3' Analysis of Variance II Anlisis de Varianza II

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3. Analysis of Variance II Análisis de
Varianza II

• Profesor Simon Wilson
• Departamento de Estadística y Econometría

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Properties of the ML Estimates (1)

• is normally distributed with mean mi
• By properties of the variance
• Estimate of s2. Observe that we can write
• and that the expressions inside the are c2
  with ni-1 degrees of freedom

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Properties of the ML Estimates (2)

• The sum of independent c2 distributions is also a
  c2 distribution, with degrees of freedom the sum
  of the degrees of freedom.
• The sum of degrees of freedom is n I
• So,
• We know (!!) the mean of a c2 deg. of freedom,
  so

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Properties of the ML Estimates (3)

• Thus
• So expected value of estimate of s2 is not s2
• This is called a biased estimator/estimador no
  centrado
• If we replace n by n-I in the definition of ,
  we have a new estimator
• This is unbiased/centrado. It is called the
  residual variance estimator. We use this estimate
  in this course

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The Residual Variance Estimator

• Also, in this course we use ni-1 and not ni when
  calculating the variance, because this is
  unbiased
• After a bit of algebra, observe that we can also
  write
• This is the easiest formula to use if we know the
  

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Class Example

• Three whiskies Jameson, JB and Glenfiddich
  are tested for alcohol content.
• Seven samples of Jameson, 8 of JB and 6 of
  Glenfiddich are taken. The alcoholic content of
  each sample is measured (as a percentage)
• The measurements are given on the next slide,
  with some summary statistics
• We want to know if there is any difference in the
  mean alcoholic content of the whiskies

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Class Example - Data
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To Do (1)

• What are the estimates of
• The mean alcoholic content of each whisky?
• The variance in alcoholic content for each whisky?

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To Do (2)

• Can you remember from Introduction to Statistics?
• Calculate a 95 confidence interval for each
  mean.
• Calculate a 95 confidence interval for the
  variance.
• You may need to know
• t 5,0.975 2.57, t 6,0.975 2.45, t 7,0.975
  2.36
• c218,0.975 31.5, c218,0.025 8.23

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T-test to compare two means (1)

• Lets take 2 of the whiskies (Jameson and JB)
• Is there a difference in mean alcohol content?
• Null hypothesis H0 m1 m2
• Remember how to do this?

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T-test to compare two means (2)

• So if the means are equal then
• 95 of the probability of N(0,1) is in (-1.96,
  1.96).
• So if our value is gt 1.96 or lt -1.96 we conclude
  that the means are different.

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T-test to compare two means (3)

• However we dont know s2 so we replace it with
  its estimate and use the t-distribution
• Our estimate for s2 comes from only the Jameson
  and JB data. We have n1 n2 2 degrees of
  freedom from these 2 levels. So

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T-test to compare 2 means (4)

• In this case n1 n2 2 13 degrees of freedom
• 95 of the probability of the t13 lies between
  2.16 and 2.16
• So if our value is gt 2.16 or lt -2.16 then we
  reject H0 conclude the means are different
• Data
• So

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Summary of test to compare two means

• To test if m1 m2, calculate
• If t gt tn1n2-2, 0.975 or t lt - tn1n2-2, 0.975
  then conclude that the means are different, else
  conclude that they are equal.

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To Do

• Test to see if there is a difference in mean
  alcoholic content between the other 2 possible
  pairs
• Jameson and Glenfiddich
• JB and Glenfiddich
• You may need the following information
• t11,0.975 2.20
• t12,0.975 2.18