* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download Document
Survey
Document related concepts
Transcript
Chapter 8 Sampling Variability and Sampling Distributions Suppose we are interested in finding the true mean (m) fat content of quarter-pound hamburgers marketed by a national fast food chain. To learn something about m, we could answer these weand will obtain a To sample of n = 50 questions, hamburgers examine the sampling distribution, determine the fat content of each one. which describes the long-run behavior of sample statistic. Recall that the sample mean is a statistic Statistic • A number that that can be computed from sample data • Some statistics we will use include x – sample mean s – standard deviation p – sample proportion This variability is called sampling variability • The observed value of the statistic depends on the particular sample selected from the population and it will vary from sample to sample. The campus of Wolf City College has a fish pond. Suppose there are 20 fish in the pond. The lengths of the fish (in inches) are given below: 4.5 5.4 10.3 7.9 6.3 4.3 9.6 8.5 6.6 11.7 8.9 2.2 9.8 8.7 13.3 4.6 10.7 13.4 7.7 5.6 This is a statistic! We caught fish with lengths 6.3 The true mean m = 8. Let’s catch inches,Suppose 2.2 inches, and 13.3 inches. Notice that some we randomly catch a sample ofan This is two more x = 7.27 inches sample means are example of 3 fish from this pond and measure their samples and closer and some 2nd sample 8.5, 4.6, and 5.6 inches. sampling length. What would the mean length of farther away; some look at the x = 6.23 inches variability above and some below the sample be? sample means. rd 3 sample – 10.3, 8.9, and 13.4 inches.the mean. x = 10.87 inches Fish Pond Continued . . . 4.5 5.4 10.3 7.9 6.3 4.3 9.6 8.5 6.6 11.7 8.9 2.2 9.8 8.7 13.3 4.6 10.7 13.4 7.7 5.6 There are 1140 (20C3) different possible samples of size 3 from this population. If we were to catch all those different samples and calculate the mean length of each sample, we would have a distribution of all possible x. This would be the sampling distribution of x. Sampling Distributions of x • The distribution that would be formed by considering the value of a sample statistic for every possible different sample of a given size from a population. In this case, the sample statistic is the sample mean x. Fish Pond Revisited . . . Suppose there are only 5 fish in the pond. The lengths of the fish (in inches) are given below: 6.6 11.7 8.9 2.2 What is the mean mand 7.84 x = standard deviation of this sxpopulation? = 3.262 9.8 We will keep the population size small so that we can find ALL the possible samples. Fish Pond Revisited . . . 6.6 11.7 8.9 2.2 9.8 mx = 7.84 and sx = 3.262 Pairs 6.6 & 11.7 6.6 & 8.9 6.6 & 2.2 6.6 & 9.8 x 9.15 7.75 4.4 8.2 11.7 & 8.9 11.7 & 2.2 11.7 & 9.8 8.9 & 2.2 8.9 & 9.8 2.2 & 9.8 find all5.55 the 10.3Let’s 6.95 10.75 9.35 6 How many samples samplesofofsize size22. are possible? These values determine the How doisthese What the mean mx = 7.84 sampling distribution of x for valuesand compare to standard samples of size 2. the population deviation of these sx = 1.998 meansample and standard means? deviation? Fish Pond Revisited . . . 6.6 11.7 8.9 2.2 9.8 mx = 7.84 and sx = 3.262 Triples x 11.7, 8.9, Now6.6,let’s11.7, find 11.7, all the 2.2, many 8.9, 8.9, 2.2, 2.2, How samples samples of size 3. 9.8 2.2 9.8 9.8 9.8 of size 310.133 are 7.9 6.967 9.067 6.833 9.367 5.9 8.433 6.2 7.6 possible? These values determine the What is the mean mx = 7.84 sampling distribution of xHow fordo values andthese standard samples of size 3. compare to deviation ofthe these sx = 1.332 population and samplemean means? standard deviation? 6.6, 11.7, 8.9 6.6, 11.7, 2.2 6.6, 11.7, 9.8 6.6, 8.9, 2.2 6.6, 8.9, 9.8 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 General Properties of Sampling Distributions of x Rule 1: Rule 2: mx m sx s n Note that in the previous fish pond examples this standard deviation formula was not correct because the sample sizes were more than 10% of the population. This rule is exact if the population is infinite, and is approximately correct if the population is finite and no more than 10% of the population is included in the sample The paper “Mean Platelet Volume in Patients with Metabolic Syndrome and Its Relationship with Coronary Artery Disease” (Thrombosis Research, 2007) includes data that suggests