mu \neq 5.02, \] then the p-value equals
Let a random sample of size 100 from a normal population with unknown mean \( \mu \) and variance 9 give the sample mean 5.608. Let \( \Phi(\cdot) \) denote the distribution function of the standard normal random variable. If \( \Phi(1.96) = 0.975 \), \( \Phi(1.64) = 0.95 \), and the uniformly most powerful unbiased test based on sample mean is used to test \[ H_0: \mu = 5.02 \quad \text{against} \quad H_1: \mu \neq 5.02, \] then the p-value equals
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For hypothesis testing using sample mean, calculate the test statistic and compare it with critical values or use the distribution to find the p-value.
Updated On: Dec 29, 2025
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Correct Answer: 0.045
Solution and Explanation
The p-value is calculated using the test statistic based on the sample mean and the standard normal distribution. Given the sample mean and the known variance, the test statistic is computed and the p-value is found to be:
\[
\text{p-value} \approx 0.045.
\]
Thus, the value is \( 0.045 \).
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