Empirical Rule Confidence Intervals Finding a good sample size. Outline. 1 Empirical Rule. 2 Confidence Intervals. 3 Finding a good sample size

Transcription

1 Outline 1 Empirical Rule 2 Confidence Intervals 3 Finding a good sample size

2 Outline 1 Empirical Rule 2 Confidence Intervals 3 Finding a good sample size

3 Question How much of the probability lies within 1 standard deviation of the mean?

4 Question How much of the probability lies within 1 standard deviation of the mean?

5 Question How much of the probability lies within 1 standard deviation of the mean? Answer Pr( 1 Z 1)

6 Question How much of the probability lies within 1 standard deviation of the mean? Answer Pr( 1 Z 1) = Pr(Z 1) Pr(Z 1)

7 Question How much of the probability lies within 1 standard deviation of the mean? Answer Pr( 1 Z 1) = Pr(Z 1) Pr(Z 1) =

8 Question How much of the probability lies within 1 standard deviation of the mean? Answer Pr( 1 Z 1) = Pr(Z 1) Pr(Z 1) = =

9 68.27% Question How much of the probability lies within 1 standard deviation of the mean? Answer Pr( 1 Z 1) = Pr(Z 1) Pr(Z 1) = =

10 Question How much of the probability lies within 2 standard deviations of the mean?

11 Question How much of the probability lies within 2 standard deviations of the mean?

12 Question How much of the probability lies within 2 standard deviations of the mean? Answer Pr( 2 Z 2) = Pr(Z 2) Pr(Z 2) = =

13 95.45% Question How much of the probability lies within 2 standard deviations of the mean? Answer Pr( 2 Z 2) = Pr(Z 2) Pr(Z 2) = =

14 Question How much of the probability lies within 3 standard deviations of the mean?

15 Question How much of the probability lies within 3 standard deviations of the mean?

16 Question How much of the probability lies within 3 standard deviations of the mean? Answer Pr( 3 Z 3) = Pr(Z 3) Pr(Z 3) = =

17 99.73% Question How much of the probability lies within 3 standard deviations of the mean? Answer Pr( 3 Z 3) = Pr(Z 3) Pr(Z 3) = =

18 The standard normal distribution Z 68% % %

19 Any normal distribution X 68% µ 3σ µ 2σ µ σ µ µ + σ µ + 2σ µ + 3σ 95% µ 3σ µ 2σ µ σ µ µ + σ µ + 2σ µ + 3σ 99.7% µ 3σ µ 2σ µ σ µ µ + σ µ + 2σ µ + 3σ

20 Empirical Rule If X is a normal random variable with standard deviation σ, 68% of the probability is within σ of the mean. 95% of the probability is within 2σ of the mean. 99.7% of the probability is within 3σ of the mean.

21 Example Problem We now can get approximate answers to some problems without the z-table! Example The heights of men aged 20 to 29 are approximately normal with mean 69 and standard deviation About what percentage of these men measure between 66 and 72? 2 Find a range of heights that contains 95% of these men.

22 Example Problem We now can get approximate answers to some problems without the z-table! Example The heights of men aged 20 to 29 are approximately normal with mean 69 and standard deviation About what percentage of these men measure between 66 and 72? 2 Find a range of heights that contains 95% of these men. Solution 1 µ = 69 and σ = 2.8, so 66 is about one standard deviation below the mean and 72 is about one standard deviation above the mean. Thus, by the Empirical Rule, a little more than 68% of the men fall within that range. (Exact answer: 71.5%)

23 Example Problem We now can get approximate answers to some problems without the z-table! Example The heights of men aged 20 to 29 are approximately normal with mean 69 and standard deviation About what percentage of these men measure between 66 and 72? 2 Find a range of heights that contains 95% of these men. Solution 1 µ = 69 and σ = 2.8, so 66 is about one standard deviation below the mean and 72 is about one standard deviation above the mean. Thus, by the Empirical Rule, a little more than 68% of the men fall within that range. (Exact answer: 71.5%)

24 Example Example Problem The heights of men aged 20 to 29 are approximately normal with mean 69 and standard deviation About what percentage of these men measure between 66 and 72? 2 Find a range of heights that contains 95% of these men. Solution, cont. 2 By the Empirical Rule, 95% of the men fall within the range from µ 2σ to µ + 2σ. Thus the range from = 63.4 to = 74.6 contains 95% of the men, so our answer is 63.4 to 74.6 inches. (More exact answer: to inches.)