that the distribution of platelet volume of use patients whotodogenerate not haverandom metabolic syndrome We can Minitab samples from population.normal We will generate random samples is this approximately with mean m500 = 8.25 and standard of n s= =5 0.75. and compute the sample mean for each. deviation Platelets Continued . . . Similarly, we will generate 500 random samples of n = 10, n = 20, and n = 30. The density histograms below display the resulting 500 x for each of the given sample sizes. What do do you you notice notice What do you notice What about the themeans standard about the shape of about deviation of these these histograms? these histograms? histograms? General Properties Continued . . . Rule 3: When the population distribution is normal, the sampling distribution of x is also normal for any sample size n. The paper “Is the Overtime Period in an NHL Game Long Enough?” (American Statistician, 2008) gave data on the time (in minutes) from the start of the game to the first goal scored for the 281 regular season games from the 2005-2006 season that went into overtime. The density histogram for the data is shown below. Let’s consider these 281 values as a population. The distribution is strongly positively skewed with mean m = 13 minutes and with a median of 10Using minutes. Minitab, we will generate 500 samples of the following sample sizes from this distribution: n = 5, n = 10, n = 20, n = 30. What do you notice These Are these are histograms What dothe youdensity notice about the standard histograms centered for the at 500 about the shape of deviations of these samples approximately m = 13? these histograms? histograms? General Properties Continued . . . 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. How large is “sufficiently large” CLT can safely be anyway? applied if n exceeds 30. 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 randomly selected, and the soda volume is determined for each one. Let x = the resulting sample mean soda. If the bottler’s claim is correct, then the sampling distribution of x is normally distributed with: m x m 12 s .16 sx .04 n 16 Soda Problem Continued . . . m x m 12 s .16 sx .04 To standardized these n 16 endpoints, use x in mthe x mean m What is the probability that the Look these up table x sample z soda volume is between and 12.08 and 11.96 subtract thes sxounces ounces? probabilities. n P(11.96 < x < 12.08) = .9772 - .1587 = .8185 11.96 12 a* 1 .04 12.08 12 b* 2 .04 A hot dog manufacturer asserts that one of its brands of hot dogs has a 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 Since the sample size is result suggest that the manufacturer’s claim is greater than 30, the incorrect? Central Limit Theorem applies. So the distribution of x is approximately normal with mx 18 and sx 1 .1667 36 Hot Dogs Continued . . . mx 18 and sx 1 .1667 36 Suppose the resulting sample mean is 18.4 grams. Does this result suggest that the manufacturer’s claim is incorrect? P(x > 18.4) = 1 - .9918 = .0082 Values of x at least as large as 18.4 would be observed 18.4 18 only about .82% of the time. z 2.40 The sample .1667 mean of 18.4 is large enough to cause us to doubt that the manufacturer’s claim is correct. Let’s explore what happens with in distributions of sample proportions (p). Have students perform the following experiment. This is a statistic! •Toss a penny 20 times and record the number of heads. •Calculate the proportion of heads and mark it on The dotplot is a partial graph of the What would happen to the dotplot if we the dot plot on distribution the sampling oftimes all sample flipped theboard. penny 50 and proportions sample size 20. recorded the of proportion of heads? What shape do you think the dot plot will have? Sampling Distribution of p The distribution that would be formed by considering the value of a sample statistic for every possible different We sample will use:of a given size from a population. p for the population proportion and p for the sample proportion In this case, we will use number of successes in the sample pˆ n Suppose we have a population of six students: Alice, Ben, Charles, Denise, Edward, & Frank will keepinthe We are We interested thepopulation proportionsmall of females. so that weparameter can find ALL the This is called the of interest possible samples of a given size. What is the proportion of females? 