Confidence Intervals
When estimating the value of some parameter, the maximum likelihood estimation provides the single estimate with highest possible likelihood.
But we might prefer to have a range of values, with some quantitative measure of how certain we are that the true parameter value could be in the range.
Suppose some measurement has a mean \mu and variance \sigma^2 across the entire population.
Independent measurements X_1, X_2, \dotsc, X_n and calculate the sample mean
A_n = \frac{X_1 + \dotsb + X_n}{n}.
Then, according to , A_n \approx \Norm\left(\mu, \frac{\sigma^2}{n}\right) as long as we take a sufficient number of measurements.
However, we don't have access to the true values of \mu, \sigma^2.
We can make an estimate s^2 of \sigma^2 from the collected data, so we know how spread out the normal distribution should be.
Let X_1, X_2, \dotsc, X_n be independent samples taken from a population.
Let A_n = \frac{X_1 + \dotsb + X_n}{n} be the sample mean.
The sample variance is:
s^2 = \frac{\sum \left(X_i - A_n\right)^2}{n - 1} = \frac{\left(\sum X_i^2\right) - n \left(A_n\right)^2}{n - 1}.
The expression:
\sqrt{\frac{s^2}{n}} = \frac{s}{\sqrt{n}}
is called the standard error of the mean.
Note that s^2 is an estimate of the variance of each individual measurement X_i taken from the population.
Therefore, \frac{s^2}{n} is an estimate of the variance of the mean A_n.
Hence "standard error of the mean" is an estimate of the standard deviation of the mean.
Next, we want to pick a distance c so that our measured sample mean A_n is within c of the true population mean \mu with a high probability.
We can pick some value \alpha \in (0, 1), and let c_{\alpha} = \Phi^{-1}\left(1 - \frac{\alpha}{2}\right).
Then the interval:
\left[A_n - c_{\alpha} \left(\frac{s}{\sqrt{n}}\right), A_n + c_{\alpha} \left(\frac{s}{\sqrt{n}}\right)\right]
is called a confidence interval with confidence level 1 - \alpha.
Let X_1, X_2, \dotsc, X_n be independent samples taken from a population.
Suppose A_n is the sample mean and s^2 the sample variance.
Then the 95% confidence limits are:
\mu_{\ell} \amp = A_n - 1.96 \left(\frac{s}{\sqrt{n}}\right)
\mu_{h} \amp = A_n + 1.96 \left(\frac{s}{\sqrt{n}}\right)
Suppose measurements are taken of the heights of several plants in a garden plot.
The measurements are given below.
(Note: for the sake of a managable calculation, only five measurements are given.
However, for to apply, the number of measurements would need to be greater.)
Plant Height Measurements
| Plant i |
Height H_i (in) |
| 1 |
9.2 |
| 2 |
9.8 |
| 3 |
10.1 |
| 4 |
9.5 |
| 5 |
10.4 |
Then the sample mean is:
A_5 = \frac{H_1 + \dotsb + H_5}{5} \amp = \frac{9.2 + \dotsb + 10.4}{5} = 9.8.
The sum of the squares is:
\sum H_i^2 \amp = H_1^2 + \dotsb + H_5^2 = 9.2^2 + \dotsb + 10.4^2 = 481.1
So the sample variance is:
s^2 = \frac{\left(\sum H_i\right)^2 - 5 (A_5)^2}{5 - 1} = \frac{481.1 - 5(9.8)^2}{4} \approx 0.225
Now, the standard error of the mean is:
\frac{s}{\sqrt{n}} = \sqrt{\frac{0.225}{5}} \approx 0.212
Finally, we can give 95% confidence limits:
\mu_{\ell} \amp = 9.8 - 1.96(0.212) \approx 9.38
\mu_{h} \amp = 9.8 + 1.96(0.212) \approx 10.22
If the individual measurements X_1, \dotsc, X_n are indicator random variables, then the mean A_n represents the proportion which take on the value 1.
For example, if the X_i's indicate heads on coin flips, then A_n represents the proportion which come up heads.
In this setting, we have a nice shortcut formula () for the variance of S_n = X_1 + \dotsc + X_n.
We can make use of this formula to get the variance of A_n without going through the sample variance calculation above:
\Var(S_n) \amp = np(1 - p)
\Var(A_n) \amp = \Var\left(\frac{S_n}{n}\right)
\amp = \frac{1}{n^2} \Var\left(S_n\right)
\amp = \frac{1}{n^2} (np(1 - p))
\amp = \frac{p(1 - p)}{n}.
Since we won't have access to the true value of p, we can use the maximum likelihood estimate \widehat{p}.
Suppose in a sample of 100 people, 40 have a particular trait.
The maximum likelihood estimate for the probability p of a random person having the trait is \widehat{p} = \frac{40}{100} = 0.4.
So the standard error will be:
\sqrt{\frac{(0.4)(1 - 0.4)}{100}} = \sqrt{\frac{0.24}{100}} \approx 0.049.
Therefore, 95% confidence limits for p are:
p_{\ell} \amp = 0.4 - 1.96(0.049) \approx 0.304
p_{h} \amp = 0.4 + 1.96(0.049) \approx 0.496
Suppose we flip a coin 100 times and count 60 heads.
Let p be the (unknown) probability that the coin comes up heads on a flip.
Give an approximate 95% confidence interval for the value of p.
[0.504, 0.696].
Suppose in a sample of 100 people, 12 are left-handed.
Give a 95% confidence interval for the proportion p of left-handed people.
[0.056, 0.184].
Give a 90% confidence interval.
[0.066, 0.174].
The weights of five mice are measured and recorded below.
Give a 95% confidence interval for the sample mean weight of mice.
(Pretend 5 measurements is large enough for the CLT to apply.)
| mouse i |
weight W_i (g) |
| 1 |
26 |
| 2 |
32 |
| 3 |
33 |
| 4 |
20 |
| 5 |
29 |
[23.39, 32.61].
The heights of five plants are measured and recorded below.
Give a 95% confidence interval around the sample mean for the heights of the plants.
(Pretend 5 measurements is large enough for the CLT to apply.)
| plant i |
height H_i (in) |
| 1 |
15 |
| 2 |
14 |
| 3 |
18 |
| 4 |
21 |
| 5 |
17 |
[14.61, 19.39].