From c800eaee2bc5e82273f225233fb0222701fa2c11 Mon Sep 17 00:00:00 2001 From: andyeisenberg Date: Tue, 24 Feb 2026 06:59:44 -0500 Subject: [PATCH] First pass at CI section --- source/sec-Confidence-Intervals.ptx | 271 ++++++++++++++++++++++------ 1 file changed, 219 insertions(+), 52 deletions(-) diff --git a/source/sec-Confidence-Intervals.ptx b/source/sec-Confidence-Intervals.ptx index 2e85b13..8616362 100644 --- a/source/sec-Confidence-Intervals.ptx +++ b/source/sec-Confidence-Intervals.ptx @@ -1,54 +1,221 @@
- Confidence Intervals + Confidence Intervals -

- Text of section. -

+

+ 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. +

+ + + + + +

+ Suppose in a sample of 100 people, 12 are left-handed. +

+ + + +

- 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. + Give a 95% confidence interval for the proportion p of left-handed people.

- - - - -

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

- + - - -

- Give a 95% confidence interval for the proportion p of left-handed people. -

- - - - - - -

- Give a 90% confidence interval. -

- - - - - +

- 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.) + Give a 90% confidence interval.

+ + + + + + +

+ 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.) +

@@ -73,16 +240,16 @@
- - + + - - -

- 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.) -

+ + +

+ 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.) +

@@ -107,7 +274,7 @@
- - - -
\ No newline at end of file + + + + \ No newline at end of file