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Statistics Quick Overview. Class #2. Thought Exercise with Our Packaging Example. Original Case (mean = 290, sd = 53). If a store manager came to you and said, “what will my sales be?” how would you answer?.
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Statistics Quick Overview Class #2
Thought Exercise with Our Packaging Example Original Case (mean = 290, sd = 53) If a store manager came to you and said, “what will my sales be?” how would you answer? If CEO came to you and said, “what will average sales be?” how would you answer? Less Variability (m = 290, sd = 5) More Variability (m = 290, sd = 186)
Thought Exercise II- We Doubled The Samples (mean = 290, sd = 53) (mean = 290, sd = 53) What do you think of these questions now? If a store manager came to you and said, “what will my sales be?” how would you answer? If CEO came to you and said, “what will average sales be?” how would you answer?
Sampling Distribution • is approximately normally distributed with a mean of µ and stdev of • Since we never know the actual σ, we approximate it with the sample standard deviation, s. • is commonly used in statistics • We call this term the standard error of the mean Let’s see how this applies to our examples
We Have 3 Measures for a Sample of Data • Mean (average) • Standard Deviation (sample standard deviation) • Standard Error of the Mean
Central Limit Theorem– General Idea • is approximately normally distributed with a mean of µ and stdev of • In other words, as you take various samples, the collection of these samples will be approximately normally distributed • The larger the value of n, the closer to normally distributed • The population data does not have to be normally distributed
A New Game 1 2 3
Basic Probability Solution space = 1 A and B
Car Example- Neither Will Start? Solution space = 1 A and B (75%) = 90% + 80% - 75%=95% 1-95% = 5% chance neither will start
Neither Will Start?A Table Can Be Helpful Acura Starts Doesn’t Starts Doesn’t 80% 75% 5% BMW 20% 15% 5% 10% 90%
Car Example- A Starting if B Starts? Solution space = 1 A and B (75%) = 75% / 80% = 93.75%
Conditional Probability:A Table Can Be Helpful Acura Starts Doesn’t Normalize this row: 75% / 80% Starts Doesn’t 80% 75% 5% BMW 20% 15% 5% 10% 90%
….Have to Do with the Front Line in Hockey…. Right Center Left
Let’s Say A Coach (Maybe Mr. Brown?) Had to Pick 3 Players for Hockey and Then Quiditch • Let’s start with hockey… • Here, order matters • The person on the left must stay on the left • The person on the right must stay on the right • So, how many different potential line-ups does Mr. Brown have to consider? • Choices are: Mr Blonde, Mr White, Mr Orange, Mr Pink, and Mr Blue 5 x 4 x 3 = 60 Where n is the number of choices, and k is the number picked. In Excel, this is PERMUT(n,k)
Let’s Carefully Write Out the Permutations Note: Each column is a unique combination of players Note: The entries within each row are different permutations of the players. This is our same problem again where n= 3 and k = 3 ==6
Mr. Brown’s Choices for a Quiditch Front Line • Here, order does not matter • He just needs a front line • All that matters is the number of unique combinations • What observation from the permutation table helps us determine the unique combinations
Figuring out the Combinations When calculating the permutations, we naturally determined the unique combinations (the columns) and then ran the permutations for each combination. If we divide out that last step, we will have just the combinations: = 60 / 6=10
Binomial Distribution • Sample Size of 10 • Case #1: Assume that the lot is good with 5% defectives • When will you reject because you find 3 or more defectives • Case #2: Assume that the lot has 40% defectives • When will you accept because you find 2 or less defectives • Let’s assume: • s is the probability of success • f is the probability of failure
Case #1 (only 5% of the lot is defective) • Example of getting 3 Failures • fssfsssssf • Probability of this is (5%)3(95%)7 • Example of getting 4 Failures • fssfsfsssf • Probability of this is (5%)4(95%)6 • What are we missing? The number of combinations
Bayes’ Rule • A1 uses drugs P(A1) = 5% • A2 does not use drugs P(A2) = 95% • B tests shows drug use • P(B | A1) = 98% • P(B | A2) = 2% What we want….
Bayes Rule- Calculation = = 5%*(98%) / ((5%*98%)+(95%*2%)) =72%