Mean Diameter of Ball Bearings in a Hypothesis Test - Prof. N. Phillips, Study notes of Probability and Statistics

Download Mean Diameter of Ball Bearings in a Hypothesis Test - Prof. N. Phillips and more Study notes Probability and Statistics in PDF only on Docsity! Math 243: Lecture File 9 N. Christopher Phillips 28 April 2009 N. Christopher Phillips () Math 243: Lecture File 9 28 April 2009 1 / 60 How big does the sample have to be? Example: As before, spiral-horned snorkacks have horn lengths (in cm) distributed N(µ, 12) with µ unknown. A simple random sample of 100 spiral-horned snorkacks has sample mean horn length 121.502 (again, in cm). We found a 95% confidence interval for the true mean horn length of spiral-horned snorkacks. The margin of error for a confidence interval is m = z∗ ( σ√ n ) . For this case, we had σ = 12, x = 121.502, n = 100, and z∗ ≈ 1.960. We got m = (1.960) ( 12√ 100 ) = (1.960)(1.2) ≈ 2.352. The resulting confidence interval was 121.502± 2.352. N. Christopher Phillips () Math 243: Lecture File 9 28 April 2009 2 / 60 How big does the sample have to be? (continued) Your boss isn’t satisfied. He wants a margin of error of at most 2.000 for the 95% confidence interval. What to do? First, the sample size n must be larger. This reduces the sampling standard deviation σ/ √ n. Caution: In situations with unknown population standard deviation σ, we will have to use the standard deviation s of the sample instead of σ (and make other changes—we will do this soon). Since the standard deviation of the sample depends on the sample, you are not guaranteed that larger sample sizes reduce s/ √ n. But it usually happens. N. Christopher Phillips () Math 243: Lecture File 9 28 April 2009 3 / 60 How big does the sample have to be? (continued) Your boss isn’t satisfied. He wants a margin of error of at most 2.000 for the 95% confidence interval. What to do? First, the sample size n must be larger. This reduces the sampling standard deviation σ/ √ n. Caution: In situations with unknown population standard deviation σ, we will have to use the standard deviation s of the sample instead of σ (and make other changes—we will do this soon). Since the standard deviation of the sample depends on the sample, you are not guaranteed that larger sample sizes reduce s/ √ n. But it usually happens. N. Christopher Phillips () Math 243: Lecture File 9 28 April 2009 3 / 60 How big does the sample have to be? (continued) Your boss isn’t satisfied. He wants a margin of error of at most 2.000 for the 95% confidence interval. What to do? First, the sample size n must be larger. This reduces the sampling standard deviation σ/ √ n. Caution: In situations with unknown population standard deviation σ, we will have to use the standard deviation s of the sample instead of σ (and make other changes—we will do this soon). Since the standard deviation of the sample depends on the sample, you are not guaranteed that larger sample sizes reduce s/ √ n. But it usually happens. N. Christopher Phillips () Math 243: Lecture File 9 28 April 2009 3 / 60 How big does the sample have to be? (continued) The margin of error for a confidence interval is m = z∗ ( σ√ n ) . We had σ = 12, x = 121.502, n = 100, and z∗ ≈ 1.960. We want a margin of error of at most 2.000 for the 95% confidence interval. Substitute 2.000 for m, and the known values for σ and z∗: 2.000 ≈ (1.960) ( 12√ n ) . Now solve for n: (2.000) √ n ≈ (1.960)(12) √ n ≈ (1.960)(12) 2.000 n ≈ ( (1.960)(12) 2.000 )2 ≈ 138.3. N. Christopher Phillips () Math 243: Lecture File 9 28 April 2009 4 / 60 How big does the sample have to be? (continued) The margin of error for a confidence interval is m = z∗ ( σ√ n ) . We had σ = 12, x = 121.502, n = 100, and z∗ ≈ 1.960. We want a margin of error of at most 2.000 for the 95% confidence interval. Substitute 2.000 for m, and the known values for σ and z∗: 2.000 ≈ (1.960) ( 12√ n ) . Now solve for n: (2.000) √ n ≈ (1.960)(12) √ n ≈ (1.960)(12) 2.000 n ≈ ( (1.960)(12) 2.000 )2 ≈ 138.3. N. Christopher Phillips () Math 243: Lecture File 9 28 April 2009 4 / 60 How big does the sample have to be? (continued) The margin