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AP Statistics Thursday, 22 January 2015 OBJECTIVE TSW investigate sampling distributions. –Everyone needs a calculator. TESTS are not graded. REMINDERS –Sign up for AP Exams by Sunday night, 04 March 2015. –If you plan on taking the AMC test on Tuesday, 03 February 2015, I need to have your money ($2.00) by tomorrow, Friday, 23 January 2015.

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Chapter 8 Sampling Distributions

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Parameter A number that describes the population Symbols we will use for parameters include - mean – standard deviation – proportion (p) – y-intercept of LSRL – slope of LSRL

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Statistic A number 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 a – y-intercept of LSRL b – slope of LSRL

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Identify the boldface values as parameter or statistic. 2.5003 2.5009 A carload lot of ball bearings has mean diameter 2.5003 cm. This is within the 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 this is outside the specified limits, the lot is mistakenly rejected. parameter statistic

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Why do we take samples instead of taking a census? A census is not always accurate. Censuses are difficult or impossible to do. Censuses are very expensive to do.

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distribution A distribution is all the values that a variable can be. (Write this definition above the box on your notes sheet.)

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sampling distribution 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.

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Consider the population – the length of fish (in inches) in my pond - consisting of the values 2, 7, 10, 11, 14 x = 8.8 x = 4.0694 What is the mean and standard deviation of this population?

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Let’s take samples of size 2 (n = 2) from this population: How many samples of size 2 are possible? x = 8.8 inches x = 2.4919 inches 5 C 2 = 10 Find all 10 of these samples and record the sample means. What is the mean and standard deviation of the sample means?

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Repeat this procedure with sample size n = 3 How many samples of size 3 are possible? 5 C 3 = 10 x = 8.8 inches x = 1.66132 inches What is the mean and standard deviation of the sample means? Find all of these samples and record the sample means.

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What do you notice? EQUALSThe mean of the sampling distribution EQUALS the mean of the population. As the sample size increases, the standard deviation of the sampling distribution decreases. x = as n xx

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unbiased 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.

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Rule 1: Rule 2: This rule is approximately correct as long as no more than 5% (10%) of the population is included in the sample. General Properties: x = n x = standard error The standard error of a statistic is the estimated standard deviation of the statistic.

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General Properties Rule 3: When the population distribution is normal, the sampling distribution of x is also normal for any sample size n.

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General Properties Rule 4: 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.

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(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) = normalcdf (23.5, 10 99, 22.8, 1.1) = 0.2623

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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? What normal curve are you now working with? Do you expect the probability to be more or less than the answer to part (a)? Explain P(X > 23.5) = = 0.0774

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If n is large or the population distribution is normal, then has approximately a standard normal distribution.

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(Example 2) 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) = 0.8427

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What is the probability that the mean length of five trout taken at random is between 8 and 12 inches long? What sample mean would be at the 95 th percentile? (Assume n = 5) P(8< x <12) = 0.9978 Do you expect the probability to be more or less than the answer to part (a)? Explain x = 11.23 inches

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(Example 3) 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 = 0.16 oz. Sixteen cans are selected with a mean of 12.1 oz. What is the probability that the average of 16 cans will exceed 12.1 oz? Do you think the bottler’s claim is correct? No, since it is not likely to happen by chance alone & the sample did have this mean, I do not think the claim that the average of 12 oz. is correct. P(x >12.1) = 0.0062

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(Example 4) A hot dog manufacturer asserts that one of its brands of hot dogs has an average fat content of 18 grams per hot dog with standard deviation of 1 gram. Consumers of this brand would probably not be disturbed if the mean was less than 18 grams, but would be unhappy if it exceeded 18 grams. An independent testing organization is asked to analyze a random sample of 36 hot dogs. Suppose the resulting sample mean is 18.4 grams. Does this result indicate that the manufacturer’s claim is incorrect? What if the sample mean was 18.2 grams, would you think the claim was incorrect? Yes, not likely to happen by chance alone. No

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Assignment WS Sample Means –Due tomorrow, Friday, 23 January 2015.

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