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# Nossi ch 11

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Contemporary math chapter 11 Power Point.

Contemporary math chapter 11 Power Point.

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### Transcript

• 1. Chapter 11 Inferential Statistics
• 2. Section 11.1 Normal Distributions
• Goals
• Study normal distributions
• Study standard normal distributions
• Find the area under a standard normal curve
• 3. 11.1 Initial Problem
• A class of 90 students had a mean test score of 74, with a standard deviation of 8.
• If the professor curves the scores, how many students will get As and how many will get Fs?
• The solution will be given at the end of the section.
• 4. Statistical Inference
• The process of making predictions about an entire population based on information from a sample is called statistical inference .
• 5. Data Distributions
• For large data sets, a smooth curve can often be used to approximate the histogram.
• 6. Data Distributions
• The larger the data set and the smaller the bin size, the better the approximation of the smooth curve.
• 7. Example 1
• The distribution of weights for a large sample of college men is shown.
• 8. Example 1
• What percent of the men have weights between:
• 167 and 192 pounds?
• 137 and 192 pounds?
• 137 and 222 pounds?
• 9. Example 1
• Solution:
• 167 and 192 pounds?
• The area under the curve is 0.2, so 20% of the men are in this weight range.
• 10. Example 1
• Solution:
• 137 and 192 pounds?
• The area under the curve is 0.4, so 40% of the men are in this weight range.
• 11. Example 1
• Solution:
• 137 and 222 pounds?
• The area under the curve is 0.6, so 60% of the men are in this weight range.
• 12. Normal Distributions
• Data that has a symmetric, bell-shaped distribution curve is said to have a normal distribution .
• The mean and standard deviation determine the exact shape and position of the curve.
• 13. Example 2
• Which normal curve has the largest mean?
• Which normal curve has the largest standard deviation?
• 14. Example 2
• Solution: The data sets are already labeled in order of smallest mean to largest mean.
• Data Set III has the largest mean.
• 15. Example 2
• Solution: Data Set III has the largest standard deviation because it is the shortest, widest curve.
• The order of the standard deviations is II, I, III.
• 16. Normal Distributions
• Normal distributions with various means and standard deviations are shown on the following slides.
• 17. Normal Distributions
• 18. Normal Distributions
• 19. Normal Distributions
• 20. Standard Normal Distribution
• The normal distribution with a mean of 0 and a standard deviation of 1 is called the standard normal distribution .
• 21. Standard Normal Distribution
• The areas under any normal distribution can be compared to the areas under the standard normal distribution, as shown in the figure on the next slide.
• 22. Standard Normal Distribution
• 23. Area
• One way to find the area under a region of the standard normal curve is to use a table.
• Tables of values for the standard normal curve are printed in textbooks to eliminate the need to do repeated complicated calculations.
• 24. Example 3
• What fraction of the total area under the standard normal curve lies between a = -0.5 and b = 1.5?
• 25. Example 3
• Solution: Find a = -0.5 and b = 1.5 in the table.
• 26. Example 3
• The value in the table is 0.6247.
• A total of 62.47% of the area is shaded.
• In any normal distribution, 62.47% of the data lies between 0.5 standard deviations below the mean and 1.5 standard deviations above the mean.
• The probability a randomly selected data value will lie between -0.5 and 1.5 is 62.47%
• 27. Example 4
• What percent of the data in a standard normal distribution lies between 0.5 and 2.5?
• Solution: The value in the table for a = 0.5 and b = 2.5 is 0.3023.
• So 30.23% of the data in a standard normal distribution lies between 0.5 and 2.5. (see previous table)
• 28. Areas
• Because the normal curve is symmetric, the areas in the previous table are repeated.
• 29. Areas
• Figure 11.11 and table 11.2
• 30. Area
• A more common type of table:
• 31. Example 5
• Find the percent of data points in a standard normal distribution that lie between z = -1.8 and z = 1.3.
• Find the two areas in Table 11.3 and add them together.
• The area from 0 to 1.3 is 0.4032.