of error for a confidence interval is m = z∗ ( σ√ n ) . We had σ = 12, x = 121.502, n = 100, and z∗ ≈ 1.960. We want a margin of error of at most 2.000 for the 95% confidence interval. Substitute 2.000 for m, and the known values for σ and z∗: 2.000 ≈ (1.960) ( 12√ n ) . Now solve for n: (2.000) √ n ≈ (1.960)(12) √ n ≈ (1.960)(12) 2.000 n ≈ ( (1.960)(12) 2.000 )2 ≈ 138.3. N. Christopher Phillips () Math 243: Lecture File 9 28 April 2009 4 / 60 How big does the sample have to be? (continued) Substitute 2.000 for m, and the known values for σ and z∗: 2.000 ≈ (1.960) ( 12√ n ) . Now solve for n: (2.000) √ n ≈ (1.960)(12) √ n ≈ (1.960)(12) 2.000 n ≈ ( (1.960)(12) 2.000 )2 ≈ 138.3. I have rounded to four significant digits, since that is all we have for z∗. N. Christopher Phillips () Math 243: Lecture File 9 28 April 2009 5 / 60 How big does the sample have to be? (continued) Substitute 2.000 for m, and the known values for σ and z∗: 2.000 ≈ (1.960) ( 12√ n ) . Now solve for n: (2.000) √ n ≈ (1.960)(12) √ n ≈ (1.960)(12) 2.000 n ≈ ( (1.960)(12) 2.000 )2 ≈ 138.3. I have rounded to four significant digits, since that is all we have for z∗. N. Christopher Phillips () Math 243: Lecture File 9 28 April 2009 5 / 60 How big does the sample have to be? (continued) Substitute 2.000 for m, and the known values for σ and z∗: 2.000 ≈ (1.960) ( 12√ n ) . Now solve for n: (2.000) √ n ≈ (1.960)(12) √ n ≈ (1.960)(12) 2.000 n ≈ ( (1.960)(12) 2.000 )2 ≈ 138.3. I have rounded to four significant digits, since that is all we have for z∗. N. Christopher Phillips () Math 243: Lecture File 9 28 April 2009 5 / 60 How big does the sample have to be? (continued) We want a margin of error of at most 2.000 for the 95% confidence interval. We got n ≈ 138.3. What is wrong with this? Even in the world of Harry Potter, it seems unlikely that one can choose a simple random sample of 138.3 spiral-horned snorkacks, and measure 138.3 horn lengths. The sample size must be an integer, and must give a margin of error of at most 2.000, so we choose the next larger integer. So actually take n = 139. (Do not round down, even if the number you get is closer.) N. Christopher Phillips () Math 243: Lecture File 9 28 April 2009 6 / 60 How big does the sample have to be? (continued) We want a margin of error of at most 2.000 for the 95% confidence interval. We got n ≈ 138.3. What is wrong with this? Even in the world of Harry Potter, it seems unlikely that one can choose a simple random sample of 138.3 spiral-horned snorkacks, and measure 138.3 horn lengths. The sample size must be an integer, and must give a margin of error of at most 2.000, so we choose the next larger integer. So actually take n = 139. (Do not round down, even if the number you get is closer.) N. Christopher Phillips () Math 243: Lecture File 9 28 April 2009 6 / 60 How big does the sample have to be? (continued) We want a margin of error of at most 2.000 for the 95% confidence interval. We got n ≈ 138.3. What is wrong with this? Even in the world of Harry Potter, it seems unlikely that one can choose a simple random sample of 138.3 spiral-horned snorkacks, and measure 138.3 horn lengths. The sample size must be an integer, and must give a margin of error of at most 2.000, so we choose the next larger integer. So actually take n = 139. (Do not round down, even if the number you get is closer.) N. Christopher Phillips () Math 243: Lecture File 9 28 April 2009 6 / 60 How big does the sample have to be? (general case) The margin of error for a confidence interval is m = z∗ ( σ√ n ) . Solve for n: m √ n = z∗σ √ n = z∗σ m n = ( z∗σ m )2 . If you use this formula, you must remember not to use the value of n it gives. Always round up. You do not need to memorize this formula. Just solve for n in the equation for the margin of error. N. Christopher Phillips () Math 243: Lecture File 9 28 April 2009 7 / 60 How big does the sample have to be? (general case) The margin of error for a confidence interval is m = z∗ ( σ√ n ) . Solve for n: m √ n = z∗σ √ n = z∗σ m n = ( z∗σ m )2 . If you use this formula, you must remember not to use the value of n it gives. Always round