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Chapter 15

Chapter 15. Week 5, Friday. Population Cross-tabs.

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Chapter 15

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  1. Chapter 15 Week 5, Friday

  2. Population Cross-tabs Suppose there are 1000 employees invited to a company party. The company has locations in Ohio, Texas, and New York. Corporate is considering the song “My Kind of Party” by Brantley Gilbert for the Welcome Video and wants to know how the 1000 employees in attendance feel about this song. The cross-tab below describes the 1000 attending employees’ feelings toward this song. Why is this not a sample? If I randomly choose 1 person from the company, what is the likelihood that the employee will like the song? 305/1000 = 30.5% If I randomly choose 1 person from Texas, what is the likelihood that the employee will like the song? 100/225 = 44.4%

  3. Population Cross-tabs Suppose there are 1000 employees invited to a company party. The company has locations in Ohio, Texas, and New York. Corporate is considering the song “My Kind of Party” by Brantley Gilbert for the Welcome Video and wants to know how the 1000 employees in attendance feel about this song. The cross-tab below describes the 1000 attending employees’ feelings toward this song. If I randomly choose 1 person from Texas, what is the likelihood that the employee will like the song? 100/225 = 44.4% P[Like the song | Texas] = • = P[Like the song AND Texas] / P[Texas] • = (100/1000) / (225/1000) • = 100/225

  4. Population Cross-tabs Suppose there are 1000 employees invited to a company party. The company has locations in Ohio, Texas, and New York. Corporate is considering the song “My Kind of Party” by Brantley Gilbert for the Welcome Video and wants to know how the 1000 employees in attendance feel about this song. The cross-tab below describes the 1000 attending employees’ feelings toward this song. If I randomly choose 1 person from Texas, what is the likelihood that the employee will like the song? 100/225 = 44.4% The general formula for events A and B: P[ A | B ] = P[ A and B ] / P[B]

  5. Recall the Coin Example Consider two fair coins: Heads, Tails Tails, Heads Tails, Tails Heads, Heads P[ at least one heads ] = 3/4 P[ two heads ] = 1/4 P[ at least one heads AND two heads ] = 1/4 P[ two heads | At least one heads ] = ? = P[ 2 heads and At least one heads ] / P[at least one heads] = (1/4) / (3/4) = 1/3 = 33%

  6. Definition: Mutually Exclusive Two events are mutually exclusive if they have nothing in common. That is P[ A and B ] = 0

  7. Definition: Mutually Exclusive Consider two fair coins: Heads, Tails Tails, Heads Tails, Tails Heads, Heads Consider the events: 1 heads and 1 tails 2 heads They are mutually exclusive

  8. Definition: Mutually Exclusive Consider two fair coins: Heads, Tails Tails, Heads Tails, Tails Heads, Heads As a result: (1) P[ {HT,TH} and HH ] = 0 (2) P[ {HT,TH} or HH ] = P[HT, TH] + P[HH] - P[{HT,TH} and HH] = P[HT, TH] + P[HH]

  9. Definition: Mutually Exclusive Consider two fair coins: Heads, Tails Tails, Heads Tails, Tails Heads, Heads General Equation: For Mutually Exclusive events, A and B: P[ A or B ] = P[A] + P[B]

  10. Definition: Independent Two events are independent if the outcome of one event does not affect the outcome of the other. That is P[ A | B ] = P[A] Also: P[A and B] = P[A] * P[B]

  11. Example 1 Example: Flip three coins. Suppose these coins are independent of each other. Calculate the probability of flipping three heads. Solution: Let H1 be the event that the first flip is heads Let H2 be the event that the second flip is heads Let H3 be the event that the third flip is heads P[H1 and H2 and H3] = P[H1]*P[H2]*P[H3] = (1/2)*(1/2)*(1/2) = 1/8

  12. Example 2 (chapter 14 #32B) Example: Consider the blood types:P[type O] = 45%, P[type B] = 11% P[type A] = 40%, P[type AB] = 4% Consider four independent donors Question 1: P[ all are type O ] = P[p1 is O and p2 is O and p3 is O and p4 is O] = P[p1 is O]*P[p2 is O]*P[p3 is O]*P[p4 is O] = 45% * 45% * 45% * 45% = 4.1%

  13. Example 2 (chapter 14 #32B) Example: Consider the blood types:P[type O] = 45%, P[type B] = 11% P[type A] = 40%, P[type AB] = 4% Consider four independent donors Question 2: P[ no one is AB ] = P[p1 not AB and p2 not AB and p3 not AB and p4 not AB] = P[p1 not AB]*P[p2 not AB]*P[p3 not AB]*P[p4 not AB] = 96% * 96% * 96% * 96% = 84.9%

  14. Example 2 (chapter 14 #32B) Example: Consider the blood types:P[type O] = 45%, P[type B] = 11% P[type A] = 40%, P[type AB] = 4% Consider four independent donors Question 3: P[ not everyone is type A ] = 1 – P[ everyone is type A] = 1 - P[p1 is A and p2 is A and p3 is A and p4 is A] = 1 - P[p1 is A]*P[p2 is A]*P[p3 is A]*P[p4 is A] = 100% - (40% * 40% * 40% * 40%) = 97.4%

  15. Example 2 (chapter 14 #32B) Example: Consider the blood types:P[type O] = 45%, P[type B] = 11% P[type A] = 40%, P[type AB] = 4% Consider four independent donors Question 4: P[ at least one person is type B ] = 1 – P[ nobody is type B ] = 1 - P[p1 not B and p2 not B and p3 not B and p4 not B] = 1 - P[p1 not B]*P[p2 not B]*P[p3 not B]*P[p4 not B] = 100% - (89% * 89% * 89% * 89%) = 37.3%

  16. Example 3 Question: Are “liking the song” and “being from Texas” independent events? Answer: P[ like the song ] = 305/1000 = 30.5% P[ like the song | from texas ] = 100/225 = 44.44% Since 44.44% is not equal to 30.5%, these events are NOT independent.

  17. Example 4: A different song Question: Are “liking the song” and “being from Ohio” independent events? Answer: P[ like the song ] = 200/1000 = 20% P[ like the song | from ohio ] = 100/500 = 20% Since both values are 20%, these events ARE independent.

  18. Example 4: A different song Question: Are “liking the song” and “being from Ohio” independent events? Alternative Answer: P[ from ohio ] = 500/1000 = 50% P[ from ohio | like the song ] = 100/200 = 50% Since both values are 50%, these events ARE independent.

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