• The area from 0 to -1.8 is 0.4641.
• The total shaded area is 0.4032 + 0.4641 = 0.8673.
• 32. Example 6
• Find the percent of data points in a standard normal distribution that lie between z = 1.2 and z = 1.7.
• Find the two areas in Table 11.3 and subtract them.
• The area from 0 to 1.2 is 0.3849.
• The area from 0 to 1.7 is 0.4554.
• The total shaded area is 0.4554 - 0.3849 = 0.0705.
• 33. 11.1 Initial Problem Solution
• A class of 90 students had a mean test score of 74 with a standard deviation of 8 points.
• The test will be curved so that all students whose scores are at least 1.5 standard deviations above or below the mean will receive As and Fs, respectively.
• How many students will get As and how many will get Fs?
• 34. Initial Problem Solution
• Because the class is large, it is likely the scores have a normal distribution.
• If the scores are curved:
• The mean of 74 will correspond to a score of 0 in the standard normal distribution.
• A score that is 1.5 standard deviations above the mean will correspond to a score of +1.5 in the standard normal distribution, while a score that is 1.5 standard deviations below the mean will correspond to a score of -1.5.
• 35. Initial Problem Solution
• The percentage of As is the same as the area to the right of z = 1.5 in the standard normal distribution.
• Approximately 43.32% of the area is between 0 and 1.5.
• Since 50% of the area is to the right of 0, the area above 1.5 is 50% - 43.32% = 6.68%
• Thus, 6.68% of the students, or approximately 6 students, will receive As.
• 36. Initial Problem Solution
• The percentage of Fs is the same as the area to the left of z = -1.5 in the standard normal distribution.
• Because of the symmetry of the normal distribution, this is the same as the area above z = 1.5, so the calculations are the same as in the last step.
• Thus, 6.68% of the students, or approximately 6 students, will receive Fs.
• 37. Section 11.2 Applications of Normal Distributions
• Goals
• Study normal distribution applications
• Use the 68-95-99.7 Rule
• Use the population z -score
• 38. 11.2 Initial Problem
• Two suppliers make an engine part.
• Supplier A charges \$120 for 100 parts which have a standard deviation of 0.004 mm from the mean size.
• Supplier B charges \$90 for 100 parts which have a standard deviation of 0.012 mm from the mean size.
• Which supplier is a better choice?
• The solution will be given at the end of the section.
• 39. Example 1
• Approximately 10% of the data in a standard normal distribution lies within 1/8 of a standard deviation from the mean.
• Within 1/8 means between -0.125 and 0.125.
• Suppose the measurements of a certain population are normally distributed with a mean of 112 and standard deviation of 24. What values correspond to the interval given above?
• 40. Example 1
• Solution: In the standard normal distribution we are considering the interval from r = -0.125 to s = 0.125.
• For the nonstandard distribution, the interval will be 112 + (-0.125)(24) = 109 to 112 + (0.125)(24) = 115.
• We know that 10% of the data values will lie between 112 and 115.
• 41. 68-95-99.7 Rule
• For all normal distributions:
• Approximately 68% of the measurements lie within 1 standard deviation of the mean.
• Approximately 95% of the measurements lie within 2 standard deviations of the mean.
• Approximately 99.7% of the measurements lie within 3 standard deviations of the mean.
• 42. 68-95-99.7 Rule
• 43. Example 2
• The HDL cholesterol levels for a group of women are approximately normally distributed with a mean of 64 mg/dL and a standard deviation of 15 mg/dL.
• Determine the percentage of these women that have HDL cholesterol levels between 19 and 109 mg/dL.
• 44. Example 2
• The mean of 64 mg/dL corresponds to 0 in the standard normal curve. Now study the numbers: 19 - 64 - 109
• 64 - 19 = 45
• 45 = 15 (the standard deviation) times 3
• 109 = 64 + 45
• 45 = 15 (the standard deviation) times 3
• 19 and 109 are each 3 s.d.’s from the mean of 64
• 45. Example 2
• The area under the standard normal curve between z = -3 and z = 3 is found:
• From 0 to 3, there is 49.85% of the area.