25 Example Example Problem The heights of men aged 20 to 29 are approximately normal with mean 69 and standard deviation About what percentage of these men measure between 66 and 72? 2 Find a range of heights that contains 95% of these men. Solution, cont. 2 By the Empirical Rule, 95% of the men fall within the range from µ 2σ to µ + 2σ. Thus the range from = 63.4 to = 74.6 contains 95% of the men, so our answer is 63.4 to 74.6 inches. (More exact answer: to inches.)

26 Example Example Problem The heights of men aged 20 to 29 are approximately normal with mean 69 and standard deviation About what percentage of these men measure between 66 and 72? 2 Find a range of heights that contains 95% of these men. Solution, cont. 2 By the Empirical Rule, 95% of the men fall within the range from µ 2σ to µ + 2σ. Thus the range from = 63.4 to = 74.6 contains 95% of the men, so our answer is 63.4 to 74.6 inches. (More exact answer: to inches.)

27 Outline 1 Empirical Rule 2 Confidence Intervals 3 Finding a good sample size

28 How good is a sample mean x? Say we have a population and we want to know its mean µ. So we choose a sample, measure it, and compute the sample mean x. Of course x could be larger or smaller than µ. How likely is it to be close to µ?

29 Recall the Central Limit Theorem Population Distribution Mean µ Standard deviation σ µ Central Limit Theorem X is normally distributed with mean µ, and with standard deviation σ n.

30 Recall the Central Limit Theorem Population Distribution Mean µ Standard deviation σ µ Distribution of X Mean µ Standard deviation σ n µ Central Limit Theorem X is normally distributed with mean µ, and with standard deviation σ n.

31 Recall the Central Limit Theorem Population Distribution Mean µ Standard deviation σ µ Distribution of X Mean µ Standard deviation σ n µ Central Limit Theorem X is normally distributed with mean µ, and with standard deviation σ n.

32 Recall the Central Limit Theorem Population Distribution Mean µ Standard deviation σ µ Distribution of X Mean µ Standard deviation σ n µ σ n µ µ + σ n Central Limit Theorem X is normally distributed with mean µ, and with standard deviation σ n.

33 Recall the Central Limit Theorem Population Distribution Mean µ Standard deviation σ µ Distribution of X Mean µ Standard deviation σ n µ σ n µ µ + σ n Central Limit Theorem X is normally distributed with mean µ, and with standard deviation σ n. We call this the standard error.

34 So our sample mean x is chosen at random from this distribution: µ 3 σ n µ 2 σ n µ σ n µ µ + σ n µ + 2 σ n µ + 3 σ n

35 So our sample mean x is chosen at random from this distribution: 95% µ 3 σ n µ 2 σ n µ σ n µ µ + σ n µ + 2 σ n µ + 3 σ n That means 95% of the time, our sample mean will be less than 2 standard errors from the population mean µ.

36 So our sample mean x is chosen at random from this distribution: 95% µ 3 σ n µ 2 σ n µ σ n µ µ + σ n µ + 2 σ n µ + 3 σ n That means 95% of the time, our sample mean will be less than 2 standard errors from the population mean µ.

37 95% of the time... x is between µ 2 σ n and µ + 2 σ n. µ 2 σ n µ + 2 σ n µ x

38 95% of the time... x is between µ 2 σ n and µ + 2 σ n. µ 2 σ n µ + 2 σ n µ x I.e., x is no farther away from µ than 2 σ n. < 2 σ n µ x

39 95% of the time... x is between µ 2 σ n and µ + 2 σ n. µ 2 σ n µ + 2 σ n µ x I.e., x is no farther away from µ than 2 σ n. < 2 σ n µ x I.e., µ is no farther away from x than 2 σ n.

