How do I find the new mean and the new variance? (N=14 with a variance of 2464,38 and a mean of 413,2)

[gsamsa00]

First of all I'm sorry if i butcher some terms, English is not my language and statistics is far from it.
Let's say i have a N=14 with a variance of 2464,38 and a mean of 413,2 (i don't have any individual values, just the N; variance and mean).
Now i add another value, let's say 625,4.
How do i calculate the new mean and the new variance for N=15?
For the mean i guess it should be "413*14"=5782 ---- 5782+625=6407 ------ 6407/15= 427 (oldmean*oldN + newvalue) / newN
But what about the variance? I'm confused about it.

 

[Dr.Peterson]

First of all I'm sorry if i butcher some terms, English is not my language and statistics is far from it.
Let's say i have a N=14 with a variance of 2464,38 and a mean of 413,2 (i don't have any individual values, just the N; variance and mean).
Now i add another value, let's say 625,4.
How do i calculate the new mean and the new variance for N=15?
For the mean i guess it should be "413*14"=5782 ---- 5782+625=6407 ------ 6407/15= 427 (oldmean*oldN + newvalue) / newN
But what about the variance? I'm confused about it.

You're right about the mean (except that, since you're given the mean to one decimal place, I would do the same with the answer).

For the standard deviation, I would similarly work backward. In order to help, I need to see what formula(s) you have learned for standard deviation. In particular, are we talking about a sample or a population; and which of several forms have you learned? Each will result in somewhat different work at the detailed level.

Can you also show some attempt? You may be more right than you realize.

 

[gsamsa00]

You're right about the mean (except that, since you're given the mean to one decimal place, I would do the same with the answer).

For the standard deviation, I would similarly work backward. In order to help, I need to see what formula(s) you have learned for standard deviation. In particular, are we talking about a sample or a population; and which of several forms have you learned? Each will result in somewhat different work at the detailed level.

Can you also show some attempt? You may be more right than you realize.

This is the formula I'm working with.
The thing that I tried is to multiply the given variance for the oldN, which should give me the total sum of the xi^2-m^2 and then adding to that the result of (newvalue^2 - newmean^2). I don't know if this is correct, also I don't know if I should use the new or old mean in this.
Also we usually just use the population standard deviation in our exercises.

 

[Dr.Peterson]

This is the formula I'm working with.
The thing that I tried is to multiply the given variance for the oldN, which should give me the total sum of the xi^2-m^2 and then adding to that the result of (newvalue^2 - newmean^2). I don't know if this is correct, also I don't know if I should use the new or old mean in this.
Also we usually just use the population standard deviation in our exercises.

This is the form I was hoping for; there are uglier ones.

But it sounds as if you're thinking that the division by N is the last thing in the formula. I would first add the (old) mean squared to the (given) variance, then multiply by N to get the old sum of squares. Then add the new value squared to that to get the new sum of squares, and recalculate the variance using all new values (including the new N).

Show us your actual work, and we can make sure it's done correctly.

 

[gsamsa00]

This is the form I was hoping for; there are uglier ones.

But it sounds as if you're thinking that the division by N is the last thing in the formula. I would first add the (old) mean squared to the (given) variance, then multiply by N to get the old sum of squares. Then add the new value squared to that to get the new sum of squares, and recalculate the variance using all new values (including the new N).

Show us your actual work, and we can make sure it's done correctly.

So, I had this set of values from my textbook where I already have the sum of squares. I did as you told me so (oldM^2 + given variance)*N(20) and I got the sum of squares (ssO2) exactly as I was given in the example. So I added the new value squared to that and got the new sum of squares ssN2. Then finally I calculated the variance by adding a new value xn of 2000. So I did (ssN2/NewN(21))- new mean squared.
The old variance was 111882469
The new one 107744755

Is everything correct?

 

[Dr.Peterson]

So, I had this set of values from my textbook where I already have the sum of squares. I did as you told me so (oldM^2 + given variance)*N(20) and I got the sum of squares (ssO2) exactly as I was given in the example. So I added the new value squared to that and got the new sum of squares ssN2. Then finally I calculated the variance by adding a new value xn of 2000. So I did (ssN2/NewN(21))- new mean squared.
The old variance was 111882469
The new one 107744755

Is everything correct?

It looks like you did the right thing; did you try calculating the actual new variance and check that it is the same?

 

[gsamsa00]

It looks like you did the right thing; did you try calculating the actual new variance and check that it is the same?

Yes, it gives me the same result. Thank you very much, it was very clear