A Mathematical View of Our World

A Mathematical View A Mathematical View of Our Worldof Our World

11stst ed. ed.

Parks, Musser, Trimpe, Parks, Musser, Trimpe, Maurer, and MaurerMaurer, and Maurer

Chapter 11Chapter 11

Inferential StatisticsInferential Statistics

Section 11.1Section 11.1

Normal DistributionsNormal Distributions

• GoalsGoals

• Study normal distributionsStudy normal distributions• Study standard normal distributionsStudy standard normal distributions• Find the area under a standard normal Find the area under a standard normal

curvecurve

11.1 Initial Problem11.1 Initial Problem

• A class of 90 students had a mean test A class of 90 students had a mean test score of 74, with a standard deviation of score of 74, with a standard deviation of 8.8.

• If the professor curves the scores, how If the professor curves the scores, how many students will get As and how many many students will get As and how many will get Fs? will get Fs? • The solution will be given at the end of the section.The solution will be given at the end of the section.

Statistical InferenceStatistical Inference

• The process of making predictions The process of making predictions about an entire population based on about an entire population based on information from a sample is called information from a sample is called statistical inferencestatistical inference. .

Data DistributionsData Distributions

• For large data sets, a smooth curve can For large data sets, a smooth curve can often be used to approximate the histogram. often be used to approximate the histogram.

Data Distributions, cont’dData Distributions, cont’d

• The larger the data set and the smaller the The larger the data set and the smaller the bin size, the better the approximation of the bin size, the better the approximation of the smooth curve. smooth curve.

Example 1Example 1• The distribution of weights for a large The distribution of weights for a large

sample of college men is shown.sample of college men is shown.

Example 1, cont’dExample 1, cont’d

• What percent of the men have weights What percent of the men have weights between:between:

a)a)167 and 192 pounds?167 and 192 pounds?

b)b)137 and 192 pounds?137 and 192 pounds?

c)c) 137 and 222 pounds?137 and 222 pounds?


• Solution:Solution:

a)a)167 and 192 pounds?167 and 192 pounds?

• The area under the curve is 0.2, so 20% The area under the curve is 0.2, so 20% of the men are in this weight range.of the men are in this weight range.



b)b)137 and 192 pounds?137 and 192 pounds?




c)c) 137 and 222 pounds?137 and 222 pounds?


Normal DistributionsNormal Distributions• Data that has a symmetric, bell-shaped Data that has a symmetric, bell-shaped

distribution curve is said to have a distribution curve is said to have a normal normal distributiondistribution..• The mean and standard deviation determine the The mean and standard deviation determine the

exact shape and position of the curve.exact shape and position of the curve.

Example 2Example 2

a)a) Which normal curve has the largest mean?Which normal curve has the largest mean?

b)b) Which normal curve has the largest Which normal curve has the largest standard deviation? standard deviation?


a)a) Solution: The data sets are already labeled Solution: The data sets are already labeled in order of smallest mean to largest mean.in order of smallest mean to largest mean.• Data Set III has the largest mean.Data Set III has the largest mean.


b)b) Solution: Data Set III has the largest Solution: Data Set III has the largest standard deviation because it is the standard deviation because it is the shortest, widest curve.shortest, widest curve.• The order of the standard deviations is II, I, III.The order of the standard deviations is II, I, III.

Normal Distributions, cont’dNormal Distributions, cont’d

• Normal distributions with various means Normal distributions with various means and standard deviations are shown on and standard deviations are shown on the following slides.the following slides.




Standard Normal DistributionStandard Normal Distribution• The normal distribution with a mean of 0 and The normal distribution with a mean of 0 and

a standard deviation of 1 is called the a standard deviation of 1 is called the standard normal distributionstandard normal distribution. .

Standard Normal Distribution, cont’dStandard Normal Distribution, cont’d

• The areas under any normal distribution The areas under any normal distribution can be compared to the areas under can be compared to the areas under the standard normal distribution, as the standard normal distribution, as shown in the figure on the next slide. shown in the figure on the next slide.

Standard Normal Distribution, cont’dStandard Normal Distribution, cont’d

AreaArea

• One way to find the area under a region One way to find the area under a region of the standard normal curve is to use a of the standard normal curve is to use a table.table.• Tables of values for the standard normal Tables of values for the standard normal

curve are printed in textbooks to eliminate curve are printed in textbooks to eliminate the need to do repeated complicated the need to do repeated complicated calculations. calculations.

Example 3Example 3• What fraction of the total area under the What fraction of the total area under the

standard normal curve lies between standard normal curve lies between a = -0.5 and b = 1.5? a = -0.5 and b = 1.5?


• Solution: Find a = -0.5 and b = 1.5 in the Solution: Find a = -0.5 and b = 1.5 in the table.table.

