# Confidence Intervals
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# Section 4.3 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.
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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
\begin{gather*} A_n = \frac{X_1 + \dotsb + X_n}{n}. \end{gather*}
Then, according to Theorem 4.2.1, \(A_n \approx \Norm\left(\mu, \frac{\sigma^2}{n}\right)\) as long as we take a sufficient number of measurements. However, we don’t have access to the true values of \(\mu, \sigma^2\text{.}\)
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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.
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## Definition 4.3.1.

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:
\begin{gather*} 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}. \end{gather*}
The expression:
\begin{gather*} \sqrt{\frac{s^2}{n}} = \frac{s}{\sqrt{n}} \end{gather*}
is called the standard error of the mean.
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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 \(A_n\text{.}\) Hence "standard error of the mean" is an estimate of the standard deviation of the mean.
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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 \(\alpha \in (0, 1)\text{,}\) and let \(c_{\alpha} = \Phi^{-1}\left(1 - \frac{\alpha}{2}\right)\text{.}\) Then the interval:
\begin{gather*} \left[A_n - c_{\alpha} \left(\frac{s}{\sqrt{n}}\right), A_n + c_{\alpha} \left(\frac{s}{\sqrt{n}}\right)\right] \end{gather*}
is called a confidence interval with confidence level \(1 - \alpha\text{.}\)
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## Theorem 4.3.2.

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:
\begin{align*} \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) \end{align*}
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## Example 4.3.3.

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 Theorem 4.2.1 to apply, the number of measurements would need to be greater.)
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Table 4.3.4. Plant Height Measurements
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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:
\begin{align*} A_5 = \frac{H_1 + \dotsb + H_5}{5} \amp = \frac{9.2 + \dotsb + 10.4}{5} = 9.8. \end{align*}
The sum of the squares is:
\begin{align*} \sum H_i^2 \amp = H_1^2 + \dotsb + H_5^2 = 9.2^2 + \dotsb + 10.4^2 = 481.1 \end{align*}
So the sample variance is:
\begin{gather*} 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 \end{gather*}
Now, the standard error of the mean is:
\begin{gather*} \frac{s}{\sqrt{n}} = \sqrt{\frac{0.225}{5}} \approx 0.212 \end{gather*}
Finally, we can give 95% confidence limits:
\begin{align*} \mu_{\ell} \amp = 9.8 - 1.96(0.212) \approx 9.38 \\ \mu_{h} \amp = 9.8 + 1.96(0.212) \approx 10.22 \end{align*}
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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 (Example 3.2.7) for the variance of \(S_n = X_1 + \dotsc + X_n\text{.}\) We can make use of this formula to get the variance of \(A_n\) without going through the sample variance calculation above:
\begin{align*} \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}. \end{align*}
Since we won’t have access to the true value of \(p\text{,}\) we can use the maximum likelihood estimate \(\widehat{p}\text{.}\)
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## Example 4.3.5.

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 \(\widehat{p} = \frac{40}{100} = 0.4\text{.}\) So the standard error will be:
\begin{gather*} \sqrt{\frac{(0.4)(1 - 0.4)}{100}} = \sqrt{\frac{0.24}{100}} \approx 0.049. \end{gather*}
Therefore, 95% confidence limits for \(p\) are:
\begin{align*} p_{\ell} \amp = 0.4 - 1.96(0.049) \approx 0.304 \\ p_{h} \amp = 0.4 + 1.96(0.049) \approx 0.496 \end{align*}
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## Exercises Exercises

### 1.

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 \(p\text{.}\)
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Answer.
\([0.504, 0.696]\text{.}\)
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### 2.

Suppose in a sample of 100 people, 12 are left-handed.
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#### (a)

Give a 95% confidence interval for the proportion \(p\) of left-handed people.
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Answer.
\([0.056, 0.184]\text{.}\)
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#### (b)

Give a 90% confidence interval.
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Answer.
\([0.066, 0.174]\text{.}\)
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### 3.

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.)
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Table 4.3.6.
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mouse \(i\) | weight \(W_i\) (g)
--- | ---
1 | 26
2 | 32
3 | 33
4 | 20
5 | 29

Answer.
\([23.39, 32.61]\text{.}\)
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### 4.

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.)
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Table 4.3.7.
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plant \(i\) | height \(H_i\) (in)
--- | ---
1 | 15
2 | 14
3 | 18
4 | 21
5 | 17

Answer.
\([14.61, 19.39]\text{.}\)
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