A credit card company operates two customer service centers. Callers to the service centers dial a single​ number, and a computer program routes callers to the center having the fewest calls waiting. As part of a customer service review​ program, the credit card center would like to determine whether the average length of a call​ (not including hold​ time) is different for the two centers. The managers of the customer service centers are willing to assume that the populations of interest are normally distributed with equal variances. Suppose a random sample of phone calls to the two centers is selected and the results that were reported are shown below. Complete parts a and b.
Center A
Center B
Sample Size
110110
140140
Sample Mean​ (seconds)
58.458.4
65.465.4
Sample St. Dev.​ (seconds)
4.84.8
3.43.4
Question content area bottom
Part 1
a. Using the sample​ results, develop aa 9898​% confidence interval estimate for the difference between the two population means. Let sample 1 be the sample from Center A and let sample 2 be the sample from Center B.
enter your response hereless than or equals≤left parenthesis mu 1 minus mu 2 right parenthesisμ1−μ2less than or equals≤enter your response here

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.

---

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.

A credit card company operates two customer service centers. Callers to the service centers dial a single number, and a computer program routes callers to the center having the fewest calls waiting. As part of a customer service review program, the credit card center would like to determine whether the average length of a call (not including hold time) is different for the two centers. The managers of the customer service centers are willing to assume that the populations of interest are normally distributed with equal variances. Suppose a random sample of phone calls to the two centers is selected and the results that were reported are shown below:

Center A
Sample Size: 110
Sample Mean: 58.4 seconds
Sample St. Dev.: 4.8 seconds

Center B
Sample Size: 140
Sample Mean: 65.4 seconds
Sample St. Dev.: 3.4 seconds

Part 1: Using the sample results, develop a 98% confidence interval estimate for the difference between the two population means. Let sample 1 be the sample from Center A and sample 2 be from Center B.

Learn how to calculate a 98% confidence interval to compare the average call lengths between two customer service centers. This problem involves statistical concepts like pooled variance, t-distribution, and inferential statistics. A sample from Center A (mean: 58.4 sec, n = 110) and Center B (mean: 65.4 sec, n = 140) is analyzed with detailed steps.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation