To construct a 98% confidence interval estimate for the difference between the two population means $(\mu_1 - \mu_2)$, we use the formula for a confidence interval for the difference between two means assuming equal variances.

Formula:

$CI = (\overline{X_1} - \overline{X_2}) \pm t_{\alpha/2} \times \sqrt{s_p^2 \left( \frac{1}{n_1} + \frac{1}{n_2} \right)}$

Where:

• $\overline{X_1}$ and $\overline{X_2}$ are the sample means,
• $n_1$ and $n_2$ are the sample sizes,
• $s_1$ and $s_2$ are the sample standard deviations,
• $s_p^2$ is the pooled variance,
• $t_{\alpha/2}$ is the critical value from the t-distribution at the 98% confidence level,
• $df = n_1 + n_2 - 2$ is the degrees of freedom.

Step 1: Calculate the pooled variance $s_p^2$

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

Given:

• $n_1 = 110$, $n_2 = 140$
• $s_1 = 4.8$, $s_2 = 3.4$

$s_p^2 = \frac{(110 - 1)(4.8)^2 + (140 - 1)(3.4)^2}{110 + 140 - 2}$ $s_p^2 = \frac{109 \times 23.04 + 139 \times 11.56}{248}$ $s_p^2 = \frac{2501.76 + 1606.84}{248} = \frac{4108.6}{248} \approx 16.56$

Now, the pooled standard deviation is: $s_p = \sqrt{16.56} \approx 4.07$

Step 2: Find the critical value $t_{\alpha/2}$

For a 98% confidence interval with $df = 110 + 140 - 2 = 248$, using a t-table or calculator, the critical value $t_{\alpha/2}$ at a 98% confidence level and 248 degrees of freedom is approximately $t_{0.01} \approx 2.33$.

Step 3: Compute the margin of error

$ME = t_{\alpha/2} \times s_p \times \sqrt{\frac{1}{n_1} + \frac{1}{n_2}}$ $ME = 2.33 \times 4.07 \times \sqrt{\frac{1}{110} + \frac{1}{140}}$ $ME = 2.33 \times 4.07 \times \sqrt{0.00909 + 0.00714}$ $ME = 2.33 \times 4.07 \times \sqrt{0.01623} \approx 2.33 \times 4.07 \times 0.1273 \approx 1.21$

Step 4: Compute the confidence interval

The difference in sample means is: $\overline{X_1} - \overline{X_2} = 58.4 - 65.4 = -7$

Now, the 98% confidence interval is: $CI = (-7) \pm 1.21$ $CI = [-8.21, -5.79]$

Conclusion:

The 98% confidence interval for the difference between the two population means is: $-8.21 \leq \mu_1 - \mu_2 \leq -5.79$

This suggests that the average call length in Center A is significantly shorter than in Center B at the 98% confidence level.

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Would you like further details or have any questions about this process? Here are five related questions for further exploration:

1. How would the confidence interval change if the sample sizes were smaller?
2. What happens to the confidence interval if the assumption of equal variances is violated?
3. How does increasing the confidence level affect the width of the confidence interval?
4. How would you interpret a confidence interval that contains zero?
5. What is the impact of larger standard deviations on the confidence interval?

Tip: The margin of error increases as the confidence level increases, meaning a wider interval gives more assurance but less precision.