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Section 5.1 Review

Section 5.1 Review. The Binomial Setting. 1) There are a fixed number n of observations. 2) The n observations are independent. 3) Each observation falls into just one of two categories, which, for convenience, we call “success” and “failure”.

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Section 5.1 Review

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  1. Section 5.1 Review

  2. The Binomial Setting 1) There are a fixed number n of observations 2) The n observations are independent 3) Each observation falls into just one of two categories, which, for convenience, we call “success” and “failure”. 4) The probability of a success, call it p, is the same for each observation

  3. Binomial Distributions The distribution of the count X of successes in the binomial setting is called the binomial distribution with parameters n and p. The parameter n is the number of observations The parameter p is is the probability of a success on any one observation. The possible values of X are the whole numbers from 0 to n. We say that X is B(n,p)

  4. Binomial Probability Tables (Table C, page T-7 to T-10) Consider the previous transistor problem using the B(10,.1) distribution. Q: What is the probability that we will pull an SRS from this distribution with exactly one bad transistor? A: P(X = 1) = 0.3874

  5. Binomial Mean and Standard Deviation X = np (1-p) Q: If a count X is B(n,p), what are the mean X and the standard deviation X ? A: If a count X has the B(n,p) distribution, then : X = np

  6. Count of successes in sample = Size of sample p X = n Sample Proportions In statistical sampling, we often want to estimate the proportion p of “successes” in a population. Our estimator is the sample proportion of successes :

  7. Mean and Standard Deviation of a Sample Proportion Let be the sample proportion of successes in an SRS of size n drawn from a large population having population proportion p of successes. The mean and standard deviation of are the following :  = p =  p(1-p) p p p p n

  8. Normal Approximations for Counts and Proportions • Let X be the count of successes in the sample, and • = X / n be the sample proportion of successes. • X is approximately N( np, np(1-p) ) ( p, ) p(1-p) p p • is approximately N n • Draw an SRS of size n from a large population having • population proportion p of successes. • When n is large, the sampling distributions of these • statistics are approximately normal. Note: Use this when np 10 and n(1-p)  10

  9. End O’ Review

  10. The Sampling Distribution of a Sample Mean Recall : 1) Counts and proportions are discrete random variables that describe categorical data. 2) Measured data is usually described with continuous random variables.

  11. The Sampling Distribution of a Sample Mean Recall : 3) Averages are less variable than individual observations

  12. The Sampling Distribution of a Sample Mean Recall : 4) Averages are more normal than individual observations 2 obs. 1 obs. 10 obs. 25 obs.

  13. Sample Mean Q: What is a sample mean? A: This is an estimate of the mean of the underlying population Q: How do we find the sample mean? A: 1) Select an SRS of size n from a population 2) Measure a variable X on each individual in the sample 3) Let each observation be labeled X1, X2, … , Xn 4) The sample mean is the average of the Xi’s Note: If the sample is large, each Xi can be thought of as an independent random variable

  14. Mean and Standard Deviation of a Sample Mean Let be the mean of an SRS of size n from a population having mean  and standard deviation . The mean and standard deviation of are:  =   x x  = n x x

  15. Mean and Standard Deviation of a Sample Mean  =  = 2.8 = = 2.8 10  100  = n x x The height of a randomly chosen Jedi varies according to the N(73, 2.8) distribution. Example: If Yoda asked the height of an SRS of 100 Jedi, what is the mean and height of this sampling distribution? 73 inches = 0.28

  16. The Sampling Distribution of x /n Q: What is the shape of the distribution? A: It depends on the population. So, if the population distribution is normal, then so is the distribution of the sample mean. This leads to the following: If a population has the N(, ) distribution, then the sample mean has the N( ) distribution. , In the previous example, the sample mean of the heights of 100 Jedi has the N(73, 0.28) distribution.

  17. The Sampling Distribution of x • So, the sample mean from a normal distribution is normal. • We can extend these ideas to the following : Any linear combination of independent normal random variables is also normally distributed. • In other words, if X and Y are independent normal random • variables, then so is aX + bY. • This means the following are normal: 2X + 7Y, 3X - 78Y, OX + 2Y, X - Y, etc.

  18. Example: Ben and Anakin are playing in the local Jedi golf tournament. Ben’s golf game X has the N(60, 4) distribution. Anakin’s golf game Y has the N(76, 12) distribution. Q: What is the probability that Anakin will score lower than Ben in the tourney? In other words, what is P(Y < X) ? A: Notice that Y - X is a linear combination of two independent random normal variables, so Y - X is normal. What is the mean of the variable Y - X ? Y-X = Y - X = 76 - 60 = 16

  19. 160 Example: Ben and Anakin are playing in the local Jedi golf tournament. Ben’s golf game X has the N(60, 4) distribution. Anakin’s golf game Y has the N(76, 12) distribution. Q: What is the probability that Anakin will score lower than Ben in the tourney? In other words, what is P(Y < X) ? A: Notice that Y - X is a linear combination of two independent random normal variables, so Y - X is normal. What is the standard deviation of the variable Y - X ? 2Y-X = 2Y + 2X = 122 + 42 = 160 Y-X = = 12.65 So, Y - X has the N(16, 12.65) distribution.

  20. = P ( ) (Y - X) - 16 < 0 - 16 12.65 12.65 Example: Ben and Anakin are playing in the local Jedi golf tournament. Ben’s golf game X has the N(60, 4) distribution. Anakin’s golf game Y has the N(76, 12) distribution. Q: What is the probability that Anakin will score lower than Ben in the tourney? In other words, what is P( Y < X) ? So, Y - X has the N(16, 12.65) distribution. P(Y < X) = P( Y - X < 0) = P ( Z < -1.26 ) = .1038 So, Anakin will score lower about 10% of the time.

  21. approximately N (, ) /n Central Limit Theorem • Draw an SRS of size n from any population with mean  • and finite standard deviation .When n is large, the • sampling distribution of the sample mean is : Notes: • The distribution of a sum or average of many small random • quantities is close to normal. • This is true if the quantities are not independent or even • if they have different distributions. • How large does n have to be depends on the distribution.

  22. The Sampling Distribution of x (Revisited) How do we find the sampling distribution ? 1) Take repeated random samples of size n from a population with mean  2) Find the sample mean for each sample 3) Collect all the sample means and display their distribution.

  23. The Sampling Distribution of x (Revisited)

  24. Homework 28, 31, 32, 37, 45

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