• From 0 to -3, there is also 49.85%.
• Approximately, 2(49.85%) = 99.7% of the women will have a HDL level between 19 and 109 mg/dL.
• 46. Example 3
• Designers of a new computer mouse have learned that the lengths of women’s hands are normally distributed with a mean of 17 cm and a standard deviation of 1 cm.
• What percentage of women have hands in the range from 15 cm to 19 cm?
• 47. Example 3
• Solution:
• A length of 15 cm is 2 standard deviations below the mean of 17 cm.
• A length of 19 cm is 2 standard deviations above the mean of 17 cm.
• According to the 68-95-99.7 Rule, the percent of women whose hands are within 2 standard deviations of the mean length is 95%.
• 48. Population z -scores
• The formula for converting a normal distribution value to a standard normal distribution value is called a population z -score .
• The population z -score of a measurement, x , is given by:
• 49. Example 5
• Suppose a normal distribution has a mean of 4 and a standard deviation of 3.
• Find the z -scores of the measurements -1, 2, 3, 5, and 9.
• 50. Example 5
• A normal distribution has a mean of 4 and a standard deviation of 3.
• 51. Example 5
• The relationship between the normal values and the standard normal values is illustrated.
• 52. Example 6
• In 1996, the finishing times for the New York City Marathon were approximately normal, with a mean of 260 minutes and a standard deviation of about 50 minutes.
• What percentage of the finishers that year had times between 285 minutes and 335 minutes.
• 53. Example 6
• Solution: Find the z -scores. (a mean of 260 minutes and a standard deviation of about 50 minutes)
• For a time of 285 minutes,
• For a time of 335 minutes,
• 54. Example 6
• Find the areas
• The area from 0 to 0.5 is 0.1915.
• The area from 0 to 1.5 is 1.4332.
• Subtract the areas to find 0.4332 – 0.1915 = 0.2417.
• The conclusion is that 24.17% of the finishing times were between 285 and 335 minutes.
• 55. Example 7
• Recall the distribution of HDL cholesterol levels from the previous example, with a mean of 64 mg/dL and a standard deviation of 15 mg/dL.
• If an HDL level of 40 mg/dL or below signals an increased risk for coronary heart disease, what percentage of the women studied are at increased risk?
• 56. Example 7
• Solution: Find the z- score for an HDL level of 40 mg/dL:
• The area between 0 and -1.6 is 0.4452.
• 57. Example 7
• But our question asked if an HDL level of 40 mg/dL or below signals an increased risk for coronary heart disease, what percentage of the women studied are at increased risk? Solution: We need the area to the left of -1.6 is 0.5 – 0.4452 = 0.0548.
• In this group of women, 5.48% of them are at increased risk for coronary heart disease because of low HDL levels.
• 58. 11.2 Initial Problem Solution
• Two suppliers make an engine part.
• Supplier A charges \$120 for 100 parts which have a standard deviation of 0.004 mm from the mean size.
• Supplier B charges \$90 for 100 parts which have a standard deviation of 0.012 mm from the mean size.
• If parts must be within 0.012 mm to be acceptable, which supplier is a better choice?
• 59. Initial Problem Solution
• Determine the cost for each acceptable part from each supplier.
• Supplier A: Since σ = 0.004 mm, all parts within 3 standard deviations will be acceptable.
• We know that 99.7% of the parts are within 3 standard deviations of the mean.
• Each acceptable part costs
• 60. Initial Problem Solution
• Determine the cost for each acceptable part from each supplier.
• Supplier B: Since σ = 0.012 mm, all parts within 1 standard deviation will be acceptable.
• We know that 68% of the parts are within 1 standard deviation of the mean.
• Each acceptable part costs
• 61. Initial Problem Solution
• Overall each part from supplier B costs less than each part from supplier A, but more parts from B will have to be thrown away.
• Each acceptable part from supplier B costs more than each acceptable part from supplier A.
