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Normal and t Distributions
Bret Hanlon and Bret Larget
Department of Statistics University of Wisconsin—Madison
October 11–13, 2011
Normal 1 / 33
Case Study
Case Study Body temperature varies within individuals over time (it can be higher when one is ill with a fever, or during or after physical exertion). However, if we measure the body temperature of a single healthy person when at rest, these measurements vary little from day to day, and we can associate with each person an individual resting body temperture. There is, however, variation among individuals of resting body temperture. A sample of n = 130 individ- uals had an average resting body temperature of 98.25 degrees Fahrenheit and a standard deviation of 0.73 degrees Fahrenheit. The next slide shows an estimated density plot from this sample.
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Density Plot
Resting Body Temperature (F)
Density
96 97 98 99 100 101
l (^) l llllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllll^ llllllllllll lll^ l^ l^ l
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Normal Distributions
The estimated density has these features: I (^) it is bell-shaped; I (^) it is nearly symmetric. Many (but not all) biological variables have similar shapes. One reason is a generalized the central limit theorem: random variables that are formed by adding many random effects will be approximately normally distributed. Important for inference, even when underlying distributions are not normal, the sampling distribution of the sample mean is approximately normal.
Normal Case Study Body Temperature 4 / 33
Example: Population
A population that is skewed.
Population
x
Density
0 200 400 600
Normal Case Study Body Temperature 5 / 33
Example: Sampling Distribution
Sampling distribution of the sample mean when n = 130.
Sampling Distribution, n=
x
Density
80 90 100 110 120 130
Normal Case Study Body Temperature 6 / 33
Case Study: Questions
Case Study How can we use the sample data to estimate with confidence the mean resting body temperture in a population? How would we test the null hypothesis that the mean resting body temperture in the population is, in fact, equal to the well-known 98. degrees Fahrenheit? How robust are the methods of inference to nonnormality in the underlying population? How large of a sample is needed to ensure that a confidence interval is no larger than some specified amount?
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The Big Picture
Many inference problems with a single quantitative, continuous variable may be modeled as a large population (bucket) of individual numbers with a mean μ and standard deviation σ. A random sample of size n has a sample mean ¯x and sample standard deviation s. Inference about μ based on sample data assumes that the sampling distribution of ¯x is approximately normal with E(¯x) = μ and SD(¯x) = σ/
n. To prepare to understand inference methods for single samples of quantitative data, we need to understand: I (^) the normal and related distributions; I (^) the sampling distribution of ¯x.
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Standard Normal Density
Standard Normal Density
Possible Values
Density
−3 −2 −1 0 1 2 3
Area within 1 = 0. Area within 2 = 0. Area within 3 = 0.
Normal Standard Normal Distribution Probability Calculations 13 / 33
General Areas
There is no formula to calculate general areas under the standard normal curve. (The integral of the density has no closed form solution.) We prefer to use R to find probabilities. You also need to learn to use normal tables for exams.
Normal Standard Normal Distribution Probability Calculations 14 / 33
R
The function pnorm() calculates probabilities under the standard normal curve by finding the area to the left. For example, the area to the left of − 1 .57 is
pnorm(-1.57) [1] 0. and the area to the right of 2.12 is 1 - pnorm(2.12) [1] 0.
Normal Standard Normal Distribution Probability Calculations 15 / 33
Tables
The table on pages 672–673 displays right tail probabilities for z = 0 to z = 4.09. A point on the axis rounded to two decimal places a.bc corresponds to a row for a.b and a column for c. The number in the table for this row and column is the area to the right. Symmetry of the normal curve and the fact that the total area is one are needed. The area to the left of − 1 .57 is the area to the right of 1.57 which is 0.05821 in the table. The area to the right of 2.12 is 0.01711. When using the table, it is best to draw a rough sketch of the curve and shade in the desired area. This practice allows one to approximate the correct probability and catch simple errors. Find the area between z = − 1 .64 and z = 2.55 on the board. Normal Standard Normal Distribution Probability Calculations 16 / 33
R
The function qnorm() is the inverse of pnorm() and finds a quantile, or location where a given area is to the right. For example, the 0.9 quantile of the standard normal curve is
qnorm(0.9) [1] 1. and the number z so that the area between −z and z is 0.99 is qnorm(0.995) [1] 2. since the area to the left of −z and to the right of z must each be (1 − 0 .99)/2 = 0.005 and 1 − 0 .005 = 0.995. Draw a sketch!
