Two independent samples from populations that are normally distributed produced the following statistics: for sample 1 the sample size was 19, the sample mean was 34.2 and the sample standard deviation was 12.6. For sample 2, the sample size was 23, the sample mean was 49.1 and the sample standard deviation was 19.4. Assume that population variances are equal. Construct a 95% confidence interval for the difference between the population means. (μ_1-μ_2)

1 answer

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:

\[ (-25.38, -4.42) \]