Suppose we have taken independent, random samples of sizes n1 = 7 and n2 = 7 from two normally distributed populations having means µ1 and µ2, and suppose we obtain  x⎯⎯1= 240
  ,  x⎯⎯2=210
  , s1 = 5, s2 = 6. Use critical values to test the null hypothesis H0: µ1 − µ2 < 20 versus the alternative hypothesis Ha: µ1 − µ2 > 20 by setting α equal to .10, .05, .01 and .001. Using the equal variance procedure, how much evidence is there that the difference between µ1 and µ2 exceeds 20? (Round your answer to 3 decimal places.)

 





To test the null hypothesis \( H_0: \mu_1 - \mu_2 \leq 20 \) against the alternative hypothesis \( H_a: \mu_1 - \mu_2 > 20 \), we will use a two-sample t-test with equal variances. The test statistic for this hypothesis test is given by:

\[ t = \frac{(\bar{x}_1 - \bar{x}_2) - (\mu_1 - \mu_2)}{s_p \sqrt{\frac{1}{n_1} + \frac{1}{n_2}}} \]

where \( s_p \) is the pooled standard deviation:

\[ s_p = \sqrt{\frac{(n_1 - 1)s_1^2 + (n_2 - 1)s_2^2}{n_1 + n_2 - 2}} \]

Given:
- \( n_1 = 7 \)
- \( n_2 = 7 \)
- \( \bar{x}_1 = 240 \)
- \( \bar{x}_2 = 210 \)
- \( s_1 = 5 \)
- \( s_2 = 6 \)
- \( \alpha = 0.10, 0.05, 0.01, 0.001 \)
- \( \mu_1 - \mu_2 = 20 \) under the null hypothesis

First, calculate the pooled standard deviation \( s_p \):

\[ s_p = \sqrt{\frac{(7 - 1) \cdot 5^2 + (7 - 1) \cdot 6^2}{7 + 7 - 2}} \]
\[ s_p = \sqrt{\frac{6 \cdot 25 + 6 \cdot 36}{12}} \]
\[ s_p = \sqrt{\frac{150 + 216}{12}} \]
\[ s_p = \sqrt{\frac{366}{12}} \]
\[ s_p = \sqrt{30.5} \]
\[ s_p = 5.528 \]

Next, calculate the test statistic \( t \):

\[ t = \frac{(240 - 210) - 20}{5.528 \sqrt{\frac{1}{7} + \frac{1}{7}}} \]
\[ t = \frac{30 - 20}{5.528 \sqrt{\frac{2}{7}}} \]
\[ t = \frac{10}{5.528 \sqrt{0.2857}} \]
\[ t = \frac{10}{5.528 \cdot 0.5345} \]
\[ t = \frac{10}{2.956} \]
\[ t = 3.383 \]

Now, we need to compare this test statistic to the critical values for different significance levels \( \alpha \). The degrees of freedom for this test is \( n_1 + n_2 - 2 = 7 + 7 - 2 = 12 \).

We will find the critical values for \( \alpha = 0.10, 0.05, 0.01, 0.001 \) from the t-distribution table with 12 degrees of freedom.

- For \( \alpha = 0.10 \), the critical value \( t_{0.10, 12} \approx 1.356 \)
- For \( \alpha = 0.05 \), the critical value \( t_{0.05, 12} \approx 1.782 \)
- For \( \alpha = 0.01 \), the critical value \( t_{0.01, 12} \approx 2.681 \)
- For \( \alpha = 0.001 \), the critical value \( t_{0.001, 12} \approx 3.930 \)

Comparing our test statistic \( t = 3.383 \) with the critical values:

- At \( \alpha = 0.10 \): \( 3.383 > 1.356 \), so we reject \( H_0 \).
- At \( \alpha = 0.05 \): \( 3.383 > 1.782 \), so we reject \( H_0 \).
- At \( \alpha = 0.01 \): \( 3.383 > 2.681 \), so we reject \( H_0 \).
- At \( \alpha = 0.001 \): \( 3.383 < 3.930 \), so we fail to reject \( H_0 \).

**Conclusion:**
There is strong evidence at the \( \alpha = 0.10, 0.05, \) and \( 0.01 \) levels to suggest that the difference between \( \mu_1 \) and \( \mu_2 \) exceeds 20. However, at the \( \alpha = 0.001 \) level, there is not sufficient evidence to reject the null hypothesis.

Would you like further details or have any questions about this procedure?

Here are five questions you might ask next:
1. Can you explain how the pooled standard deviation is calculated in more detail?
2. What are the assumptions underlying the two-sample t-test with equal variances?
3. How would the test change if the sample sizes were different?
4. Can you provide an example with a different set of data to illustrate this test?
5. What is the interpretation of the p-value in hypothesis testing?

**Tip:** When conducting hypothesis tests, always check the assumptions of the test to ensure the validity of the results.

Learn how to conduct a two-sample t-test to compare means of two populations with equal variances. Understand the calculations involving pooled standard deviation and critical values. Suitable for undergraduate level students.

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