Suppose we have taken independent, random samples of sizes n = 6 and na = 8 from two normally distributed populations having means uj and µ2, and suppose we obtain I = 231 I, = 199 , s1 = 6, s2 = 4. Use critical values to test the null hypothesis H0: H1 - P2 < 22 versus the alternative hypothesis H2: Hı - 42 > 22 by setting a equal to .10, .05, .01 and .001. Using the equal variance procedure, how much evidence is there that the difference between H, and pz exceeds 22? (Round your answer to 3 decimal places.) 4.053 Reject H0 at a = 0.1, 0.05, and, 0.01, very strong evidence.
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![**Hypothesis Testing with Independent Samples**
**Problem Statement:**
Suppose we have taken independent, random samples of sizes \( n_1 = 6 \) and \( n_2 = 8 \) from two normally distributed populations having means \( \mu_1 \) and \( \mu_2 \), and suppose we obtain:
\[ \bar{x_1} = 231 \]
\[ \bar{x_2} = 199 \]
with sample standard deviations:
\[ s_1 = 6 \]
\[ s_2 = 4 \]
We aim to use critical values to test the null hypothesis \( H_0: \mu_1 - \mu_2 \leq 22 \) versus the alternative hypothesis \( H_a: \mu_1 - \mu_2 > 22 \) by setting \( \alpha \) equal to 0.10, 0.05, 0.01, and 0.001. Using the equal variance procedure, how much evidence is there that the difference between \( \mu_1 \) and \( \mu_2 \) exceeds 22? *(Round your answer to 3 decimal places.)*
**Procedure and Results:**
We will compute the test statistic \( t \) based on the given data.
The computed test statistic is:
\[ t = 4.053 \]
**Decision Rule:**
The decision rule involves comparing the test statistic to the critical values for the chosen significance levels (\( \alpha \)). Based on the provided information, our test results are summarized as follows:
| Test Statistic (t) | Conclusion | Significance Levels | Strength of Evidence |
|-------------------|------------|--------------------|---------------------|
| 4.053 | Reject \( H_0 \) for \( \alpha = 0.1, 0.05, 0.01, \) and \( 0.001 \) | 0.1, 0.05, 0.01, and 0.001 | Very strong evidence |
**Explanation:**
- The test statistic calculated (4.053) is significantly higher than the critical values for all given significance levels (\( \alpha = 0.1, 0.05, 0.01, \) and \( 0.001 \)).
- We reject the null hypothesis \( H_0 \) in favor of the alternative hypothesis \(](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fbb793db6-60a8-4c07-89d5-7b9abbee0f1b%2Fdcb1cad0-af09-42da-a963-3230ecdbdb5a%2Fqhidxx_processed.png&w=3840&q=75)
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