There are two call centers (A and B) having similar number of agents. Similar number of calls are dialed per month from each call center offering products to customers. We need to find out if both call centers are performing at a comparable level. The measure is sales per call over the entire month (termed as Conversion Rate). Sample size (call volume) is big (upwards of 80,000 calls per month from each call center). So

*we can assume a normal distribution instead of t-distribution*(can normal distribution be used for such large number of samples when we don't know population st. dev.)

Using 95% confidence interval here is my understanding of how will we test for significance (using just the critical values method; not the difference of means method):

- Take call center A's (calls, sales) data; calculate mean sales per call (Conversion Rate) and standard deviation of sales per call. Consider this Conversion Rate to be the Null Hypothesis, H0. The Conversion Rate in this case is 16.99 (monetory value in local currency). Hence H0 = 16.99
- Alternate Hypotesis H1 <> 16.99; Conversion Rate being the test statistic.
- Then we determine the sampling distribution of test statistic in case H0 is true. I have doubts here. As we don't have the population values,
*will our sampling distribution be normal-distribution with mean and st. error based on H0's mean and st. error? Should I interpret the deviation as standard error or standard deviation in this case?* - Take call center B's one month worth of (calls, sales) data; calculate mean sales (Conversion Rate) and standard deviation of sales. The Conversion Rate in this case is 17.14 (monetory value in local currency).
- If Conversion Rate (test statistic) from call center B does not fall in the critical region then H0 holds.

Using the following website, I know that H0 holds: https://www.medcalc.org/calc/comparison_of_means.php

I just want to know if my understanding is correct. I am taking help from his great book to build my understanding https://learningstatisticswithr.com/.