t and F distribution notes

Ch. 4 χ 2 , t, and F
• Summary of Theoretical Background
• Examples (involving one sample):
1. Confidence interval for µ using the t distribution
2. A test for µ using the F distribution
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Summary of Theoretical Background
• X 1 , X 2 , . . . , X n is a random sample from a normal population with
mean or expected value µ and variance σ 2 .
• S 2 is the sample variance and ¯X is the sample mean.
• (n − 1)S 2 /σ 2 is a χ 2 (n−1)
random variable.
• If Z is standard normal and X is χ 2 , and Z and X are independent,
(ν)
then T = √ Z is a t random variable on ν degrees of freedom.
X/ν
• If X 1 ∼ χ 2 ν 1
and X 2 ∼ χ 2 ν 2
, and X 1 and X 2 are independent, then
F = X 1/ν 1
X 2 /ν 2
has an F distribution on ν 1 and ν 2 degrees of freedom.
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A t Random Variable
• If X 1 , X 2 , . . . , X n are independent normal measurements with mean
µ and variance σ 2 , then
is standard normal.
Z = ¯X − µ
σ/ √ n
• We estimate σ with S:
is not standard normal.
T = ¯X − µ
S/ √ n
• T has a t distribution on n − 1 degrees of freedom.
• The number of degrees of freedom = denominator of S 2 .
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• The density of the t distribution is bell-shaped like the normal, but it is
flatter and more spread out.
• As the number of degrees of freedom increases, a t random variable
becomes more and more like a standard normal random variable:
t n−1 → z as n increases

Percentiles of the t distribution
• Given in a table at the back of the textbook.
• Or use qt(1-a, df)
• Example. Find the value of t .025,7 .
Look at the 7th row and 5th column: t .025,7 = 2.365. or
qt(.975, 7); [1] 2.364624
• Example. Find the value of t .025,∞ .
Look at the bottom row and 3rd column: t .025,∞ = z .025 = 1.96.
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Example: Confidence interval for µ
• Find a 95% confidence interval for the expected value of
concentration measurements taken from a chemical process. Sample
measurements are
204 190 202 207
204 202 201 195
• If X denotes a concentration measurement, then ¯x = 201.,
s = 5.50, and n = 8 so a 95% confidence interval for µ = E[X] is
¯x ± t .025,7
s
√
8
= 201 ± 2.365(5.5)/ √ 8
= 201 ± 4.60
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The F distribution
• Recall that Z 2 is χ 2 (1) and (n − 1)S2 /σ 2 is χ 2 (n−1) , and S2 and ¯X
are independent, if the underlying sample is random and normal with
mean µ and variance σ 2 .
• Therefore,
( ¯X − µ) 2
S 2 /n
has an F distribution on 1 and n − 1 degrees of freedom.
• Percentiles of the F distribution can be read from the F tables at the
back of the book or use qf(1-a, n1, n2)
• e.g. f .95,1,7 = 5.59 or qf(.95, 1, 7); [1] 5.591448
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Example: Testing µ (using the F distribution)
• Is the expected value of the concentration measurements different
from 207?
• We test this by contradiction:
Suppose µ = 207. Then
f =
(¯x − µ)2
s 2 /n
=
(201 − 207)2
5.5 2 /8
= 9.52
Conclusion: This value exceeds 5.59, so the probability of observing
an f value this large, under our assumption (called the Null
Hypothesis) is less than .05. We have evidence that the true
expected concentration differs from 207.
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