40 95% of the time... x is between µ 2 σ n and µ + 2 σ n. µ 2 σ n µ + 2 σ n µ x I.e., x is no farther away from µ than 2 σ n. < 2 σ n µ x I.e., µ is no farther away from x than 2 σ n. I.e., µ is between x 2 σ n and x + 2 σ n. x 2 σ n x + 2 σ n µ x

41 Our first confidence interval x Now the perspective from the ground! I took my sample and got my x. I know that 95% of the times I do that, the real µ is somewhere between x 2 σ n and x + 2 σ n. So I am 95% confident that µ is somewhere in that interval! This is my 95% confidence interval.

42 Our first confidence interval x 2 σ n x + 2 σ n x Now the perspective from the ground! I took my sample and got my x. I know that 95% of the times I do that, the real µ is somewhere between x 2 σ n and x + 2 σ n. So I am 95% confident that µ is somewhere in that interval! This is my 95% confidence interval.

43 Our first confidence interval x 2 σ n x + 2 σ n µ x Now the perspective from the ground! I took my sample and got my x. I know that 95% of the times I do that, the real µ is somewhere between x 2 σ n and x + 2 σ n. So I am 95% confident that µ is somewhere in that interval! This is my 95% confidence interval.

44 Our first confidence interval x 2 σ n x + 2 σ n µ x Now the perspective from the ground! I took my sample and got my x. I know that 95% of the times I do that, the real µ is somewhere between x 2 σ n and x + 2 σ n. So I am 95% confident that µ is somewhere in that interval! This is my 95% confidence interval.

45 The 95% Confidence Interval Definition Suppose we have a population with a standard deviation σ. We take a random sample of size n and compute the sample mean x. Then (x 2 σ n, x + 2 σ n ) which means the range of numbers from x 2 σ n to x + 2 σ n, is the 95% confidence interval for µ. That means we are 95% confident that the true population mean µ is somewhere in that range of numbers.,

46 The 95% Confidence Interval Definition Suppose we have a population with a standard deviation σ. We take a random sample of size n and compute the sample mean x. Then (x 2 σ n, x + 2 σ n ) which means the range of numbers from x 2 σ n to x + 2 σ n, is the 95% confidence interval for µ. That means we are 95% confident that the true population mean µ is somewhere in that range of numbers.,

47 What about σ? (x 2 σ n, x + 2 σ n ), You might object at this point, because our formula still has σ in it! We do know the sample mean x, and we do know the sample size n. However, if we don t know the population mean µ, how on earth would we know the population standard deviation σ? We ll learn in the next class period how to get rid of σ. For now, we ll just pretend.

48 Example Example A magazine surveyed 1200 students from 100 colleges about how much time they spend on the Internet per week. The magazine reported that the average was 15.1 hours. Assuming the population standard deviation is 5 hours, construct a 95% confidence interval for this statistic.

49 Example Example A magazine surveyed 1200 students from 100 colleges about how much time they spend on the Internet per week. The magazine reported that the average was 15.1 hours. Assuming the population standard deviation is 5 hours, construct a 95% confidence interval for this statistic. Solution We see that n = 1200, x = 15.1, and σ = 5, so the standard error is σ n = = We know a 95% confidence interval is which is (x 2 σ n, x + 2 σ n ), (15.1 2(0.1444), (0.1444)) or just (14.81, 15.39). Thus the magazine can be 95% confident that the true population mean is somewhere between hours and hours.

50 Example Example A magazine surveyed 1200 students from 100 colleges about how much time they spend on the Internet per week. The magazine reported that the average was 15.1 hours. Assuming the population standard deviation is 5 hours, construct a 95% confidence interval for this statistic. Solution We see that n = 1200, x = 15.1, and σ = 5, so the standard error is σ n = = We know a 95% confidence interval is (x 2 σ n, x + 2 σ n ), which is (15.1 2(0.1444), (0.1444)) or just (14.81, 15.39). Thus the magazine can be 95% confident that the true population mean is somewhere between hours and hours.