Normal Standard Normal Distribution Quantile Calculations 17 / 33
Tables
Finding quantiles from the normal table almost always requires some round off error. To find the number z so that the area between −z and z is 0. requires finding the probability 0.00500 in the middle of the table. We see z = 2.57 has a right tail area of 0.00508 and z = 2.58 has a right ail area of 0.00494, so the value of z we seek is between 2. and 2.58. For exam purposes, it is okay to pick the closest, here 2.57. Use the table to find the 0.03 quantile as accurately as possible. Draw a sketch!
Normal Standard Normal Distribution Quantile Calculations 18 / 33
General Normal Density
The general normal density with mean μ and standard deviation σ is a symmetric, bell-shaped probability density with equation:
f (x) =
2 πσ
e
− (^12)
x−μ σ
, (−∞ < x < ∞)
Sketches of general normal curves have the same shape as standard normal curves, but have rescaled axes.
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General Normal Density
Normal Density
Possible Values
Density
μ − 3 σ μ − 2 σ μ − σ μ μ + σ μ + 2 σ μ + 3 σ
Area within 1 SD = 0. Area within 2 SD = 0. Area within 3 SD = 0.
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Answers (with R, table will be close)
(^4) 97.76 and 98.
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The χ^2 Distribution
The χ^2 distribution is used to find p-values for the test of independence and the G-test we saw earlier for contingency tables. Now that the normal distribution has been introduced, we can better motivate the χ^2 distribution.
Definition
If Z 1 ,... , Zk are independent standard normal random variables, then
X 2 = Z 12 + · · · + Z (^) k^2
has a χ^2 distribution with k degrees of freedom.
Normal Other Distributions Chi-square Distributions 26 / 33
The χ^2 Distribution
The functions pchisq() and qchisq() find probabilities and quantiles, respectively, from the χ^2 distributions. The table on pages 669–671 has the same information for limited numbers of quantiles for each χ^2 distribution with 100 or fewer degrees of freedom. Unlike the normal distributions where all normal curves are just rescalings of the standard normal curve, each χ^2 distribution is different.
Normal Other Distributions Chi-square Distributions 27 / 33
t Distribution
Definition
If Z is a standard normal random variable and if X 2 is a χ^2 random variable with k degrees of freedom, then
T =
Z
X 2 /k
has a t distribution with k degrees of freedom.
t densities are symmetric, bell-shaped, and centered at 0 just like the standard normal density, but are more spread out (higher variance). As the degrees of freedom increases, the t distributions converge to the standard normal. t distributions will be useful for statistical inference for one or more populations of quantitative variables.
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The Central Limit Theorem
The Central Limit Theorem
If X 1 ,... , Xn are an independent sample from a common distribution F with mean E(Xi ) = μ and variance Var(Xi ) = σ^2 , (which need not be normal), then X¯ =
∑n i=1 Xi n is approximately normal with E( ¯X ) = μ and Var( ¯X ) = σ 2 n if the sample size n is sufficiently large.
The central limit theorem (and its cousins) justifies almost all inference methods the rest of the semester.
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Mean of the Sampling Distribution of ¯X
The mean of the sampling distribution of ¯X is found using the linearity properties of expectation.
E( ¯X ) = E
( ∑n i=1 Xi n
n
E(X 1 + · · · + Xn)
=
n
E(X 1 ) + · · · + E(Xn)
n
nμ = μ
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Variance of the Sampling Distribution of ¯X
The variance of the sampling distribution of ¯X is found using the properties of variances of sums.
Var( ¯X ) = Var
( ∑n i=1 Xi n
n
Var(X 1 + · · · + Xn)
=
n
Var(X 1 ) + · · · + Var(Xn)
n
nσ^2
σ^2 n
Also, SE( ¯X ) = √σn.
Normal The Central Limit Theorem 31 / 33
Case Study
Example
In a population, suppose that: the mean resting body temperature is 98.25 degrees Fahrenheit; the standard deviation is 0.73 degrees Fahrenheit; resting body temperatures are normally distributed. Let X 1 ,... , X 40 be the resting body temperatures of 40 randomly chosen individuals from the population. Find: (^1) P( ¯X < 98), the probability that the sample mean is less than 98. (^2) P(98 < X¯ < 100), the probability that the sample mean is between 98 and 100. (^3) the 0.90 quantile of the sampling distribution of ¯X. (^4) The cutoff values for the middle 50% of the sampling distribution of X¯.
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