To construct a 95% confidence interval for the difference between the population means (\( \mu_1 - \mu_2 \)), we will use the following formula for the confidence interval when the population variances are assumed to be equal:
\[ (\bar{X}_1 - \bar{X}_2) \pm t^* \times SE \]
Where:
- \( \bar{X}_1 \) and \( \bar{X}_2 \) are the sample means.
- \( t^* \) is the t-value for the desired confidence level.
- \( SE \) is the standard error of the difference between the means.
Step 1: Calculate the difference between the sample means
\[ \bar{X}_1 - \bar{X}_2 = 34.2 - 49.1 = -14.9 \]
Step 2: Calculate the pooled standard deviation
Since we assume that the population variances are equal, we need to calculate the pooled standard deviation (\( S_p \)):
\[ S_p = \sqrt{\frac{(n_1 - 1) S_1^2 + (n_2 - 1) S_2^2}{n_1 + n_2 - 2}} \]
Where:
- \( n_1 = 19 \), \( S_1 = 12.6 \)
- \( n_2 = 23 \), \( S_2 = 19.4 \)
Calculate \( S_1^2 \) and \( S_2^2 \):
\[ S_1^2 = (12.6)^2 = 158.76 \] \[ S_2^2 = (19.4)^2 = 376.36 \]
Now substitute into the formula for \( S_p \):
\[ S_p = \sqrt{\frac{(19 - 1)(158.76) + (23 - 1)(376.36)}{19 + 23 - 2}} \] \[ = \sqrt{\frac{(18)(158.76) + (22)(376.36)}{40}} \] \[ = \sqrt{\frac{2857.68 + 8280.92}{40}} \] \[ = \sqrt{\frac{11138.6}{40}} = \sqrt{278.465} \approx 16.69 \]
Step 3: Calculate the standard error (SE)
The standard error of the difference in means is calculated as follows:
\[ SE = S_p \sqrt{\frac{1}{n_1} + \frac{1}{n_2}} \] \[ = 16.69 \sqrt{\frac{1}{19} + \frac{1}{23}} \] \[ = 16.69 \sqrt{0.05263 + 0.04348} = 16.69 \sqrt{0.09611} \approx 16.69 \times 0.310 = 5.18 \]
Step 4: Determine the t-value
To find the appropriate \( t^* \) value, we need the degrees of freedom (\( df \)):
\[ df = n_1 + n_2 - 2 = 19 + 23 - 2 = 40 \]
Using a t-distribution table, we find the two-tailed critical value for \( df = 40 \) at a 95% confidence level:
\[ t^* \approx 2.021 (you can use a t-table or calculator for more accurate values) \]
Step 5: Construct the confidence interval
Now we can construct the confidence interval:
\[ (\bar{X}_1 - \bar{X}_2) \pm t^* \times SE \] \[ = -14.9 \pm 2.021 \times 5.18 \] Calculating the margin of error:
\[ 2.021 \times 5.18 \approx 10.48 \]
Now compute the confidence interval:
\[ -14.9 - 10.48 \quad \text{to} \quad -14.9 + 10.48 \] \[ = -25.38 \quad \text{to} \quad -4.42 \]
Final Result
The 95% confidence interval for the difference between the population means (\( \mu_1 - \mu_2 \)) is approximately:
\[ (-25.38, -4.42) \]