51 Example Example A magazine surveyed 1200 students from 100 colleges about how much time they spend on the Internet per week. The magazine reported that the average was 15.1 hours. Assuming the population standard deviation is 5 hours, construct a 95% confidence interval for this statistic. Solution We see that n = 1200, x = 15.1, and σ = 5, so the standard error is σ n = = We know a 95% confidence interval is (x 2 σ n, x + 2 σ n ), which is (15.1 2(0.1444), (0.1444)) or just (14.81, 15.39). Thus the magazine can be 95% confident that the true population mean is somewhere between hours and hours.

52 Example Example A magazine surveyed 1200 students from 100 colleges about how much time they spend on the Internet per week. The magazine reported that the average was 15.1 hours. Assuming the population standard deviation is 5 hours, construct a 95% confidence interval for this statistic. Solution We see that n = 1200, x = 15.1, and σ = 5, so the standard error is σ n = = We know a 95% confidence interval is (x 2 σ n, x + 2 σ n ), which is (15.1 2(0.1444), (0.1444)) or just (14.81, 15.39). Thus the magazine can be 95% confident that the true population mean is somewhere between hours and hours.

53 Example Example A magazine surveyed 1200 students from 100 colleges about how much time they spend on the Internet per week. The magazine reported that the average was 15.1 hours. Assuming the population standard deviation is 5 hours, construct a 95% confidence interval for this statistic. Solution We see that n = 1200, x = 15.1, and σ = 5, so the standard error is σ n = = We know a 95% confidence interval is (x 2 σ n, x + 2 σ n ), which is (15.1 2(0.1444), (0.1444)) or just (14.81, 15.39). Thus the magazine can be 95% confident that the true population mean is somewhere between hours and hours.

54 Example Example A magazine surveyed 1200 students from 100 colleges about how much time they spend on the Internet per week. The magazine reported that the average was 15.1 hours. Assuming the population standard deviation is 5 hours, construct a 95% confidence interval for this statistic. Solution We see that n = 1200, x = 15.1, and σ = 5, so the standard error is σ n = = We know a 95% confidence interval is (x 2 σ n, x + 2 σ n ), which is (15.1 2(0.1444), (0.1444)) or just (14.81, 15.39). Thus the magazine can be 95% confident that the true population mean is somewhere between hours and hours.

55 What about other numbers? What if we want to be more confident?

56 What about other numbers? What if we want to be more confident? We chose 2 σ n because 95% of the time, x is that close to µ. 95% µ 3 σ n µ 2 σ n µ σ n µ µ + σ n µ + 2 σ n µ + 3 σ n

57 What about other numbers? What if we want to be more confident? We chose 2 σ n because 95% of the time, x is that close to µ. 95% µ 3 σ n µ 2 σ n µ σ n µ µ + σ n µ + 2 σ n µ + 3 σ n What if we wanted to be 99.7% sure?

58 What about other numbers? What if we want to be more confident? We chose 2 σ n because 95% of the time, x is that close to µ. 99.7% µ 3 σ n µ 2 σ n µ σ n µ µ + σ n µ + 2 σ n µ + 3 σ n What if we wanted to be 99.7% sure? Use 3 standard errors.

59 What about other numbers? What if we want to be more confident? We chose 2 σ n because 95% of the time, x is that close to µ. 99.7% µ 3 σ n µ 2 σ n µ σ n µ µ + σ n µ + 2 σ n µ + 3 σ n What if we wanted to be 99.7% sure? Use 3 standard errors. What if we wanted to be 68% sure?

60 What about other numbers? What if we want to be more confident? We chose 2 σ n because 95% of the time, x is that close to µ. 68% µ 3 σ n µ 2 σ n µ σ n µ µ + σ n µ + 2 σ n µ + 3 σ n What if we wanted to be 99.7% sure? Use 3 standard errors. What if we wanted to be 68% sure? Use 1 standard error.

61 What about other numbers? What if we want to be more confident? We chose 2 σ n because 95% of the time, x is that close to µ. 68% µ 3 σ n µ 2 σ n µ σ n µ µ + σ n µ + 2 σ n µ + 3 σ n What if we wanted to be 99.7% sure? Use 3 standard errors. What if we wanted to be 68% sure? Use 1 standard error. What if we wanted a different percentage?

