Title: The accuracy of sample estimates from populations
Course: UGQuants (SSPS) > Section: Probability

The accuracy of sample estimates from populations

We are now ready for the last link in this long logical chain of reasoning.

Imagine that we don’t have an infinite number of repeated samples but only one random sample. Imagine too that we do not know the population parameter that we are trying to estimate.  This should be pretty easy to imagine, as it is precisely the situation researchers are usually in!

Using the properties of the normal distribution – since the estimate for a population parameter obtained via repeated random samples would be normally distributed - we can now calculate exactly what the probability is that a finding from our single random sample lies within a certain distance of the true-but-unknown value for the population.

We know, by virtue of the properties of the normal curve, that our finding has a 68% chance of lying within one standard error above or below the true population mean and about a 95% chance of lying within two, as illustrated earlier. The actual size of this standard error will depend on the size of our random sample (the greater it is the lower the error) and the amount of variation the values the variable we are interested in take (the more spread out the values the greater the error).

Usually, by convention, we work with a z value of 1.96. This is chosen since it represents the critical region where values are more than 1.96 standard deviations or standard errors away from the mean equal to 0.025 + 0.025 = 0.05 or 5%. We can therefore know that for any one sample, our result (the mean value in the sample) will have a 95% chance of lying within 1.96 standard errors of the population parameter (the mean value in the population).

Confidence intervals

Let us return to the earlier example of mean female life expectancy in our population comprising the 214 cases from the WDI 2011 data. Imagine we obtained a result of 71.0 in a random sample of 50 cases. We would know that we had a 95% chance that the true population value for female life expectancy will be 71 ± two standard errors. We can estimate the standard deviation of the population by using the standard deviation from our sample – which is 10.5.

Thus the calculation becomes:

sample mean ± z value * (population standard deviation / square root of sample size)

71.0 ± 1.96 * (10.5 /7.1) [7.1 is the square root of 50, the size of our sample]

71.0 ± 1.96 * 1.48

71.0 ±  2.9 years

With a sample of only 50, we can be 95% sure that the true population mean female life expectancy is between 68.1 and 73.9 years. Say we increased our sample size to 10,000, the estimate for the mean that we obtain is 72.1, and the standard deviation is again 10.5. The calculation would become:

72.1 ± 1.96 * (10.5 / 100)

72.1 ± 1.96 * 0.105

72.1 ± 0.21

= 71.83 to 72.31 years

Note that we are 95% confident, which is not the same as 100% confident, or being ‘certain’. This is because we can never know when the actual random sample we had drawn was one of the 5% of possible samples lying more than 1.96 standard deviations from the mean.

Confidence intervals for a proportion

So far we have assumed that the population parameter we are interested in is an interval level variable such as age or height: one, that is, that can take any of a continuous range of values. However, we are often interested in nominal and ordinal level data where our characteristic of interest takes the form of the distribution of cases or observations across a limited range of values. Here, therefore, we are interested in the proportion of observations in our population taking a particular value. For example we might want to know the proportion of women in a population, the proportion who have a degree, or the proportion who smoke. We therefore need a way to calculate the standard error of a distribution of sample proportions rather than means.

This formula is very similar to that for means, except that instead of calculating the standard deviation of the mean of the variable to measure its spread, we multiply the proportion by (1 - the proportion) and take the square root of the result. As in the case of the formula for means, we divide the result by the square root of the sample size: standard error of the proportion = square root of (proportion * (1-proportion) / square root of sample size  SE(p) = √(p*(1-p)) / √n

We can then use this standard error in exactly the same way as before to calculate confidence intervals for the proportion using the appropriate z score for the degree of confidence we wish to obtain.

Some examples.

In a random sample of 100 adults, 30 had taken a foreign holiday in the previous year. What is the 95% confidence interval for the adult population?

= √0.3 * (1-0.3) / √100

= √ 0.21 / 10

= 0.46/ 10

= 0.046

For a 95% level of confidence we can take all results within 1.96 standard errors so our interval will be

0.3 - (1.96 *0.046) to 0.3 + (1.96 *0.046)

= 0.3 - 0.089 to 0.3 + 0.089

= 0.211 to 0.389

So we can be 95% sure that between 21 and 39 adults in the population from which the random sample was drawn will have taken a holiday.

Example 2

In a random sample of 100 adults, 6 were found to have a higher educational degree. What is the 95% confidence interval for the adult population?

= √(.06*.94) / 10

= √0.0564 /10

= 0.237/10

= 0.0237

Our Confidence Interval will be:

0.06 - (1.96 * 0.0237) to 0.06 + (1.96 * 0.0237)

= 0.06 - 0.047 to 0.06 + 0.047

= 0.013 to 0.107

So we can be 95% sure that between 1 and 11 adults in the population from which the random sample was drawn will have a higher degree.