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Sampling Distributions Parameter • A number that describes the population • Symbols we will use for parameters include m - mean s – standard deviation p – proportion Statistic • A number that that can be computed from sample data without making use of any unknown parameter • Symbols we will use for statistics include x – mean s – standard deviation p – proportion Identify the boldface values as parameter or statistic. A carload lot of ball bearings has mean diameter 2.5003 cm. This is within the Parameter specifications for acceptance of the lot by the purchaser. By chance, an inspector chooses 100 bearings from the lot that have mean diameter 2.5009 cm. Because Statistic this is outside the specified limits, the lot is mistakenly rejected. Why do we take samples instead of taking a census? • A census is not always accurate. • Census are difficult or impossible to do. • Census are very expensive to do. A distribution is all the values that a variable can be. The sampling distribution of a statistic is the distribution of values taken by the statistic in all possible samples of the same size from the same population. Consider the population – the length of fish (in inches) in my pond - consisting of the values 2, 7, 10, 11, 14 What is the mean mand 8.8 x =standard deviation of this sx = 4.0694 population? Let’s take samples of size 2 (n = 2) from this population: How many samples of size 2 are possible? C = 10 5 2 Find all 10 of these samples and record the sample means. Permutations 7 2 10 14 11 Pairs Pairs x 2, 7 2, 10 2, 11 2, 14 7, 10 7, 11 7, 14 10, 11 10, 14 11, 14 4.5 6 6.5 8 8.5 9 mx = 8.8 sx = 2.4919 10. 10. 12 5 5 What is the mean and standard deviation of the sample means? 12. 5 Repeat this procedure with sample size n = 3 How many samples of size 3 are possible? C = 10 5 mx = 8.8 sx = 3 What mean Find allisofthe these and standard samples and deviation of the record the sample 1.66132 sample means? means. Permutations 2 2 2 2 2 2 11 10 10 7 7 7 14 11 14 10 11 14 7 7 7 10 10 10 11 11 11 14 14 14 Pairs 2,7 2,7, 2,7 2,7, 2,7 7,1 7,1 7,1 7,1 10, 10, 2,1 2,1 2,1 2,1 2,1 2,1 7,1 7,1 Pairs 2,7, x x ,14 0,1 0,1 10,10 11,11 14 11 6.3 6.6 7.6 9.3 1,1 1,1 44 10. 6 mx = 8.8 sx = 1.66132 11, 0,1 0,1 1,1 1,1 11, 14 11 14 44 11. 7.6 9 6 0, 0,1 0,1 0, 14 44 14 8.6 10. 3 What is the mean and standard deviation of the sample means? What do you notice? • The mean of the sampling distribution EQUALS the mean of the population. mx = m • As the sample size increases, the standard deviation of the sampling distribution decreases. as n sx A statistic used to estimate a parameter is unbiased if the mean of its sampling distribution is equal to the true value of the parameter being estimated. General Properties Rule 1: mx = m s Rule 2: sx = n This rule is approximately correct as long as no more than 10% of the population is included in the sample General Properties Rule 3: When the population distribution is normal, the sampling distribution of x is also normal for any sample size n. General Properties Rule 4: Central Limit Theorem When n is sufficiently large, the sampling distribution of x is well approximated by a normal curve, even when the population distribution is not itself normal. CLT can safely be applied if n exceeds 30. If n is large or the population distribution is normal, then x mx x mx z s sx n has approximately a standard normal distribution. Example 1 The army reports that the distribution of head circumference among soldiers is approximately normal with mean 22.8 inches and standard deviation of 1.1 inches. a) What is the probability that a randomly selected soldier’s head will have a circumference that is greater than 23.5 inches? P(X > 23.5) = .2623 Normcdf(23.5,100,22.8,1.1) b) What is the probability that a random sample of five soldiers will have an average head circumference that is greater than 23.5 inches? Do you expect the probability to be moreWhat or lessnormal than the answer curve are to you partnow (a)? working Explainwith? P(X > 23.5) = .0774 Normcdf(23.5,100,22.8, (1.1/sqrt5)) A soft-drink bottler claims that, on average, cans contain 12 oz of soda. Let x denote the actual volume of soda in a randomly selected can. Suppose that x is normally distributed with s = .16 oz. Sixteen cans are to selected with a mean of 12.1 oz. What is the probability that the average of 16 cans will exceed 12.1 oz? P(x >12.1) = .0062 Normcdf(12.1, 100, 12, (0.16/(sqrt 16)) Suppose a team of biologists has been studying the Pinedale children’s fishing pond. Let x represent the length of a single trout taken at random from the pond. This group of biologists has determined that the length has a normal distribution with mean of 10.2 inches and standard deviation of 1.4 inches. What is the probability that a single trout taken at random from the pond is between 8 and 12 inches long? P(8 < X < 12) = .8427 Normcdf(8, 12, 10.2, 1.4) What is the probability that the mean length of five trout taken at random is between 8 and 12 inches long? Do xyou expect the probability to P(8< <12) = .9978 be more or less than the answer to part (a)? Explain What sample mean would be at the 95th percentile? (Assume n = 5) x = 11.23 inches