62 How to find a y-confidence interval (e.g., for a 92% confidence interval, y = 0.92) 1 Draw the standard normal curve Z. 2 Draw two vertical bars symmetrically on the graph, and label the middle with y. 3 That means the remaining area is 1 y. 4 That means the left tail has area 1 y 2. 5 Use the z-table (backwards) to learn where that tail ends! 6 Use that many standard errors!

63 How to find a y-confidence interval (e.g., for a 92% confidence interval, y = 0.92) 1 Draw the standard normal curve Z. 2 Draw two vertical bars symmetrically on the graph, and label the middle with y. 3 That means the remaining area is 1 y. 4 That means the left tail has area 1 y 2. 5 Use the z-table (backwards) to learn where that tail ends! 6 Use that many standard errors!

64 How to find a y-confidence interval (e.g., for a 92% confidence interval, y = 0.92) 1 Draw the standard normal curve Z. y 2 Draw two vertical bars symmetrically on the graph, and label the middle with y. 3 That means the remaining area is 1 y. 4 That means the left tail has area 1 y 2. 5 Use the z-table (backwards) to learn where that tail ends! 6 Use that many standard errors!

65 How to find a y-confidence interval (e.g., for a 92% confidence interval, y = 0.92) 1 Draw the standard normal curve Z. y 1 y 2 Draw two vertical bars symmetrically on the graph, and label the middle with y. 3 That means the remaining area is 1 y. 4 That means the left tail has area 1 y 2. 5 Use the z-table (backwards) to learn where that tail ends! 6 Use that many standard errors!

66 How to find a y-confidence interval (e.g., for a 92% confidence interval, y = 0.92) 1 Draw the standard normal curve Z. 1 y 2 y 1 y 2 Draw two vertical bars symmetrically on the graph, and label the middle with y. 3 That means the remaining area is 1 y. 4 That means the left tail has area 1 y 2. 5 Use the z-table (backwards) to learn where that tail ends! 6 Use that many standard errors!

67 How to find a y-confidence interval (e.g., for a 92% confidence interval, y = 0.92) 1 Draw the standard normal curve Z. 1 y 2 y 1 y z 2 Draw two vertical bars symmetrically on the graph, and label the middle with y. 3 That means the remaining area is 1 y. 4 That means the left tail has area 1 y 2. 5 Use the z-table (backwards) to learn where that tail ends! 6 Use that many standard errors!

68 How to find a y-confidence interval (e.g., for a 92% confidence interval, y = 0.92) 1 Draw the standard normal curve Z. 1 y 2 y 1 y z 2 Draw two vertical bars symmetrically on the graph, and label the middle with y. 3 That means the remaining area is 1 y. 4 That means the left tail has area 1 y 2. 5 Use the z-table (backwards) to learn where that tail ends! 6 Use that many standard errors!

69 Example: Pregnancy Example Dr. McKnight is studying the length of human pregnancies, which have a standard deviation of 14 days. She studies a random sample of 50 pregnancies, and discovers that the sample mean is 274 days. What is the 98% confidence interval?

70 Example: Pregnancy Example Dr. McKnight is studying the length of human pregnancies, which have a standard deviation of 14 days. She studies a random sample of 50 pregnancies, and discovers that the sample mean is 274 days. What is the 98% confidence interval? Solution Let s follow our procedure!...

71 Example: Pregnancy We need a 98% confidence interval. 1 Draw the standard normal curve Z. 2 Draw vertical bars and label the middle with That means the remaining area is = That means the left tail has area = The z-table (backwards) tells us the tail ends at So we need 2.33 standard errors!

72 Example: Pregnancy We need a 98% confidence interval. 1 Draw the standard normal curve Z. 2 Draw vertical bars and label the middle with That means the remaining area is = That means the left tail has area = The z-table (backwards) tells us the tail ends at So we need 2.33 standard errors!

73 Example: Pregnancy We need a 98% confidence interval. 1 Draw the standard normal curve Z Draw vertical bars and label the middle with That means the remaining area is = That means the left tail has area = The z-table (backwards) tells us the tail ends at So we need 2.33 standard errors!

74 Example: Pregnancy We need a 98% confidence interval. 1 Draw the standard normal curve Z Draw vertical bars and label the middle with That means the remaining area is = That means the left tail has area = The z-table (backwards) tells us the tail ends at So we need 2.33 standard errors!

