Part 1A 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.[table]
| Center A | Center B
Sample Size | 115115 | 140140 Sample Mean (seconds) | 57.657.6 | 66.766.7 Sample St. Dev. (seconds) | 6.46.4 | 6.46.4
[/table] Part 1a. Using the sample results, develop an 8080% 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.[input]enter your response here less than or equals≤left parenthesis mu 1 minus mu 2 right parenthesisμ1−μ2less than or equals≤[input]enter your response here (Round to two decimal places as needed.)

Question

Part 1A 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.[table]
| Center A | Center B
Sample Size | 115115 | 140140 Sample Mean​ (seconds) | 57.657.6 | 66.766.7 Sample St. Dev.​ (seconds) | 6.46.4 | 6.46.4
[/table] Part 1a. Using the sample​ results, develop an 8080​% 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.[input]enter your response here less than or equals≤left parenthesis mu 1 minus mu 2 right parenthesisμ1−μ2less than or equals≤[input]enter your response here ​(Round to two decimal places as​ needed.)

BlackTom AI · Accepted Answer

Because the two populations are assumed normal with equal variances, a pooled‑variance t confidence interval is appropriate for μ1 − μ2.

1) Relevant concepts
- When population variances are assumed equal, use the pooled sample standard deviation s_p and the t distribution with df = n1 + n2 − 2.
- 80% confidence means α = 0.20 and the two‑tailed critical value is t_{α/2, df} = t_{0.10, 253}.

2) Given data and preliminary computations
- n1 = 115, x̄1 = 57.6, s1 = 6.4
- n2 = 140, x̄2 = 66.7, s2 = 6.4
- Sample difference: x̄1 − x̄2 = 57.6 − 66.7 = −9.1
- Degrees of freedom: df = 115 + 140 − 2 = 253

3) Pooled standard deviation
Because s1 = s2, s_p^2 = ((n1−1)s1^2 + (n2−1)s2^2)/(n1+n2−2) = (114·6.4^2 + 139·6.4^2)/253 = 6.4^2 = 40.96, so s_p = 6.4.

4) Standard error of the difference
SE = s_p * sqrt(1/n1 + 1/n2)
= 6.4 * sqrt(1/115 + 1/140)
= 6.4 * sqrt(0.00869565 + 0.00714286)
= 6.4 * sqrt(0.01583851)
= 6.4 * 0.125899 ≈ 0.80575 ≈ 0.806 (intermediate values shown).

5) Critical t and margin of error
For df = 253 and 80% CI, t_{0.10,253} ≈ 1.282.
Margin = t * SE ≈ 1.282 * 0.80575 ≈ 1.0339 ≈ 1.03.

6) Confidence interval
(x̄1 − x̄2) ± margin = −9.1 ± 1.0339
Lower = −9.1 − 1.0339 = −10.1339 → −10.13 (rounded)
Upper = −9.1 + 1.0339 = −8.0661 → −8.07 (rounded)

Why this is correct and other approaches
- A z‑interval would use the normal critical value (≈1.282) which is nearly identical here because df is large; however the t procedure is theoretically correct given unknown σ and finite samples.
- Using an unpooled (Welch) interval would require a different df and produce a very similar but not identical interval; pooling is justified because s1 = s2 and the problem statement assumes equal variances.

Conclusion
The 80% confidence interval for μ1 − μ2 is (−10.13, −8.07), which matches the provided answer.

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