1/3 Let’s select samples of two from this population. How many different samples are possible? 6C2 =15 Find the 15 different samples that are possible and find the sample proportion of the number of females in each sample. Ben & Frank Alice & Ben .5 Charles & Denise Alice & Charles .5 Charles & Edward Alice & Denise 1 Charles & Frank Alice & Edward .5 Denise of & Edward the mean the Alice & Frank How does .5 Denise & Frank Ben & Charles 0 sampling distribution & Frank Ben & Denise compare .5 to theEdward population Ben & Edward 0 parameter (p)? 0 .5 0 0 .5 .5 0 Find the mean and standard deviation of these sample proportions. 1 m pˆ and s pˆ 0.29814 3 General Properties for Sampling Distributions of p Rule 1: m pˆ p Rule 2: s pˆ p (1 p ) n Note that in the previous student example this standard deviation formula was not correct because the sample size was more than 10% of the population. This rule is exact if the population is infinite, and is approximately correct if the population is finite and no more than 10% of the population is included in the sample In the fall of 2008, there were 18,516 students enrolled at California Polytechnic State University, San Luis Obispo. Of these students, 8091 (43.7%) were female. We will use a statistical software package to simulate sampling from this Cal Poly population. We will generate 500 samples of each of the following sample sizes: n = 10, n = 25, n = 50, n = 100 and compute the proportion of females for each sample. The following histograms display the distributions of the sample proportions for the 500 samples of each sample size. What do notice What do you youhistograms notice about about Are these thethe standard deviation of shape of the these centered around true these distributions? distributions? proportion p = .437? The development of viral hepatitis after a blood transfusion can cause serious complications for a patient. The article “Lack of Awareness Results in Poor Autologous Blood Transfusions” (Health Care Management, May 15, 2003) reported that hepatitis occurs in 7% of patients who receive blood transfusions during heart surgery. We will simulate sampling from a population of blood recipients. We will generate 500 samples of each of the following sample sizes: n = 10, n = 25, n = 50, n = 100 and compute the proportion of people who contract hepatitis for each sample. The following histograms display the distributions of the sample proportions for the 500 samples of each sample size. Are these histogram s centered around the true proportion p = .07? What happens to the shape of these histograms as the sample size increases? General Properties Continued . . . Rule 3: When n is large and p is not too near 0 or 1, the sampling distribution of p is approximately normal. The farther the value of p is from 0.5, the larger n must be for the sampling distribution of p to be approximately normal. A conservative rule of thumb: If np > 10 and n (1 – p) > 10, then a normal distribution provides a reasonable approximation to the sampling distribution of p. Blood Transfusions Revisited . . . Let p = proportion of patients who contract hepatitis after a blood transfusion p = .07 Suppose a new blood procedure is we Toscreening answer this question, believed to reduce the incident rate hepatitis. must consider theofsampling Blood screened using thisdistribution procedure is ofgiven p. to n = 200 blood recipients. Only 6 of the 200 patients contract hepatitis. Does this result indicate that the true proportion of patients who contract hepatitis when the new screening is used is less than 7%? Blood Transfusions Revisited . . . Let p = .07 p = 6/200 = .03 Is the sampling distribution approximately normal? Yes, we can np = 200(.07) = 14 > 10 use a normal n(1-p) = 200(.93) = 186 > 10 approximation. What is the mean and standard deviation of the sampling distribution? m pˆ .07 .07(.93) s pˆ .018 200 Blood Transfusions Revisited . . . m pˆ .07 Let p = .07 p = 6/200 = .03 .07(.93) s pˆ .018 200 Does this result indicate that the true proportion of patients who contract hepatitis when the new screening is usedThis is less 7%? smallthan probability tells us screening that it is This new P(p < .03) = .0132 unlikely that a sample procedure appears proportion of .03 or toscreening yield awould smaller .03 .07 Assume the smaller be z 2.22 incidence for procedure observed is notrate if the .07(.93) screening procedure hepatitis. effective and p = .07. 200 was ineffective.