75 Example: Pregnancy We need a 98% confidence interval. 1 Draw the standard normal curve Z Draw vertical bars and label the middle with That means the remaining area is = That means the left tail has area = The z-table (backwards) tells us the tail ends at So we need 2.33 standard errors!

76 Example: Pregnancy We need a 98% confidence interval. 1 Draw the standard normal curve Z Draw vertical bars and label the middle with That means the remaining area is = That means the left tail has area = The z-table (backwards) tells us the tail ends at So we need 2.33 standard errors!

77 Example: Pregnancy We need a 98% confidence interval. 1 Draw the standard normal curve Z Draw vertical bars and label the middle with That means the remaining area is = That means the left tail has area = The z-table (backwards) tells us the tail ends at So we need 2.33 standard errors!

78 Example Example: Pregnancy Dr. McKnight is studying the length of human pregnancies, which have a standard deviation σ = 14 days. She studies a random sample of 50 pregnancies, and discovers the sample mean is 274 days. What is the 98% confidence interval? Solution, cont. Thus we need 2.33 standard errors to get 98% confidence. So the confidence interval is (x 2.33 σ n, x σ n ) = ( , ) = (269.4, 278.6). Thus Dr. McKnight can be 98% sure that the true population mean is between days and days.

79 Example Example: Pregnancy Dr. McKnight is studying the length of human pregnancies, which have a standard deviation σ = 14 days. She studies a random sample of 50 pregnancies, and discovers the sample mean is 274 days. What is the 98% confidence interval? Solution, cont. Thus we need 2.33 standard errors to get 98% confidence. So the confidence interval is (x 2.33 σ n, x σ n ) = ( , ) = (269.4, 278.6). Thus Dr. McKnight can be 98% sure that the true population mean is between days and days.

80 Example Example: Pregnancy Dr. McKnight is studying the length of human pregnancies, which have a standard deviation σ = 14 days. She studies a random sample of 50 pregnancies, and discovers the sample mean is 274 days. What is the 98% confidence interval? Solution, cont. Thus we need 2.33 standard errors to get 98% confidence. So the confidence interval is (x 2.33 σ n, x σ n ) = ( , ) = (269.4, 278.6). Thus Dr. McKnight can be 98% sure that the true population mean is between days and days.

81 Example Example: Pregnancy Dr. McKnight is studying the length of human pregnancies, which have a standard deviation σ = 14 days. She studies a random sample of 50 pregnancies, and discovers the sample mean is 274 days. What is the 98% confidence interval? Solution, cont. Thus we need 2.33 standard errors to get 98% confidence. So the confidence interval is (x 2.33 σ n, x σ n ) = ( , ) = (269.4, 278.6). Thus Dr. McKnight can be 98% sure that the true population mean is between days and days.

82 Example Example: Pregnancy Dr. McKnight is studying the length of human pregnancies, which have a standard deviation σ = 14 days. She studies a random sample of 50 pregnancies, and discovers the sample mean is 274 days. What is the 98% confidence interval? Solution, cont. Thus we need 2.33 standard errors to get 98% confidence. So the confidence interval is (x 2.33 σ n, x σ n ) = ( , ) = (269.4, 278.6). Thus Dr. McKnight can be 98% sure that the true population mean is between days and days.

83 Outline 1 Empirical Rule 2 Confidence Intervals 3 Finding a good sample size

84 Finding a good sample size When you re planning to conduct an experiment, it s important to get a large enough sample size. Example An environmental scientist is studying the pollution levels in a stream. He wants to know how many samples he should take to be 90% sure that his estimate x differs from the true value µ by no more than 1.5 µg/liter. (The standard deviation of the pollution levels is 6 µg/liter.) How many samples should he take?

85 Finding a good sample size When you re planning to conduct an experiment, it s important to get a large enough sample size. Example An environmental scientist is studying the pollution levels in a stream. He wants to know how many samples he should take to be 90% sure that his estimate x differs from the true value µ by no more than 1.5 µg/liter. (The standard deviation of the pollution levels is 6 µg/liter.) How many samples should he take? Solution First, we need to find out what a 90% confidence interval looks like.

