The sampling error or sample error In statistics, it is the difference between the mean value of a sample and the mean value of the total population. To illustrate the idea, let's imagine that the total population of a city is one million people, of which you want its average shoe size, for which a random sample of one thousand people is taken.
The average size that emerges from the sample will not necessarily coincide with that of the total population, although if the sample is not biased the value must be close. This difference between the mean value of the sample and that of the total population is the sampling error.
In general, the mean value of the total population is unknown, but there are techniques to reduce this error and formulas to estimate the sampling error margin that will be exposed in this article.
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Let's say that you want to know the mean value of a certain measurable characteristic x in a population of size N, but how N is a large number, it is not feasible to carry out the study on the total population, then a aleatory sample of size n<
The mean value of the sample is denoted by
Suppose they take m samples from the total population N, all the same size n with mean values
These mean values will not be identical to each other and will all be around the population mean value μ. The sampling error margin E indicates the expected separation of the mean values
The standard margin of error ε size sample n it is:
ε = σ / √n
where σ is the standard deviation (the square root of the variance), which is calculated using the following formula:
σ = √ [(x -
The meaning of standard margin of error ε is the following:
The middle value
In the previous section the formula was given to find the error range standard of a sample of size n, where the standard word indicates that it is a margin of error with 68% confidence.
This indicates that if many samples of the same size were taken n, 68% of them will give average values
There is a simple rule, called the rule 68-95-99.7 which allows us to find the margin of sampling error E for confidence levels of 68%, 95% Y 99.7% easily, since this margin is 1⋅ε, 2⋅ε and 3⋅ε respectively.
If he confidence level γ is not any of the above, then the sampling error is the standard deviation σ multiplied by the factor Zγ, which is obtained through the following procedure:
1.- First the significance level α which is calculated from confidence level γ using the following relationship: α = 1 - γ
2.- Then you have to calculate the value 1 - α / 2 = (1 + γ) / 2, which corresponds to the accumulated normal frequency between -∞ and Zγ, in a standardized normal or Gaussian distribution F (z), whose definition can be seen in figure 2.
3.- The equation is solved F (Zγ) = 1 - α / 2 by means of the tables of the normal distribution (cumulative) F, or by means of a computer application that has the inverse standardized Gaussian function F-1.
In the latter case we have:
Zγ = G-1(1 - α / 2).
4.- Finally, this formula is applied for the sampling error with a reliability level γ:
E = Zγ⋅(σ / √n)
Calculate the standard margin of error in the average weight of a sample of 100 newborns. The calculation of the average weight was
The standard margin of error it is ε = σ / √n = (1,500 kg) / √100 = 0.15 kg. Which means that with these data it can be inferred that the weight of 68% of newborns is between 2,950 kg and 3.25 kg.
Determine the margin of sampling error E and the weight range of 100 newborns with a 95% confidence level if the mean weight is 3,100 kg with standard deviation σ = 1,500 kg.
If the rule 68; 95; 99.7 → 1⋅ε; 2⋅ε; 3⋅ε, you have:
E = 2⋅ε = 2⋅0.15 kg = 0.30 kg
That is, 95% of newborns will have weights between 2,800 kg and 3,400 kg.
Determine the range of weights of the newborns from Example 1 with a confidence margin of 99.7%.
The sampling error with 99.7% confidence is 3 σ / √n, which for our example is E = 3 * 0.15 kg = 0.45 kg. From this it is inferred that 99.7% of newborns will have weights between 2,650 kg and 3,550 kg.
Determine the factor Zγ for a reliability level of 75%. Determine the margin of sampling error with this level of reliability for the case presented in example 1.
The confidence level it is γ = 75% = 0.75 which is related to the significance level α through relationship γ= (1 - α), so that the significance level is α = 1 - 0.75 = 0.25.
This means that the cumulative normal probability between -∞ and Zγ it is:
P (Z ≤ Zγ ) = 1 - 0.125 = 0.875
What corresponds to a value Zγ 1.1503, as shown in Figure 3.
That is, the sampling error is E = Zγ⋅(σ / √n)= 1.15⋅(σ / √n).
When applied to the data from example 1, it gives an error of:
E = 1.15 * 0.15 kg = 0.17 kg
With a confidence level of 75%.
What is the confidence level if Zα / 2 = 2.4 ?
P (Z ≤ Zα / 2 ) = 1 - α / 2
P (Z ≤ 2.4) = 1 - α / 2 = 0.9918 → α / 2 = 1 - 0.9918 = 0.0082 → α = 0.0164
The significance level is:
α = 0.0164 = 1.64%
And finally, the confidence level remains:
1- α = 1 - 0.0164 = 100% - 1.64% = 98.36%
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