up. You do not need to memorize this formula. Just solve for n in the equation for the margin of error. N. Christopher Phillips () Math 243: Lecture File 9 28 April 2009 7 / 60 How big does the sample have to be? (general case) The margin of error for a confidence interval is m = z∗ ( σ√ n ) . Solve for n: m √ n = z∗σ √ n = z∗σ m n = ( z∗σ m )2 . If you use this formula, you must remember not to use the value of n it gives. Always round up. You do not need to memorize this formula. Just solve for n in the equation for the margin of error. N. Christopher Phillips () Math 243: Lecture File 9 28 April 2009 7 / 60 How big does the sample have to be? Another example. You have a scientific instrument which measures the amount of uranium hexaflouride in air samples. For any given air sample, the results of using this instrument vary normally with mean the true amount of uranium hexaflouride in the air sample and standard deviation 12.5 mg (milligrams). How many measurements do you need to make to get a 99% confidence interval with margin of error 3 mg? From Table C, z∗ ≈ 2.576. m = z∗ ( σ√ n ) . 3 ≈ (2.576) ( 12.5√ n ) 3 √ n ≈ (2.576)(12.5) N. Christopher Phillips () Math 243: Lecture File 9 28 April 2009 8 / 60 How big does the sample have to be? Another example. You have a scientific instrument which measures the amount of uranium hexaflouride in air samples. For any given air sample, the results of using this instrument vary normally with mean the true amount of uranium hexaflouride in the air sample and standard deviation 12.5 mg (milligrams). How many measurements do you need to make to get a 99% confidence interval with margin of error 3 mg? From Table C, z∗ ≈ 2.576. m = z∗ ( σ√ n ) . 3 ≈ (2.576) ( 12.5√ n ) 3 √ n ≈ (2.576)(12.5) N. Christopher Phillips () Math 243: Lecture File 9 28 April 2009 8 / 60 How big does the sample have to be? Another example. You have a scientific instrument which measures the amount of uranium hexaflouride in air samples. For any given air sample, the results of using this instrument vary normally with mean the true amount of uranium hexaflouride in the air sample and standard deviation 12.5 mg (milligrams). How many measurements do you need to make to get a 99% confidence interval with margin of error 3 mg? From Table C, z∗ ≈ 2.576. m = z∗ ( σ√ n ) . 3 ≈ (2.576) ( 12.5√ n ) 3 √ n ≈ (2.576)(12.5) N. Christopher Phillips () Math 243: Lecture File 9 28 April 2009 8 / 60 How big does the sample have to be? Another example. m = z∗ ( σ√ n ) . 3 ≈ (2.576) ( 12.5√ n ) 3 √ n ≈ (2.576)(12.5) √ n ≈ (2.576)(12.5) 3 n ≈ ( (2.576)(12.5) 3 )2 ≈ 115.2. So actually take n = 116. This is a rather large number of repetitions of a measurement. N. Christopher Phillips () Math 243: Lecture File 9 28 April 2009 9 / 60 How big does the sample have to be? Another example. m = z∗ ( σ√ n ) . 3 ≈ (2.576) ( 12.5√ n ) 3 √ n ≈ (2.576)(12.5) √ n ≈ (2.576)(12.5) 3 n ≈ ( (2.576)(12.5) 3 )2 ≈ 115.2. So actually take n = 116. This is a rather large number of repetitions of a measurement. N. Christopher Phillips () Math 243: Lecture File 9 28 April 2009 9 / 60 How big does the sample have to be? Another example. m = z∗ ( σ√ n ) . 3 ≈ (2.576) ( 12.5√ n ) 3 √ n ≈ (2.576)(12.5) √ n ≈ (2.576)(12.5) 3 n ≈ ( (2.576)(12.5) 3 )2 ≈ 115.2. So actually take n = 116. This is a rather large number of repetitions of a measurement. N. Christopher Phillips () Math 243: Lecture File 9 28 April 2009 9 / 60 Hypotheses Examples of questions: Example 1: We know US women of ages 20–29 years have mean height 64 inches. Are women in this age group in Eugene taller on average? Example 2: Do women in this age group in Eugene differ, on average, from those in the general population? Example 3: Are female intercollegiate soccer players taller than average? N. Christopher Phillips () Math 243: Lecture File 9 28 April 2009 10 / 60 Hypotheses Examples of questions: Example 1: We know US women of ages 20–29 years have mean height 64 inches. Are women in this age group in Eugene taller on average? Example 2: Do women in this age group in Eugene differ, on average, from those in the general population? Example 