### Class 12 Notes

```Class 12
The Behavior of Sums and Averages
Amore Frozen Foods
EMBS 7.3-7.6
(selections from)
Review Class 11. chi-squared test
of independence
•
•
•
•
•
H0: The Categorical variables are independent
Ha: They are not.
Pivot Table to create contingency table.
Excel to create Expected Counts
CHITEST() to calculate p-value.
– or Excel to get distances and calculated chisquared statistic, =chidist with dof = (#rows1)(#col-1) to get p-value
There is often
another way
• H0: Athlete and HSStat are independent
• H0: P1=P2
– P1 is probability an athlete took HSStat
– P2 is the probability a non-athlete took HSStat
• You can use the method of section 11.1 to test H0.
(check out equation 11.7 to see what’s involved)
– And you get exactly the same p-value!!
• So, the chi-squared test of independence tests the
equality of two probabilities (for a 2x2) and more!
That’s why we learned it.
Amore
• At a target of 8.22, Amore’s filling machine fills
mac and cheese tins with an amount N(8.22,0.22)
– Resulting in about 16% of tins (pies) being
underweight
• The FDA requires Amore to test each 20-minute
batch using a random sample of five tins. The
entire batch must be rejected if the 5-pie sample
averages less than 8 ounces.
• What is the probability a batch will be rejected?
We know how many sample pies
will be underfilled….
• Binomial(n=5,p=.16) assuming pie weights are
independent.
• E(number underfilled) = n*p = 5*.16 = 0.8
• Var(number underfilled) = n*p*(1-p) = 0.672
• Std dev (number underfilled) = 0.820
0.45
0.4
Probability
0.35
n is too small to
use the normal
approximation
0.3
0.25
0.2
0.15
0.1
0.05
0
01
21
32
43
# Sample Pies Underfilled
54
56
But the fate of the batch rests on
the total (average) weight of the
five sampled pies.
We need to figure out what the
probability distribution of either the
total or the average of five is.
For Amore….
We can work
our problem
either way.
= 1 + 2 + 3 + 4 + 5
1 + 2 + 3 + 4 + 5
=
5
“X-bar”
The sample mean
The sample
mean is just
the total
times 0.2.
The X’s are independent and identically
8.22,
distributed N(10.2,.22)
random variables
= 1 + 2 + 3 + 4 + 5
The Distribution of (Total) =
E(Total) =
Var(Total) =
Standard Deviation (Total) =
Means and
Variances
standard
deviations is
The X’s are independent and identically
distributed N(μ,σ) random variables
= 1 + 2 + 3 + … +
The Distribution of (Total) =
Normal
E(Total) =
n*μ
Var(Total) =
n*σ2
Standard Deviation (Total) =

Means and
Variances
standard
deviations is
P(reject Batch) = P(Total<40) = NORMDIST(40,5*8.22,5^.5*.22,true) = 0.013
The X’s are independent and identically
distributed N(μ,σ) random variables
1 + 2 + 3 + … +
=

The Distribution of ( ) =
E( ) =
Normal
μ
Var( ) =
σ2/n
Standard Deviation ( ) =
/
Means and
Variances
standard
deviations is
P(reject Batch) = P(<8) = NORMDIST(8,8.22,.22/5^.5,true) = 0.013
The standard deviation of a
sample mean ()
Aka “the standard
error of the sample
mean”
A standard ERROR is
some other standard
deviation divided by
somehow.

=

~(, )

(7.3)
The formulas hold
for n=1 and n=∞ and
n’s in between.
The distribution of a sample mean
Assumes
independent,
identically
distributed
Normal
random
variables.

~(, )

Illustration. The lorex fill test data
Illustration. The lorex fill test data
0.158
5
=
Small Schools are Better?
Did Bill Gates waste a billion dollars because he
failed to understand the formula for the
standard deviation of the sample mean?
The Gates Foundation certainly spent a lot of
money, along with many others, pushing for
smaller schools and a lot of the push came
because people jumped to the wrong
conclusion when they discovered that the
smallest schools were consistently among the
best performing schools.
Small School dominate the “top 25”
….and the bottom!
In recent years Bill Gates and the Gates
Foundation have acknowledged that
their earlier emphasis on small schools
was misplaced.
He should have taken
The X’s are independent and identically
distributed N(μ,σ) random variables
= 1 + 2 + 3 + … +
The Distribution of (Total) =
Normal
E(Total) =
n*μ
Var(Total) =
n*σ2
Standard Deviation (Total) =

Means and
Variances
standard
deviations is
The X’s are independent and identically
distributed N(μ,σ) random variables
= 1 + 2 + 3 + … +
The Distribution of (Total) =
E(Total) =
Var(Total) =
Standard Deviation (Total) =
Means and
Variances
standard
deviations is
The X’s are independent and identically
distributed N(μ,σ) random variables
= 1 + 2 + 3 + … +
The Distribution of (Total) =
??
E(Total) =
μ1 + μ2 + … μn
Var(Total) =
σ21+ σ22+…. σ2n
Standard Deviation (Total) =

Means and
Variances
standard
deviations is
Illustration: Suppose Wunderdog
did five sports
Sport
1
2
3
4
5
Total
n
149
85
133
209
311
887
Number correct if
guessing
X1
X2
X3
X4
X5
sum of the Xs
Means and Variances
standard deviations is
Expected #
correct Variance
74.5
37.25
42.5
21.25
66.5
33.25
104.5
52.25
155.5
77.75
443.5
221.75
Std Dev
6.10
4.61
5.77
7.23
8.82
14.89
Illustration: Suppose Wunderdog
did five sports
Sport
1
2
3
4
5
Total
n
149
85
133
209
311
887
Number correct if
guessing
X1
X2
X3
X4
X5
sum of the Xs
Means and Variances
standard deviations is
Expected #
correct Variance
74.5
37.25
42.5
21.25
66.5
33.25
104.5
52.25
155.5
77.75
443.5
221.75
Std Dev
6.10
4.61
5.77
7.23
8.82
14.89
=[887*.5*.5]^.5
=887*.5
=887*.5*.5
The Binomial Distribution “knows” that means and variances add.
Example Problem
EMBS 40 p 306.
• A large BW survey of MBA’s ten years out found
mean dining out spending to be \$115.5 (per
week) and population standard deviation of \$35.
You will do a follow-up study and survey a
random sample of 40 of the MBA alums.
– What sample mean will you get?
– What is the probability your sample mean will be
within \$10 of the population mean?
– What is the probability your sample mean will be less
than \$100?
Summary
• If we know the X’s are independent, identically
distributed N(μ,σ) random variables…
– The total of n X’s will be N(nμ, σ)
– The sample mean of n X’s is written as
–  will be N(μ, σ/ )
• Means and variances nicely add (if the X’s are
independent). Adding standard deviations is
– The BINOMIAL knows this…..
Exam1
• Complete answers have been posted
• The Binomial works for small n.
– P(Cubs Win) = BINOMDIST(3,3,.5,false)
– P(2Gin4) = BINOMIDIST(2,4,p,false)
• Whether p is 0.5 or 0.48
– P(2P l noD) = BINOMDIST(2,2,.2,false)
– P(1P l noD) = BINOMDIST(1,2,.2,false)
– P(0P l noD) = BINOMDIST(0,2,.2,false)
– #correctly filled is B(n=144,p=.92775)
```