Section 4.3 Confidence Intervals
-
+
+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 TheoremΒ 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 donβt have access to the true values of \(\mu, \sigma^2\text{.}\)
+
+
+
+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.
+
+
+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.
+
+
+ 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.
+
+
+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 \(1 - \alpha\text{.}\)
+
+
+
+
+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*}
+
+
+
+
+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 TheoremΒ 4.2.1 to apply, the number of measurements would need to be greater.)
+ Table 4.3.4. 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:
+
+\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*}
+
+
+
+
+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 \(X_i\)βs indicate heads on coin flips, then \(A_n\) represents the proportion which come up heads. In this setting, we have a nice shortcut formula (ExampleΒ 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 wonβt have access to the true value of \(p\text{,}\) we can use the maximum likelihood estimate \(\widehat{p}\text{.}\)
+
+
+
+
+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*}
+
+
+
+
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 \(p\text{.}\)
+
2.
Suppose in a sample of 100 people, 12 are left-handed.
(a)
Give a 95% confidence interval for the proportion \(p\) of left-handed people.
+
(b)
Give a 90% confidence interval.
+
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.)
- Table 4.3.1.
+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.)
- Table 4.3.2.
+
+Definition 4.3.1. +
++Theorem 4.3.2. +
++Example 4.3.3. +
+| Plant \(i\) + | +Height \(H_i\) (in) | +
| \(1\) | +\(9.2\) | +
| \(2\) | +\(9.8\) | +
| \(3\) | +\(10.1\) | +
| \(4\) | +\(9.5\) | +
| \(5\) | +\(10.4\) | +