3: Are female intercollegiate soccer players taller than average? N. Christopher Phillips () Math 243: Lecture File 9 28 April 2009 10 / 60 A warning As we will see, we will not detect very small differences. For example, it is almost certainly true that the mean height of Eugene women of ages 20–29 years is not the same as the mean height of US women of ages 20–29 years. However, if the difference is small enough, we will not detect it. N. Christopher Phillips () Math 243: Lecture File 9 28 April 2009 11 / 60 Examples of questions (continued) Example 4: Does being left handed affect IQ? Example 5: Suppose we know the mean weight of spiral-horned snorkacks. Are crumple-horned snorkacks, on average, smaller (lighter)? Example 6: You suspect that the Consolidated Generic Unhealthy Foods Corporation is, on average, underfilling its 12 ounce bags of potato chips. Example 7: You have bought a large quantity of 7 mm ball bearings from Wang’s Widgets Inc., and you are preparing to sue Wang’s Widgets Inc. because, you claim, the diameter of the balls in the ball bearings is not as advertised. N. Christopher Phillips () Math 243: Lecture File 9 28 April 2009 12 / 60 Examples of questions (continued) Example 4: Does being left handed affect IQ? Example 5: Suppose we know the mean weight of spiral-horned snorkacks. Are crumple-horned snorkacks, on average, smaller (lighter)? Example 6: You suspect that the Consolidated Generic Unhealthy Foods Corporation is, on average, underfilling its 12 ounce bags of potato chips. Example 7: You have bought a large quantity of 7 mm ball bearings from Wang’s Widgets Inc., and you are preparing to sue Wang’s Widgets Inc. because, you claim, the diameter of the balls in the ball bearings is not as advertised. N. Christopher Phillips () Math 243: Lecture File 9 28 April 2009 12 / 60 Examples of questions (continued) Example 4: Does being left handed affect IQ? Example 5: Suppose we know the mean weight of spiral-horned snorkacks. Are crumple-horned snorkacks, on average, smaller (lighter)? Example 6: You suspect that the Consolidated Generic Unhealthy Foods Corporation is, on average, underfilling its 12 ounce bags of potato chips. Example 7: You have bought a large quantity of 7 mm ball bearings from Wang’s Widgets Inc., and you are preparing to sue Wang’s Widgets Inc. because, you claim, the diameter of the balls in the ball bearings is not as advertised. N. Christopher Phillips () Math 243: Lecture File 9 28 April 2009 12 / 60 The basic idea We choose, for example, a simple random sample of balls in 7 mm ball bearings from Wang’s Widgets Inc., and find their mean diameter x (in mm). If x is far enough away from 7, then we interpret this as evidence that the true mean diameter µ is different from 7. Suppose, for example, that the standard deviation of the diameters of the balls is known to be 0.1 (in mm), and that we chose a simple random sample of size 100. Suppose we got x = 7.03. The sampling standard deviation is σ/ √ n = 0.1/ √ 100 = 0.01. If the true mean were 7, it would be unlikely that we got a sample mean as far away as x = 7.03. Indeed, the probability that the sample mean x is in (6.997, 7.003) is (by the Rule of Thumb) about 0.997, leaving only a very small probability of about 0.003 that x ≤ 6.97 or x ≥ 7.03. We interpret this as strong evidence that the true mean µ was not, after all, equal to 7. N. Christopher Phillips () Math 243: Lecture File 9 28 April 2009 13 / 60 The basic idea We choose, for example, a simple random sample of balls in 7 mm ball bearings from Wang’s Widgets Inc., and find their mean diameter x (in mm). If x is far enough away from 7, then we interpret this as evidence that the true mean diameter µ is different from 7. Suppose, for example, that the standard deviation of the diameters of the balls is known to be 0.1 (in mm), and that we chose a simple random sample of size 100. Suppose we got x = 7.03. The sampling standard deviation is σ/ √ n = 0.1/ √ 100 = 0.01. If the true mean were 7, it would be unlikely that we got a sample mean as far away as x = 7.03. Indeed, the probability that the sample mean x is in (6.997, 7.003) is (by the Rule of Thumb) about 0.997, leaving only a very small probability of about 0.003 that x ≤ 6.97 or x ≥ 7.03. We interpret this as strong evidence that the true mean µ was not, after all, equal to 7. N. Christopher Phillips () Math 243: Lecture File 9 28 April 2009 13 / 60 The basic idea We choose, for example, a simple random sample of balls in 7 mm ball bearings from Wang’s Widgets Inc., and find their mean diameter x (in mm). If x is far enough away from 7, then we interpret this as evidence that the true mean diameter µ is different from 7. Suppose, for example, that the standard deviation of the diameters of the balls is known to be 0.1 (in mm), and that we chose a simple random sample of size 100. Suppose we got x = 7.03. The sampling standard deviation is σ/ √ n = 0.1/ √ 100 = 0.01. If the true mean were 7, it would be unlikely that we got a sample mean as far away as x = 7.03. Indeed, the probability that the sample mean x is in (6.997, 7.003) is (by the Rule of Thumb) about 0.997, leaving only a very small probability of about 0.003 that x ≤ 6.97 or x ≥ 7.03. We interpret this as strong evidence that the true mean µ was not, after all, equal to 7. N. Christopher Phillips () Math 243: Lecture File 9 28 April 2009 13 / 60 The basic idea We choose, for example, a simple random sample of balls in 7 mm ball bearings from Wang’s Widgets Inc., and find their mean diameter x (in mm). If x is far enough away from 7, then we interpret this as evidence that the true mean diameter µ is different from 7. Suppose, for example, that the standard deviation of the diameters of the balls is known to be 0.1 (in mm), and that we chose a simple random sample of size 100. Suppose we got x = 7.03. The sampling standard deviation is σ/ √ n = 0.1/ √ 100 = 0.01. If the true mean were 7, it would be unlikely that we got a sample mean as far away as x = 7.03. Indeed, the probability that the sample mean x is in (6.997, 7.003) is (by the Rule of Thumb) about 0.997, leaving only a very small probability of about 0.003 that x ≤ 6.97 or x ≥ 7.03. We interpret this as strong evidence that the true mean µ was not, after all, equal to 7. N. Christopher Phillips () Math 243: Lecture File 9 28 April 2009 13 / 60 The basic idea (continued) As before, suppose that the standard deviation of the diameters of balls in 7 mm ball bearings from Wang’s Widgets Inc. is known to be 0.1 (in mm), and that we chose a simple random sample of size 100. Suppose instead we got x = 7.003. The sampling standard deviation is σ/ √ n = 0.1/ √ 100 = 0.01. If the true mean were 7, our x would be only 0.3 sampling standard deviations away from the true mean. Using Table A (details omitted for now, but see below), I found that the probability that x is that far, or farther, from µ is about 0.7642. Accordingly, this observed x would not be unreasonable is the true mean were 7. We don’t have good evidence on which to base a lawsuit against Wang’s Widgets Inc. Caution: We don’t conclude that the true mean is 7! (Maybe we were extremely lucky with our sample, and the true mean is equal to 7.003.) All we conclude is that we didn’t find good evidence that the true mean isn’t 7. N. Christopher Phillips () Math 243: Lecture File 9 28 April 2009 14 / 60 The basic idea (continued) As before, suppose that the standard deviation of the diameters of balls in 7 mm ball bearings from Wang’s Widgets Inc. is known to be 0.1 (in mm), and that we chose a simple random sample of size 100. Suppose instead we got x = 7.003. The sampling standard deviation is σ/ √ n = 0.1/ √ 100 = 0.01. If the true mean were 7, our x would be only 0.3 sampling standard deviations away from the true mean. Using Table A (details omitted for now, but see below), I found that the probability that x is that far, or farther, from µ is about 0.7642. Accordingly, this observed x would not be unreasonable is the true mean were 7. We don’t have good evidence on which to base a lawsuit against Wang’s Widgets Inc. Caution: We don’t conclude that the true mean is 7! (Maybe we were extremely lucky with our sample, and the true mean is equal to 7.003.) All we