Numerical variables from a single sample Chapter 11 μ = 67.4 σ = 3.9 μ Y = μ = 67.4 σ Y = σ n = 3.9 5 = 1.7 is normally distributed whenever: Y is normally distributed or n is large Y Inference about means Z = Y − μ σ Y = Y − μ σ n Because is normally distributed, we can convert its distribution to a standard normal distribution: Y This would give a probability distribution of the difference between a sample mean and the population mean. But... We don’t know s ... However, we do know s, the standard deviation of our sample. We can use that as an estimate of s.
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Inference about means But We don’t know swhitlock/bio300/overheads/...its distribution to a standard normal distribution: € Y This would give a probability distribution of the
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Numerical variables from a single sample
Chapter 11
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µ = 67.4
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σ = 3.9
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µY = µ = 67.4
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σY =σn
=3.95
=1.7
is normally distributed whenever:
Y is normally distributedor
n is large€
Y
Inference about means
€
Z =Y −µσY
=Y −µσ n
Because is normally distributed, we can convert its distribution to a standard normal distribution:
€
Y
This would give a probability distribution of the difference between a sample mean and the population mean.
But... We don’t know s . . .
However, we do know s, the standard deviation of our sample. We can use that as an estimate of s.
€
µ = 67.4
€
σ = 3.9
N In most cases, we don’t know the real population distribution.
We only have a sample.
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SEY =sn
=3.15
=1.4
We use this as an estimate of
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σY
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s = 3.1
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Y = 67.1
A good approximation to the standard normal is then:
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t =Y −µSEY
=Y −µs / n
t has a Student’s tdistribution
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Z =Y −µσY
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t =Y −µSEY
Discovered by William Gossett, of the Guinness Brewing Company
Z t9
Degrees of freedom
df = n - 1
We use the t-distribution to calculate a confidence interval of the mean
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Y ± SEY tα 2( ),df
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−tα 2( ),df <Y −µSEY
< tα 2( ),df
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Y − tα 2( ),df SEY < µ < Y + tα 2( ),df SEY
Another way to express this is:
We rearrange the above to generate:
95% confidence interval for a mean
Example: Paradise flying snakes
0.9, 1.4, 1.2, 1.2, 1.3, 2.0, 1.4, 1.6
Undulation rates (in Hz)
https://www.youtube.com/watch?v=16aGSx9gFO4
Estimate the mean and standard deviation
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Y =1.375s = 0.324n = 8
Find the standard error
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Y ± SEY tα 2( ),df
SEY =sn
=0.3248
= 0.115
Find the critical value of t
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df = n −1= 7
tα 2( ),df = t0.05 2( ),7
= 2.36
Table C: Student's t distributionFinding the critical value of t in R
With these data, we cannot reject the null hypothesis that the mean human body temperature is 98.6.
t.test(bodyTempSmallData$temperature, mu = 98.6)
One Sample t-test
data: bodyTempSmallData$temperaturet = -0.56065, df = 24, p-value = 0.5802alternative hypothesis: true mean is not equal to 98.695 percent confidence interval:98.24422 98.80378sample estimates:mean of x
98.524
Body temperature revisited: n = 130
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n =130Y = 98.25s = 0.733
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t =Y −µ0s / n
=98.25 − 98.60.733/ 130
= −5.44
Body temperature revisited: n = 130
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t = −5.44
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t0.05(2),129 = ±1.98
t is further out in the tail than the critical value, so we could reject the null hypothesis. Human body temperature is not 98.6ºF. P = 2.4 x 10–7