86 Example: Pollution We need a 90% confidence interval. 1 Draw the standard normal curve Z. 2 Draw vertical bars and label the middle with That means the remaining area is = That means the left tail has area = The z-table (backwards) tells us the tail ends at So we need 1.65 standard errors!

87 Example: Pollution We need a 90% confidence interval. 1 Draw the standard normal curve Z. 2 Draw vertical bars and label the middle with That means the remaining area is = That means the left tail has area = The z-table (backwards) tells us the tail ends at So we need 1.65 standard errors!

88 Example: Pollution We need a 90% confidence interval. 1 Draw the standard normal curve Z Draw vertical bars and label the middle with That means the remaining area is = That means the left tail has area = The z-table (backwards) tells us the tail ends at So we need 1.65 standard errors!

89 Example: Pollution We need a 90% confidence interval. 1 Draw the standard normal curve Z Draw vertical bars and label the middle with That means the remaining area is = That means the left tail has area = The z-table (backwards) tells us the tail ends at So we need 1.65 standard errors!

90 Example: Pollution We need a 90% confidence interval. 1 Draw the standard normal curve Z Draw vertical bars and label the middle with That means the remaining area is = That means the left tail has area = The z-table (backwards) tells us the tail ends at So we need 1.65 standard errors!

91 Example: Pollution We need a 90% confidence interval. 1 Draw the standard normal curve Z Draw vertical bars and label the middle with That means the remaining area is = That means the left tail has area = The z-table (backwards) tells us the tail ends at So we need 1.65 standard errors!

92 Example: Pollution We need a 90% confidence interval. 1 Draw the standard normal curve Z Draw vertical bars and label the middle with That means the remaining area is = That means the left tail has area = The z-table (backwards) tells us the tail ends at So we need 1.65 standard errors!

93 Example Example: Pollution An environmental scientist is studying the pollution levels in a stream. He wants to know how many samples he should take to be 90% sure that his estimate x differs from the true value µ by no more than 1.5 µg/liter. (The standard deviation of the pollution levels is 6 µg/liter.) How many samples should he take? Solution, cont. So the scientist can be 90% confident that the difference between x and µ is no more than 1.65 standard errors. He wants that maximum difference to be 1.5 µg/liter, i.e., 1.5 = 1.65 σ n = n (plugging in σ = 6). We need to find n. 1.5 n = = 9.9 n = = 6.6 n = = Thus the scientist must take at least 44 samples.

94 Example Example: Pollution An environmental scientist is studying the pollution levels in a stream. He wants to know how many samples he should take to be 90% sure that his estimate x differs from the true value µ by no more than 1.5 µg/liter. (The standard deviation of the pollution levels is 6 µg/liter.) How many samples should he take? Solution, cont. So the scientist can be 90% confident that the difference between x and µ is no more than 1.65 standard errors. He wants that maximum difference to be 1.5 µg/liter, i.e., 1.5 = 1.65 σ n = n (plugging in σ = 6). We need to find n. 1.5 n = = 9.9 n = = 6.6 n = = Thus the scientist must take at least 44 samples.

95 Example Example: Pollution An environmental scientist is studying the pollution levels in a stream. He wants to know how many samples he should take to be 90% sure that his estimate x differs from the true value µ by no more than 1.5 µg/liter. (The standard deviation of the pollution levels is 6 µg/liter.) How many samples should he take? Solution, cont. So the scientist can be 90% confident that the difference between x and µ is no more than 1.65 standard errors. He wants that maximum difference to be 1.5 µg/liter, i.e., 1.5 = 1.65 σ n = n (plugging in σ = 6). We need to find n. 1.5 n = = 9.9 n = = 6.6 n = = Thus the scientist must take at least 44 samples.

96 Example Example: Pollution An environmental scientist is studying the pollution levels in a stream. He wants to know how many samples he should take to be 90% sure that his estimate x differs from the true value µ by no more than 1.5 µg/liter. (The standard deviation of the pollution levels is 6 µg/liter.) How many samples should he take? Solution, cont. So the scientist can be 90% confident that the difference between x and µ is no more than 1.65 standard errors. He wants that maximum difference to be 1.5 µg/liter, i.e., 1.5 = 1.65 σ n = n (plugging in σ = 6). We need to find n. 1.5 n = = 9.9 n = = 6.6 n = = Thus the scientist must take at least 44 samples.