conclude is that we didn’t find good evidence that the true mean isn’t 7. N. Christopher Phillips () Math 243: Lecture File 9 28 April 2009 14 / 60 The basic idea (continued) As before, suppose that the standard deviation of the diameters of balls in 7 mm ball bearings from Wang’s Widgets Inc. is known to be 0.1 (in mm), and that we chose a simple random sample of size 100. Suppose instead we got x = 7.003. The sampling standard deviation is σ/ √ n = 0.1/ √ 100 = 0.01. If the true mean were 7, our x would be only 0.3 sampling standard deviations away from the true mean. Using Table A (details omitted for now, but see below), I found that the probability that x is that far, or farther, from µ is about 0.7642. Accordingly, this observed x would not be unreasonable is the true mean were 7. We don’t have good evidence on which to base a lawsuit against Wang’s Widgets Inc. Caution: We don’t conclude that the true mean is 7! (Maybe we were extremely lucky with our sample, and the true mean is equal to 7.003.) All we conclude is that we didn’t find good evidence that the true mean isn’t 7. N. Christopher Phillips () Math 243: Lecture File 9 28 April 2009 14 / 60 The basic idea (continued) As before, suppose that the standard deviation of the diameters of balls in 7 mm ball bearings from Wang’s Widgets Inc. is known to be 0.1 (in mm), and that we chose a simple random sample of size 100. Suppose instead we got x = 7.003. The sampling standard deviation is σ/ √ n = 0.1/ √ 100 = 0.01. If the true mean were 7, our x would be only 0.3 sampling standard deviations away from the true mean. Using Table A (details omitted for now, but see below), I found that the probability that x is that far, or farther, from µ is about 0.7642. Accordingly, this observed x would not be unreasonable is the true mean were 7. We don’t have good evidence on which to base a lawsuit against Wang’s Widgets Inc. Caution: We don’t conclude that the true mean is 7! (Maybe we were extremely lucky with our sample, and the true mean is equal to 7.003.) All we conclude is that we didn’t find good evidence that the true mean isn’t 7. N. Christopher Phillips () Math 243: Lecture File 9 28 April 2009 14 / 60 The basic idea (continued) As before, suppose that the standard deviation of the diameters of balls in 7 mm ball bearings from Wang’s Widgets Inc. is known to be 0.1 (in mm), and that we chose a simple random sample of size 100. Suppose instead we got x = 7.003. The sampling standard deviation is σ/ √ n = 0.1/ √ 100 = 0.01. If the true mean were 7, our x would be only 0.3 sampling standard deviations away from the true mean. Using Table A (details omitted for now, but see below), I found that the probability that x is that far, or farther, from µ is about 0.7642. Accordingly, this observed x would not be unreasonable is the true mean were 7. We don’t have good evidence on which to base a lawsuit against Wang’s Widgets Inc. Caution: We don’t conclude that the true mean is 7! (Maybe we were extremely lucky with our sample, and the true mean is equal to 7.003.) All we conclude is that we didn’t find good evidence that the true mean isn’t 7. N. Christopher Phillips () Math 243: Lecture File 9 28 April 2009 14 / 60 The formal setup: Hypotheses We operate in terms of two hypotheses. The null hypothesis H0 is the statement that there is no difference, no change, etc., as appropriate. You are seeking evidence against this hypothesis. The alternative hypothesis Ha is the statement that there is a difference. There two forms: one sided and two sided. Both hypotheses make assertions about the population. N. Christopher Phillips () Math 243: Lecture File 9 28 April 2009 15 / 60 The formal setup: Hypotheses We operate in terms of two hypotheses. The null hypothesis H0 is the statement that there is no difference, no change, etc., as appropriate. You are seeking evidence against this hypothesis. The alternative hypothesis Ha is the statement that there is a difference. There two forms: one sided and two sided. Both hypotheses make assertions about the population. N. Christopher Phillips () Math 243: Lecture File 9 28 April 2009 15 / 60 The formal setup: Hypotheses We operate in terms of two hypotheses. The null hypothesis H0 is the statement that there is no difference, no change, etc., as appropriate. You are seeking evidence against this hypothesis. The alternative hypothesis Ha is the statement that there is a difference. There two forms: one sided and two sided. Both hypotheses make assertions about the population. N. Christopher Phillips () Math 243: Lecture File 9 28 April 2009 15 / 60 Example 1 revisited We know US women of ages 20–29 years have mean height 64 inches. We want to know if women in Eugene of ages 20–29 years are on average taller than those in the general population. Let µ be the mean height of women of ages 20–29 years in Eugene. H0: µ = 64. Ha: µ > 64. This is a one sided hypothesis test. We can’t take the null hypothesis to be µ ≤ 64. When we carry out the hypothesis test, we assume the null hypothesis is true, so it has to provide a specific value of µ to use in our calculations. N. Christopher Phillips () Math 243: Lecture File 9 28 April 2009 17 / 60 Example 1 revisited We know US women of ages 20–29 years have mean height 64 inches. We want to know if women in Eugene of ages 20–29 years are on average taller than those in the general population. Let µ be the mean height of women of ages 20–29 years in Eugene. H0: µ = 64. Ha: µ > 64. This is a one sided hypothesis test. We can’t take the null hypothesis to be µ ≤ 64. When we carry out the hypothesis test, we assume the null hypothesis is true, so it has to provide a specific value of µ to use in our calculations. N. Christopher Phillips () Math 243: Lecture File 9 28 April 2009 17 / 60 Example 1 revisited We know US women of ages 20–29 years have mean height 64 inches. We want to know if women in Eugene of ages 20–29 years are on average taller than those in the general population. Let µ be the mean height of women of ages 20–29 years in Eugene. H0: µ = 64. Ha: µ > 64. This is a one sided hypothesis test. We can’t take the null hypothesis to be µ ≤ 64. When we carry out the hypothesis test, we assume the null hypothesis is true, so it has to provide a specific value of µ to use in our calculations. N. Christopher Phillips () Math 243: Lecture File 9 28 April 2009 17 / 60 Example 2 revisited Change the question slightly. We want to know if the mean height of women in Eugene of ages 20–29 years differs from the mean height of those in the general population. Again, let µ be the mean height of women of ages 20–29 years in Eugene. H0: µ = 64. Ha: µ 6= 64. This is a two sided hypothesis test. Note the change in Ha (but H0 is the same). N. Christopher Phillips () Math 243: Lecture File 9 28 April 2009 18 / 60 Example 2 revisited Change the question slightly. We want to know if the mean height of women in Eugene of ages 20–29 years differs from the mean height of those in the general population. Again, let µ be the mean height of women of ages 20–29 years in Eugene. H0: µ = 64. Ha: µ 6= 64. This is a two sided hypothesis test. Note the change in Ha (but H0 is the same). N. Christopher Phillips () Math 243: Lecture File 9 28 April 2009 18 / 60 Example 2 revisited Change the question slightly. We want to know if the mean height of women in Eugene of ages 20–29 years differs from the mean height of those in the general population. Again, let µ be the mean height of women of ages 20–29 years in Eugene. H0: µ = 64. Ha: µ 6= 64. This is a two sided hypothesis test. Note the change in Ha (but H0 is the same). N. Christopher Phillips () Math 243: Lecture File 9 28 April 2009 18 / 60 Caution: It is very unlikely that the mean height of women of ages 20–29 years in Eugene is exactly 64 inches. Thus, the null hypothesis is unlikely to be exactly true. However, we don’t reject it unless we have what we believe to be strong evidence against it. N. Christopher Phillips () Math 243: Lecture File 9 28 April 2009 19 / 60 Example 3 revisited Are female intercollegiate soccer players taller than average? We compare with the mean height of US women of ages 20–29 years, given as 64 inches. Let µ be the mean height of female intercollegiate soccer