How to compute sample size n, so that the SE = 4 points from the estimated mean with CI = 95%?

I

Ieva

New Member
#1
#1
I need to calculate sample size n, so that SE of the estimated mean would be 4 points. Total amount of units: N = 200. There are 55 units in total in the 1st group. N2 = 80, N3 = 65. The sample strata sizes are approximately proportional to their proportion in population, i.e., n1/N1 = n2/N2 = n3/N3 = 1/4.
Here is the data:

1st group - 80 ,68, 72, 85, 90, 62, 61, 92,
85, 87, 91, 81, 79, 83,

2nd group - 85, 82, 48, 75, 53, 73, 65,
78, 49, 69, 72, 81, 53, 59,
68, 52, 71, 61, 59, 42,

3rd group - 42, 32 ,36 ,31 ,65 ,29 ,43,
19, 53, 14, 61, 31, 42, 30,
39, 32.

Calculated mean for the stratified sample: y_strata= 59. 989. The st. dev. for the first group sd(1st) = 10.254, sd(2nd ) = 12.578, and sd(3rd) = 14.643. I need to calculate the sample size so that the standard error of the mean would equal 4 points. I need to use proportional allocation. The confidence level is 95%.

How to do this?
 
O

obh

Well-Known Member
#2
#2
I don't exactly understand your data, but the principle is very simple.

What formula are you going to use to calculate the confidence interval?

For example, if you use the Normal approximation (better use the Wilson score)

p̂±z*sqr( p̂(1-p̂)/n)

MOE=z*sqr( p̂(1-p̂)/n)

SE, standard error, is the standard deviation of the sampling distribution of the statistic, in this case, p̂
 
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I

Ieva

New Member
#3
#3
obh said:
I don't exactly understand your data, but the principle is very simple.

What formula are you going to use to calculate the confidence interval?

For example, if you use the Normal approximation (better use the Wilson score)

p±z*sqr( p̂(1-p̂)/n)

MOE=z*sqr( p̂(1-p̂)/n)

SE, standard error, is the standard deviation of the sampling distribution of the statistic, in this case, p̂
Click to expand...



So, in this case the p value will be the st. deviation for the sample size that is chosen know? This is stratified sample, and it was said, that I need to use proportional allocation to calculate the sample size.
 
I

Ieva

New Member
#7
#7
Your formula: p±z*sqr( p̂(1-p̂)/n). If p is estimated proportion, then I assume p is the real propotion, but how do I obtain it?
 
I

Ieva

New Member
#9
#9
Ok. so, in this case, where the sample is stratified, I need to approximate the needed sample for each strata and then just sum it. For example, if n1/N1 = 14/55 = 0.255. Then n_hat = 0.255 +/- 1.96 * 4 = 0.255 +/- 7.84 = (7.585; 8.095). Does it make sense. to sum all the lower bounds and upper bound to obtain the sample size interval? Also, am I right, to equal SE = sqr( p̂(1-p̂)/n)?
 
I

Ieva

New Member
#11
#11
obh said:
SE is sqr( p̂(1-p̂)/n) :)
Just to ensure, can you please tell me what is n1,N1, n2, N2, m3 , N3?


What is 80? what is 68 ?
Click to expand...
n1 - 1st strata (group) sample size, n2 - 2nd strata sample size, n3 = 3rd strata sample size. N1 - 1st strata population size, N2 - 2nd strata population size, N3 - 3rd strata population size.

So the initial problem: The school wants to determine average mark for 6th grade students. The total population N = 200. The whole of the population ( 6th grade students) is divided into three groups: I, II, III. I needed to estimate the average mark for a 6th grade student. I computed and it's equal to y_mean = 59. 989. ( I used the formula y_mean = 1/N * y_total_strata). Where y_total_strata is estimate for the total scores in the population. y_total_strata = SUM(Nh*y_strata_h), where y_strata_h is the mean of each strata in the sample and Nh is total population size for each strata. (h - strata).
I also computed the st. error for n = 50, SE = 2.299. Now, I need to find the sample size n, so that the SE would be equal to 4 (points ).
 
O

obh

Well-Known Member
#12
#12
Hi Leva,

In this case, it is a mean confidence interval (not a proportion)

What is the difference between N1 and N2? what do you mean by different strata? different student characters?
What is the difference between the 3 groups?
Why do you want to have SE=4? in my opinion, for a confidence interval, it is more tangible to define the required MOE? ( CI is: Average ± MOE)
Mean CI Formula
 
I

Ieva

New Member
#13
#13
No, it is not mean confidence interval. I need to find sample size n, so that standard error would be 4 points within the estimated mean with confidence level 95%. The problem doesn't state the differences between groups 1,2 and 3.
 
O

obh

Well-Known Member
#14
#14
Hi Leva,

Is it an assignment, or a real problem?
The confidence level is related to the confidence interval. the standard deviation of the mean (SE) is not related to the confidence level.

The SE, standard deviation of the average is S/√n
 
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