Example 3, cont’dExample 3, cont’d• Solution, cont’d: The value in the table is Solution, cont’d: The value in the table is

0.6247.0.6247.

• A total of 62.47% of the area is shaded.A total of 62.47% of the area is shaded.

• In any normal distribution, 62.47% of the data In any normal distribution, 62.47% of the data lies between 0.5 standard deviations below lies between 0.5 standard deviations below the mean and 1.5 standard deviations above the mean and 1.5 standard deviations above the mean.the mean.

• The probability a randomly selected data The probability a randomly selected data value will lie between -0.5 and 1.5 is 62.47%value will lie between -0.5 and 1.5 is 62.47%

Example 4Example 4

• What percent of the data in a standard What percent of the data in a standard normal distribution lies between 0.5 and normal distribution lies between 0.5 and 2.5? 2.5?


• Solution: The value in the table for Solution: The value in the table for a = 0.5 and b = 2.5 is 0.3023.a = 0.5 and b = 2.5 is 0.3023.• So 30.23% of the data in a standard So 30.23% of the data in a standard

normal distribution lies between 0.5 and normal distribution lies between 0.5 and 2.5. 2.5.

Areas, cont’dAreas, cont’d

• Because the normal curve is Because the normal curve is symmetric, the areas in the previous symmetric, the areas in the previous table are repeated. table are repeated.

Areas, cont’dAreas, cont’d

• Figure Figure 11.11 11.11 and and table table 11.211.2

Areas, cont’dAreas, cont’d• A more common type of table:A more common type of table:

Example 5Example 5• Find the percent of data points in a standard Find the percent of data points in a standard

normal distribution that lie between normal distribution that lie between zz = -1.8 = -1.8 and and zz = 1.3. = 1.3.


• Solution: Find the two areas in Table Solution: Find the two areas in Table 11.3 and add them together.11.3 and add them together.• The area from 0 to 1.3 is 0.4032.The area from 0 to 1.3 is 0.4032.

• The area from 0 to -1.8 is 0.4641.The area from 0 to -1.8 is 0.4641.

• The total shaded area is 0.4032 + 0.4641 The total shaded area is 0.4032 + 0.4641 = 0.8673. = 0.8673.

Example 6Example 6• Find the percent of data points in a standard Find the percent of data points in a standard

normal distribution that lie between normal distribution that lie between zz = 1.2 = 1.2 and and zz = 1.7. = 1.7.


• Solution: Find the two areas in Table Solution: Find the two areas in Table 11.3 and subtract them.11.3 and subtract them.• The area from 0 to 1.2 is 0.3849.The area from 0 to 1.2 is 0.3849.

• The area from 0 to 1.7 is 0.4554.The area from 0 to 1.7 is 0.4554.

• The total shaded area is 0.4554 - 0.3849 The total shaded area is 0.4554 - 0.3849 = 0.0705. = 0.0705.

Question:Question:The value from the table associated withThe value from the table associated with z z = 2.1 is = 2.1 is 0.4821. To find the percentage of data values less than 0.4821. To find the percentage of data values less than -2.1 in a standard normal distribution, what do you need -2.1 in a standard normal distribution, what do you need to do?to do?

a. Add the table a. Add the table value to 0.5.value to 0.5.b. Subtract the b. Subtract the table value from 0.5.table value from 0.5.c. The table value c. The table value is the answer.is the answer.d. Divide the table d. Divide the table value in half.value in half.

Question:Question:What percentage of data values lie What percentage of data values lie between between z z = -1.2 and = -1.2 and zz = -0.7 in a = -0.7 in a standard normal distribution?standard normal distribution?

a. 11.51%a. 11.51%

b. 62.49%b. 62.49%

c. 12.69%c. 12.69%

d. 24.20% d. 24.20%

11.1 Initial Problem Solution11.1 Initial Problem Solution

• A class of 90 students had a mean test score A class of 90 students had a mean test score of 74 with a standard deviation of 8 points.of 74 with a standard deviation of 8 points.

• The test will be curved so that all students The test will be curved so that all students whose scores are at least 1.5 standard whose scores are at least 1.5 standard deviations above or below the mean will deviations above or below the mean will receive As and Fs, respectively.receive As and Fs, respectively.

• How many students will get As and how many How many students will get As and how many will get Fs? will get Fs?

Initial Problem Solution, cont’dInitial Problem Solution, cont’d• Because the class is large, it is likely the Because the class is large, it is likely the

scores have a normal distribution.scores have a normal distribution.• If the scores are curved:If the scores are curved:

• The mean of 74 will correspond to a score of 0 in The mean of 74 will correspond to a score of 0 in the standard normal distribution.the standard normal distribution.

