To construct a 95% confidence interval for the difference between population means \( (\mu_1 - \mu_2) \) for the two independent samples, we follow these steps:
Step 1: Gather the data
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Sample 1:
- \( n_1 = 19 \)
- \( \bar{x}_1 = 34.2 \)
- \( s_1 = 12.6 \)
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Sample 2:
- \( n_2 = 23 \)
- \( \bar{x}_2 = 49.1 \)
- \( s_2 = 19.4 \)
Step 2: Calculate the pooled standard deviation (Sp)
Using the formula for the pooled standard deviation:
\[ s_p = \sqrt{\frac{(n_1 - 1)s_1^2 + (n_2 - 1)s_2^2}{n_1 + n_2 - 2}} \]
Calculating \( s_p \):
\[ s_1^2 = 12.6^2 = 158.76 \] \[ s_2^2 = 19.4^2 = 376.36 \] \[ s_p = \sqrt{\frac{(19 - 1) \cdot 158.76 + (23 - 1) \cdot 376.36}{19 + 23 - 2}} \] \[ = \sqrt{\frac{18 \cdot 158.76 + 22 \cdot 376.36}{40}} \] \[ = \sqrt{\frac{2857.68 + 8280.92}{40}} = \sqrt{\frac{11138.6}{40}} = \sqrt{278.465} \approx 16.69 \]
Step 3: Calculate the standard error (SE)
Using the pooled standard deviation to calculate the standard error for the difference between the means:
\[ SE = s_p \sqrt{\frac{1}{n_1} + \frac{1}{n_2}} \] \[ = 16.69 \sqrt{\frac{1}{19} + \frac{1}{23}} = 16.69 \sqrt{0.05263 + 0.04348} = 16.69 \sqrt{0.09611} \] \[ \approx 16.69 \cdot 0.3101 \approx 5.18 \]
Step 4: Calculate the difference between the means
\[ \bar{x}_1 - \bar{x}_2 = 34.2 - 49.1 = -14.9 \]
Step 5: Determine the critical value
For a 95% confidence level with \( df = n_1 + n_2 - 2 = 19 + 23 - 2 = 40 \), we can find the critical value \( t^* \) using a t-distribution table or calculator. For \( df = 40 \), \( t^* \approx 2.021 \).
Step 6: Compute the confidence interval
The confidence interval for \( (\mu_1 - \mu_2) \) is given by:
\[ (\bar{x}_1 - \bar{x}_2) \pm t^* \cdot SE \]
Substituting the values we have:
\[ -14.9 \pm 2.021 \cdot 5.18 \] \[ -14.9 \pm 10.47 \]
This results in:
\[ [-14.9 - 10.47, -14.9 + 10.47] = [-25.37, -4.43] \]
Final Result
The 95% confidence interval for the difference between the population means \( (\mu_1 - \mu_2) \) is approximately:
\[ \boxed{[-25.37, -4.43]} \]