Upcoming SlideShare
×

Nossi ch 11

1,090 views

Published on

Contemporary math chapter 11 Power Point.

Published in: Technology, Economy & Finance
0 Likes
Statistics
Notes
• Full Name
Comment goes here.

Are you sure you want to Yes No
• Be the first to comment

• Be the first to like this

Views
Total views
1,090
On SlideShare
0
From Embeds
0
Number of Embeds
13
Actions
Shares
0
37
0
Likes
0
Embeds 0
No embeds

No notes for slide

Nossi ch 11

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