97 Example Example: Pollution An environmental scientist is studying the pollution levels in a stream. He wants to know how many samples he should take to be 90% sure that his estimate x differs from the true value µ by no more than 1.5 µg/liter. (The standard deviation of the pollution levels is 6 µg/liter.) How many samples should he take? Solution, cont. So the scientist can be 90% confident that the difference between x and µ is no more than 1.65 standard errors. He wants that maximum difference to be 1.5 µg/liter, i.e., 1.5 = 1.65 σ n = n (plugging in σ = 6). We need to find n. 1.5 n = = 9.9 n = = 6.6 n = = Thus the scientist must take at least 44 samples.

98 Example Example: Pollution An environmental scientist is studying the pollution levels in a stream. He wants to know how many samples he should take to be 90% sure that his estimate x differs from the true value µ by no more than 1.5 µg/liter. (The standard deviation of the pollution levels is 6 µg/liter.) How many samples should he take? Solution, cont. So the scientist can be 90% confident that the difference between x and µ is no more than 1.65 standard errors. He wants that maximum difference to be 1.5 µg/liter, i.e., 1.5 = 1.65 σ n = n (plugging in σ = 6). We need to find n. 1.5 n = = 9.9 n = = 6.6 n = = Thus the scientist must take at least 44 samples.

99 Example Example: Pollution An environmental scientist is studying the pollution levels in a stream. He wants to know how many samples he should take to be 90% sure that his estimate x differs from the true value µ by no more than 1.5 µg/liter. (The standard deviation of the pollution levels is 6 µg/liter.) How many samples should he take? Solution, cont. So the scientist can be 90% confident that the difference between x and µ is no more than 1.65 standard errors. He wants that maximum difference to be 1.5 µg/liter, i.e., 1.5 = 1.65 σ n = n (plugging in σ = 6). We need to find n. 1.5 n = = 9.9 n = = 6.6 n = = Thus the scientist must take at least 44 samples.

100 Example Example: Pollution An environmental scientist is studying the pollution levels in a stream. He wants to know how many samples he should take to be 90% sure that his estimate x differs from the true value µ by no more than 1.5 µg/liter. (The standard deviation of the pollution levels is 6 µg/liter.) How many samples should he take? Solution, cont. So the scientist can be 90% confident that the difference between x and µ is no more than 1.65 standard errors. He wants that maximum difference to be 1.5 µg/liter, i.e., 1.5 = 1.65 σ n = n (plugging in σ = 6). We need to find n. 1.5 n = = 9.9 n = = 6.6 n = = Thus the scientist must take at least 44 samples.

101 Example Example: Pollution An environmental scientist is studying the pollution levels in a stream. He wants to know how many samples he should take to be 90% sure that his estimate x differs from the true value µ by no more than 1.5 µg/liter. (The standard deviation of the pollution levels is 6 µg/liter.) How many samples should he take? Solution, cont. So the scientist can be 90% confident that the difference between x and µ is no more than 1.65 standard errors. He wants that maximum difference to be 1.5 µg/liter, i.e., 1.5 = 1.65 σ n = n (plugging in σ = 6). We need to find n. 1.5 n = = 9.9 n = = 6.6 n = = Thus the scientist must take at least 44 samples.

102 Example Example: Pollution An environmental scientist is studying the pollution levels in a stream. He wants to know how many samples he should take to be 90% sure that his estimate x differs from the true value µ by no more than 1.5 µg/liter. (The standard deviation of the pollution levels is 6 µg/liter.) How many samples should he take? Solution, cont. So the scientist can be 90% confident that the difference between x and µ is no more than 1.65 standard errors. He wants that maximum difference to be 1.5 µg/liter, i.e., 1.5 = 1.65 σ n = n (plugging in σ = 6). We need to find n. 1.5 n = = 9.9 n = = 6.6 n = = Thus the scientist must take at least 44 samples.