Stats refresher:

I have two random variables (x1 and x2) with known mean and variances. I need to calculate the expected value and confidence intervals for the sum of a and b trials respectively. Assume both are normally distributed and independent.

For example, x1 mean = 15, var = 5, x2 mean = 10 var =3

What is the confidence interval for sum(7 trials of x1, 6 trials of x2)? I can calc the expected value easily.

In words, I know that I sell on average 15 boxes of product x1 with variance 5 and 10 boxes of product x2 with variance 3 per day. I have 7 days of selling product x1 and 6 days of product x2 in the month. What is the confidence interval of my expected total sales for the remainder of the month?

I could brute force it with a monte carlo, but would prefer to just calc it...

  • $\begingroup$ This is just really confusing. Is your measurement looking at the number of days you will sell boxes or the total number of boxes that you sold? When you say you had 7 days of selling $X_1$, does that mean your actual measurement is $7X_1$? $\endgroup$ – Daniel Mar 8 '18 at 16:09
  • $\begingroup$ Correct - the question is how many boxes will I sell over the remainder of the month and within what confidence interval. There are 7 selling days for product 1 and 6 for product 2. I know my average sales per day and the variance of that average, so the expected value is easy - having a harder time with the confidence interval. $\endgroup$ – flyingmeatball Mar 8 '18 at 16:20
  • $\begingroup$ If by "known mean" you mean "known expectation", than looking for a confidence interval for the mean is meaningless (pun intended...). You can calculate the exact value of the expectation (as you said) and no confidence interval is needed. $\endgroup$ – Zahava Kor Mar 8 '18 at 20:24
  • $\begingroup$ ...But I do need to know the confidence interval, that's the whole point. I have historical mean shipments per day and historical standard deviation for shipments per day by product. I want to know a range around the expected total shipments for the remaining shipping days in the month. $\endgroup$ – flyingmeatball Mar 8 '18 at 21:00
  • $\begingroup$ You are not phrasing the problem correctly - you cannot have a confidence interval around an expected value, only around an average. Unfortunately, the confidence interval is one of the most misunderstood concepts in statistics. If you insist on calculating an interval, than the variance of the total sales will be 7xVar1 + 6xVar2 = 7x5+6x3=53 and the standard deviation will be sqrt(53)=7.28. You now take the average of total sales (165) ± Zx7.28 where Z is determined by your confidence level. Again, this does not make much sense because 165 is already the expected value. $\endgroup$ – Zahava Kor Mar 9 '18 at 2:28

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