• A score that is 1.5 standard deviations above the A score that is 1.5 standard deviations above the mean will correspond to a score of +1.5 in the mean will correspond to a score of +1.5 in the standard normal distribution, while a score that is standard normal distribution, while a score that is 1.5 standard deviations below the mean will 1.5 standard deviations below the mean will correspond to a score of -1.5.correspond to a score of -1.5.

Initial Problem Solution, cont’dInitial Problem Solution, cont’d

• The percentage of As is the same as the area The percentage of As is the same as the area to the right of to the right of zz = 1.5 in the standard normal = 1.5 in the standard normal distribution.distribution.• Approximately 43.32% of the area is between 0 Approximately 43.32% of the area is between 0

and 1.5.and 1.5.

• Since 50% of the area is to the right of 0, the Since 50% of the area is to the right of 0, the area above 1.5 is 50% - 43.32% = 6.68%area above 1.5 is 50% - 43.32% = 6.68%

• Thus, 6.68% of the students, or approximately 6 Thus, 6.68% of the students, or approximately 6 students, will receive As. students, will receive As.


• The percentage of Fs is the same as the area The percentage of Fs is the same as the area to the left of to the left of zz = -1.5 in the standard normal = -1.5 in the standard normal distribution.distribution.• Because of the symmetry of the normal Because of the symmetry of the normal

distribution, this is the same as the area above distribution, this is the same as the area above zz = 1.5, so the calculations are the same as in the = 1.5, so the calculations are the same as in the last step.last step.

• Thus, 6.68% of the students, or approximately 6 Thus, 6.68% of the students, or approximately 6 students, will receive Fs. students, will receive Fs.


Applications of Normal Applications of Normal

DistributionsDistributions

• GoalsGoals

• Study normal distribution applicationsStudy normal distribution applications• Use the 68-95-99.7 RuleUse the 68-95-99.7 Rule• Use the population Use the population zz-score-score

11.2 Initial Problem11.2 Initial Problem• Two suppliers make an engine part.Two suppliers make an engine part.

• Supplier A charges $120 for 100 parts which Supplier A charges $120 for 100 parts which have a standard deviation of 0.004 mm from have a standard deviation of 0.004 mm from the mean size.the mean size.

• Supplier B charges $90 for 100 parts which Supplier B charges $90 for 100 parts which have a standard deviation of 0.012 mm from have a standard deviation of 0.012 mm from the mean size.the mean size.

• Which supplier is a better choice?Which supplier is a better choice?• The solution will be given at the end of the The solution will be given at the end of the

section.section.

Normal DistributionsNormal Distributions

• If a data set is represented by a If a data set is represented by a normal distribution with mean normal distribution with mean μμ and and standard deviation standard deviation σσ, the percentage , the percentage of the data between of the data between μμ ++ r rσσ and and μμ + + ssσσ is the same as the percentage of the is the same as the percentage of the data in a standard normal distribution data in a standard normal distribution that lies between that lies between rr and and ss. .


Example 1Example 1• Approximately 10% of the data in a Approximately 10% of the data in a

standard normal distribution lies within 1/8 standard normal distribution lies within 1/8 of a standard deviation from the mean.of a standard deviation from the mean.• Within 1/8 means between -0.125 and 0.125.Within 1/8 means between -0.125 and 0.125.

• Suppose the measurements of a certain Suppose the measurements of a certain population are normally distributed with a population are normally distributed with a mean of 112 and standard deviation of 24. mean of 112 and standard deviation of 24. What values correspond to the interval What values correspond to the interval given above?given above?


• Solution: In the standard normal distribution Solution: In the standard normal distribution we are considering the interval from we are considering the interval from r r = -= -0.125 to 0.125 to ss = 0.125. = 0.125.• For the nonstandard distribution, the interval will For the nonstandard distribution, the interval will

be 112 + (-0.125)(24) = 109 to be 112 + (-0.125)(24) = 109 to 112 + (0.125)(24) = 115.112 + (0.125)(24) = 115.

• We know that 10% of the data values will lie We know that 10% of the data values will lie between 112 and 115.between 112 and 115.

Example 2Example 2

• The HDL cholesterol levels for a group of The HDL cholesterol levels for a group of women are approximately normally women are approximately normally distributed with a mean of 64 mg/dL and a distributed with a mean of 64 mg/dL and a standard deviation of 15 mg/dL.standard deviation of 15 mg/dL.

• Determine the percentage of these women Determine the percentage of these women that have HDL cholesterol levels between that have HDL cholesterol levels between 19 and 109 mg/dL. 19 and 109 mg/dL.

Example 2, cont’dExample 2, cont’d• Solution: The mean of 64 mg/dL Solution: The mean of 64 mg/dL

corresponds to 0 in the standard normal corresponds to 0 in the standard normal curve.curve.

• The value of 19 is 45 less than the mean, The value of 19 is 45 less than the mean, corresponding to 3 standard deviations corresponding to 3 standard deviations below the mean.below the mean.

