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**The Behavior of Sums and Averages Amore Frozen Foods**

Class 12 The Behavior of Sums and Averages Amore Frozen Foods EMBS (selections from)

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**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 chi-squared statistic, =chidist with dof = (#rows-1)(#col-1) to get p-value

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**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.

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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?

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**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 n is too small to use the normal approximation

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**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.

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**For Amore…. 𝑇𝑜𝑡𝑎𝑙= 𝑋 1 + 𝑋 2 + 𝑋 3 + 𝑋 4 + 𝑋 5**

We can work our problem either way. 𝑇𝑜𝑡𝑎𝑙= 𝑋 1 + 𝑋 2 + 𝑋 3 + 𝑋 4 + 𝑋 5 The sample mean is just the total times 0.2. 𝑋 = 𝑋 1 + 𝑋 2 + 𝑋 3 + 𝑋 4 + 𝑋 5 5 “X-bar” The sample mean

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**The X’s are independent and identically distributed N(10. 2,**

The X’s are independent and identically distributed N(10.2,.22) random variables 8.22, 𝑇𝑜𝑡𝑎𝑙= 𝑋 1 + 𝑋 2 + 𝑋 3 + 𝑋 4 + 𝑋 5 Means and Variances nicely Add. Adding standard deviations is very bad. The Distribution of (Total) = E(Total) = Var(Total) = Standard Deviation (Total) =

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**The X’s are independent and identically distributed N(μ,σ) random variables**

𝑇𝑜𝑡𝑎𝑙= 𝑋 1 + 𝑋 2 + 𝑋 3 + …+ 𝑋 𝑛 Means and Variances nicely Add. Adding standard deviations is very bad. The Distribution of (Total) = Normal E(Total) = n*μ Var(Total) = n*σ2 Standard Deviation (Total) = 𝑛 𝜎 P(reject Batch) = P(Total<40) = NORMDIST(40,5*8.22,5^.5*.22,true) = 0.013

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**P(reject Batch) = P( 𝑋 <8) = NORMDIST(8,8.22,.22/5^.5,true) = 0.013**

The X’s are independent and identically distributed N(μ,σ) random variables 𝑋 = 𝑋 1 + 𝑋 2 + 𝑋 3 + … + 𝑋 𝑛 𝑛 Means and Variances nicely Add. Adding standard deviations is very bad. The Distribution of ( 𝑋 ) = Normal E( 𝑋 ) = μ Var( 𝑋 ) = σ2/n Standard Deviation ( 𝑋 ) = 𝜎/ 𝑛 P(reject Batch) = P( 𝑋 <8) = NORMDIST(8,8.22,.22/5^.5,true) = 0.013

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**The standard deviation of a sample mean ( 𝑋 )**

Aka “the standard error of the sample mean” 𝜎 𝑋 = 𝜎 𝑛 (7.3) A standard ERROR is some other standard deviation divided by 𝑛 somehow. The formulas hold for n=1 and n=∞ and n’s in between. 𝑋 𝑛 ~𝑁(𝜇, 𝜎 𝑛 )

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**The distribution of a sample mean**

Assumes independent, identically distributed Normal random variables. 𝑋 𝑛 ~𝑁(𝜇, 𝜎 𝑛 )

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**Illustration. The lorex fill test data**

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**Illustration. The lorex fill test data**

=

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**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.

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**Small School dominate the “top 25”**

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….and the bottom!

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**He should have taken “Statistics in Business”**

In recent years Bill Gates and the Gates Foundation have acknowledged that their earlier emphasis on small schools was misplaced. He should have taken “Statistics in Business”

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**The X’s are independent and identically distributed N(μ,σ) random variables**

𝑇𝑜𝑡𝑎𝑙= 𝑋 1 + 𝑋 2 + 𝑋 3 + …+ 𝑋 𝑛 Means and Variances nicely Add. Adding standard deviations is very bad. The Distribution of (Total) = Normal E(Total) = n*μ Var(Total) = n*σ2 Standard Deviation (Total) = 𝑛 𝜎

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**The X’s are independent and identically distributed N(μ,σ) random variables**

𝑇𝑜𝑡𝑎𝑙= 𝑋 1 + 𝑋 2 + 𝑋 3 + …+ 𝑋 𝑛 Means and Variances nicely Add. Adding standard deviations is very bad. The Distribution of (Total) = E(Total) = Var(Total) = Standard Deviation (Total) =

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**The X’s are independent and identically distributed N(μ,σ) random variables**

𝑇𝑜𝑡𝑎𝑙= 𝑋 1 + 𝑋 2 + 𝑋 3 + …+ 𝑋 𝑛 Means and Variances nicely Add. Adding standard deviations is very bad. The Distribution of (Total) = ?? E(Total) = μ1 + μ2 + … μn Var(Total) = σ21+ σ22+…. σ2n Standard Deviation (Total) = 𝑉𝑎𝑟

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**Illustration: Suppose Wunderdog did five sports**

Number correct if guessing Expected # correct Variance Std Dev 1 149 X1 74.5 37.25 6.10 2 85 X2 42.5 21.25 4.61 3 133 X3 66.5 33.25 5.77 4 209 X4 104.5 52.25 7.23 5 311 X5 155.5 77.75 8.82 Total 887 sum of the Xs 443.5 221.75 14.89 Means and Variances nicely add. Adding standard deviations is very bad.

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**Illustration: Suppose Wunderdog did five sports**

Number correct if guessing Expected # correct Variance Std Dev 1 149 X1 74.5 37.25 6.10 2 85 X2 42.5 21.25 4.61 3 133 X3 66.5 33.25 5.77 4 209 X4 104.5 52.25 7.23 5 311 X5 155.5 77.75 8.82 Total 887 sum of the Xs 443.5 221.75 14.89 Means and Variances nicely add. Adding standard deviations is very bad. =[887*.5*.5]^.5 =887*.5 =887*.5*.5 The Binomial Distribution “knows” that means and variances add.

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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?

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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, 𝑛 σ) The sample mean of n X’s is written as will be N(μ, σ/ 𝑛 ) Means and variances nicely add (if the X’s are independent). Adding standard deviations is very bad. The BINOMIAL knows this….. 𝑋 𝑛 𝑋 𝑛

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**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)

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