players. H0: µ = 64. Ha: µ > 64. N. Christopher Phillips () Math 243: Lecture File 9 28 April 2009 20 / 60 Example 3 revisited Are female intercollegiate soccer players taller than average? We compare with the mean height of US women of ages 20–29 years, given as 64 inches. Let µ be the mean height of female intercollegiate soccer players. H0: µ = 64. Ha: µ > 64. N. Christopher Phillips () Math 243: Lecture File 9 28 April 2009 20 / 60 Example 3 revisited Are female intercollegiate soccer players taller than average? We compare with the mean height of US women of ages 20–29 years, given as 64 inches. Let µ be the mean height of female intercollegiate soccer players. H0: µ = 64. Ha: µ > 64. N. Christopher Phillips () Math 243: Lecture File 9 28 April 2009 20 / 60 Example 4 revisited Does being left handed affect IQ? We compare with the mean IQ of left handed people with the mean IQ of the general population, which is 100. Let µ be the mean IQ of left handed people. H0: µ = 100. Ha: µ 6= 100. N. Christopher Phillips () Math 243: Lecture File 9 28 April 2009 21 / 60 Example 4 revisited Does being left handed affect IQ? We compare with the mean IQ of left handed people with the mean IQ of the general population, which is 100. Let µ be the mean IQ of left handed people. H0: µ = 100. Ha: µ 6= 100. N. Christopher Phillips () Math 243: Lecture File 9 28 April 2009 21 / 60 Example 5 revisited Suppose we know the mean weight of spiral-horned snorkacks. Are crumple-horned snorkacks, on average, smaller (lighter)? To be specific, suppose that spiral-horned snorkacks are known to have a mean mass of 60 kg. Let µ be the mean mass, in kg, of crumple-horned snorkacks. H0: µ = 60. Ha: µ < 60. N. Christopher Phillips () Math 243: Lecture File 9 28 April 2009 22 / 60 Example 5 revisited Suppose we know the mean weight of spiral-horned snorkacks. Are crumple-horned snorkacks, on average, smaller (lighter)? To be specific, suppose that spiral-horned snorkacks are known to have a mean mass of 60 kg. Let µ be the mean mass, in kg, of crumple-horned snorkacks. H0: µ = 60. Ha: µ < 60. N. Christopher Phillips () Math 243: Lecture File 9 28 April 2009 22 / 60 Example 5 revisited Suppose we know the mean weight of spiral-horned snorkacks. Are crumple-horned snorkacks, on average, smaller (lighter)? To be specific, suppose that spiral-horned snorkacks are known to have a mean mass of 60 kg. Let µ be the mean mass, in kg, of crumple-horned snorkacks. H0: µ = 60. Ha: µ < 60. N. Christopher Phillips () Math 243: Lecture File 9 28 April 2009 22 / 60 Example 6 revisited You suspect that the Consolidated Generic Unhealthy Foods Corporation is, on average, underfilling its 12 ounce bags of potato chips. We compare the mean weight of the contents of bags of potato chips with 12 ounces, which is what the weight is supposed to be. Let µ be the mean weight, in ounces, of the contents of 12 ounce bags of potato chips produced by the Consolidated Generic Unhealthy Foods Corporation. H0: µ = 12. Ha: µ < 12. N. Christopher Phillips () Math 243: Lecture File 9 28 April 2009 23 / 60 Example 6 revisited You suspect that the Consolidated Generic Unhealthy Foods Corporation is, on average, underfilling its 12 ounce bags of potato chips. We compare the mean weight of the contents of bags of potato chips with 12 ounces, which is what the weight is supposed to be. Let µ be the mean weight, in ounces, of the contents of 12 ounce bags of potato chips produced by the Consolidated Generic Unhealthy Foods Corporation. H0: µ = 12. Ha: µ < 12. N. Christopher Phillips () Math 243: Lecture File 9 28 April 2009 23 / 60 Example 6 revisited You suspect that the Consolidated Generic Unhealthy Foods Corporation is, on average, underfilling its 12 ounce bags of potato chips. We compare the mean weight of the contents of bags of potato chips with 12 ounces, which is what the weight is supposed to be. Let µ be the mean weight, in ounces, of the contents of 12 ounce bags of potato chips produced by the Consolidated Generic Unhealthy Foods Corporation. H0: µ = 12. Ha: µ < 12. N. Christopher Phillips () Math 243: Lecture File 9 28 April 2009 23 / 60