Give a 99.9% confidence interval, for  
μ
1
−
μ
2
  given the following information.

n
1
=
25
 ,  
¯
x
1
=
2.65
 ,  
s
1
=
1
 
n
2
=
20
 ,  
¯
x
2
=
3.11
 ,  
s
2
=
0.71

To calculate the \( 99.9\% \) confidence interval for the difference in means, \( \mu_1 - \mu_2 \), we’ll use the following formula for a confidence interval:

\[
(\bar{x}_1 - \bar{x}_2) \pm t_{\alpha/2} \times \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}
\]

where:
- \( \bar{x}_1 = 2.65 \), \( s_1 = 1 \), and \( n_1 = 25 \)
- \( \bar{x}_2 = 3.11 \), \( s_2 = 0.71 \), and \( n_2 = 20 \)
- \( t_{\alpha/2} \) is the critical t-value for a \( 99.9\% \) confidence level, with \( df = \min(n_1 - 1, n_2 - 1) \).

### Step 1: Calculate \( \bar{x}_1 - \bar{x}_2 \)
\[
\bar{x}_1 - \bar{x}_2 = 2.65 - 3.11 = -0.46
\]

### Step 2: Determine the critical \( t \)-value
Since we’re using a 99.9% confidence level, the significance level \( \alpha = 0.001 \), so \( t_{\alpha/2} = t_{0.0005} \). With \( df = \min(24, 19) = 19 \), we’ll look up this \( t \)-value in a t-table or calculate it. The approximate value for \( t_{0.0005} \) with \( df = 19 \) is:

\[
t_{0.0005} \approx 3.883
\]

### Step 3: Calculate the standard error
\[
SE = \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}} = \sqrt{\frac{1^2}{25} + \frac{0.71^2}{20}}
\]
\[
SE = \sqrt{0.04 + 0.0252} = \sqrt{0.0652} \approx 0.2553
\]

### Step 4: Compute the margin of error
\[
\text{Margin of Error} = t_{\alpha/2} \times SE = 3.883 \times 0.2553 \approx 0.9916
\]

### Step 5: Construct the confidence interval
\[
(\bar{x}_1 - \bar{x}_2) \pm \text{Margin of Error} = -0.46 \pm 0.9916
\]

So, the \( 99.9\% \) confidence interval for \( \mu_1 - \mu_2 \) is approximately:
\[
(-1.4516, 0.5316)
\]

This interval suggests that the true difference in means, \( \mu_1 - \mu_2 \), lies between \(-1.4516\) and \(0.5316\) with 99.9% confidence.

Would you like further details on any of the steps? 

---

### Related Questions
1. How would this change if the confidence level were 95% instead of 99.9%?
2. What is the interpretation of the confidence interval in this context?
3. How would the interval change with different sample sizes?
4. Why do we use the t-distribution instead of the normal distribution here?
5. How does the variability in sample standard deviations affect the interval width?

### Tip
When interpreting confidence intervals, remember they indicate a range of values where the true parameter is likely to be found, not a probability of where a parameter lies in repeated samples.

Give a 99.9% confidence interval, for μ₁ − μ₂ given the following information.

n₁ = 25, x̄₁ = 2.65, s₁ = 1
n₂ = 20, x̄₂ = 3.11, s₂ = 0.71

This solution provides a detailed step-by-step method to calculate the 99.9% confidence interval for the difference in two means. Using sample sizes (n₁ = 25, n₂ = 20), means (x̄₁ = 2.65, x̄₂ = 3.11), and standard deviations (s₁ = 1, s₂ = 0.71), the final interval is approximately (-1.4516, 0.5316). Learn how t-values and standard error influence the calculation.

Math Problem Statement

n 1

25 ,
¯ x 1

2.65 ,
s 1

n 2

20 ,
¯ x 2

3.11 ,
s 2

Solution

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