• They should choose supplier A.
• 62. Section 11.3 Confidence Intervals
• Goals
• Study proportions
• Study population proportions
• Study sample proportions
• Study confidence intervals
• Study margin of error
• 63. Proportions
• A fraction of the population under consideration is called a population proportion .
• The notation for a population proportion is p .
• For example, if 65,000,000 of 130,000,000 people support the President’s budget, the population proportion of people who support the budget is
• 64. Proportions
• A fraction of the sample being measured is called a sample proportion .
• The notation for a sample proportion is .
• For example, if 198 of 413 people polled support the President’s budget, the sample proportion of people who support the budget is
• 65. Example 1
• A college has 3520 freshman, of which 1056 have consumed an alcoholic beverage in the last 30 days.
• Of the 50 students surveyed in a health class, 11 say they have had an alcoholic beverage in the last 30 days.
• What are the population proportion and the sample proportion?
• 66. Example 1
• The population is the 3520 freshmen at the college.
• The population proportion is
• The sample is the 50 students who were surveyed.
• The sample proportion is
• 67. Example 1
• A distribution of the sample proportions for various possible samples of this population is shown at right.
• 68. Sample Proportions Distribution
• If samples of size n are taken from a population having a population proportion p , then the set of all sample proportions has a mean and standard deviation of:
• 69. Example 2
• Suppose the population proportion of a group is 0.4, and we choose a simple random sample of size 30.
• Find the mean and standard deviation of the set of all sample proportions.
• 70. Example 2
• Solution: In this case,
• p = 0.4 and n = 30.
• The mean is
• The standard deviation is
• 71. Example 2
• The sample proportion distribution is graphed below.
• 72. Example 3
• Fox News asked 900 registered voters whether or not they would take a smallpox vaccine.
• Suppose it is known that 60% of all Americans would take the vaccine. What is the approximate percentage of samples for which between 58% and 62% of voters in the sample would take the shot?
• 73. Example 3
• Solution: We know that p = 0.6, so the mean of the sample proportion distribution is 0.6.
• The sample size is n = 900, so the standard deviation is
• 74. Example 3
• Approximately 68% of the samples would show a sample proportion of between 58% and 62%.
• (1 standard deviation to each side of the mean.)
• 75. Standard Error
• In most situations, we do not know the population proportion.
• The point of measuring the sample is to estimate the population proportion.
• The standard error is the standard deviation of the set of all sample proportions:
• 76. Example 4
• What is the standard error in a sample of size 400 if the sample proportion in one sample is 35%?
• 77. Example 4
• Solution: Use the formula from the previous slide:
• 78. Confidence Intervals
• According to the 68-95-99.7 Rule, 95% of the time the sample proportion will be within 2 standard deviations of the population proportion.
• A 95% confidence interval is the interval
• 79. Confidence Intervals
• For a 95% confidence interval, the margin of error is
• s is called the standard error and is calculated the same way as standard deviation.
• 80. Example 5
• Determine the 95% confidence interval and the margin of error for a sample size of 400 with a sample proportion of 35%.
• 81. Example 5
• Solution: In a previous example we found the standard error in this case to be 2.4%.
• Calculate the confidence interval bounds:
• The margin of error is
• 82. Example 6
• In a sample of 600 U.S. citizens, 362 people say they drive an American-built car.
• Find the 95% confidence interval and the margin of error for the proportion of the population that drive an American-built car.
• 83. Example 6
• Solution: The sample proportion is:
• The standard error is:
• 84. Example 6
• Calculate the confidence interval bounds:
• The margin of error is
• 85. Example 6
• With a confidence level of 95% we can say that 60.3% of Americans drive American-built cars, with a margin of error of ± 4%.
• 86. Chapter 11 Assignment
• Section 11.1 pg 711 (3-6, 11,12,17-22,31,32)
• Section 11.2 pg 723 (3a,b,4a,b,10a,b,11,15,23)
• Section 11.3 pg 738 (3,5,9,23,29,33)