• The value of 64 is 45 more than the mean, The value of 64 is 45 more than the mean, corresponding to 3 standard deviations corresponding to 3 standard deviations above the mean.above the mean.


• Solution, cont’d: The area under the Solution, cont’d: The area under the standard normal curve between standard normal curve between zz = -3 and = -3 and zz = 3 is found: = 3 is found:• From 0 to 3, there is 49.87% of the area.From 0 to 3, there is 49.87% of the area.

• From 0 to -3, there is also 49.87%.From 0 to -3, there is also 49.87%.

• Approximately, 2(49.87%) = 99.74% of the Approximately, 2(49.87%) = 99.74% of the women will have a HDL level between 19 and women will have a HDL level between 19 and 109 mg/dL. 109 mg/dL.

68-95-99.7 Rule68-95-99.7 Rule• For all normal distributions:For all normal distributions:

• Approximately 68% of the measurements Approximately 68% of the measurements lie within 1 standard deviation of the mean.lie within 1 standard deviation of the mean.

• Approximately 95% of the measurements Approximately 95% of the measurements lie within 2 standard deviations of the lie within 2 standard deviations of the mean.mean.

• Approximately 99.7% of the measurements Approximately 99.7% of the measurements lie within 3 standard deviations of the lie within 3 standard deviations of the mean.mean.

68-95-99.7 Rule, cont’d68-95-99.7 Rule, cont’d

Example 3Example 3• Designers of a new computer mouse Designers of a new computer mouse

have learned that the lengths of have learned that the lengths of women’s hands are normally women’s hands are normally distributed with a mean of 17 cm and a distributed with a mean of 17 cm and a standard deviation of 1 cm.standard deviation of 1 cm.

• What percentage of women have What percentage of women have hands in the range from 15 cm to 19 hands in the range from 15 cm to 19 cm? cm?

Example 3, cont’dExample 3, cont’d• Solution: Solution:

• A length of 15 cm is 2 standard deviations A length of 15 cm is 2 standard deviations below the mean of 17 cm.below the mean of 17 cm.

• A length of 19 cm is 2 standard deviations A length of 19 cm is 2 standard deviations above the mean of 17 cm.above the mean of 17 cm.

• According to the 68-95-99.7 Rule, the According to the 68-95-99.7 Rule, the percent of women whose hands are within percent of women whose hands are within 2 standard deviations of the mean length 2 standard deviations of the mean length is 95%. is 95%.

Question:Question:

Recall from the previous example Recall from the previous example that women’s hands have a mean that women’s hands have a mean length of 17 cm, with a standard length of 17 cm, with a standard deviation of 1 cm. Use the 68-95-deviation of 1 cm. Use the 68-95-99.7 Rule to determine what percent 99.7 Rule to determine what percent of women’s hands are between 14 of women’s hands are between 14 cm and 18cm long. cm and 18cm long.

a. 68.00%a. 68.00% b. 81.50%b. 81.50%c. 49.85%c. 49.85% d. 83.85%d. 83.85%

Example 4Example 4

• The lake sturgeon has a mean length of The lake sturgeon has a mean length of 114 cm and a standard deviation of 29 cm.114 cm and a standard deviation of 29 cm.

• If the lengths are normally distributed, If the lengths are normally distributed, determine:determine:

a)a) What percent of lake sturgeon had lengths What percent of lake sturgeon had lengths between 56 cm and 143 cm? between 56 cm and 143 cm?

b)b) What percent of lake sturgeon were not What percent of lake sturgeon were not between 56 cm and 143 cm in length?between 56 cm and 143 cm in length?

Example 4, cont’dExample 4, cont’d• Solution: Solution:

• Note that 114 cm is 2 standard deviations above the Note that 114 cm is 2 standard deviations above the mean.mean.

• Also, 56 cm is 1 standard deviation below the mean.Also, 56 cm is 1 standard deviation below the mean.

Example 4, cont’dExample 4, cont’d• Solution, cont’d:Solution, cont’d:

a)a) The area from The area from zz = -1 to = -1 to zz = 2 is 0.135 + 0.34 + 0.34 = = 2 is 0.135 + 0.34 + 0.34 = 0.815.0.815.

• So 81.5% of the sturgeon were between 56 cm and So 81.5% of the sturgeon were between 56 cm and 143 cm long.143 cm long.

Example 4, cont’dExample 4, cont’d• Solution, cont’d:Solution, cont’d:

b)b) This is the complement of the event in part (a).This is the complement of the event in part (a).

• So 100% - 81.5% = 18.5% of the sturgeon So 100% - 81.5% = 18.5% of the sturgeon were not between 56 cm and 143 cm long.were not between 56 cm and 143 cm long.

Population Population zz-scores-scores• The formula for converting a normal The formula for converting a normal

distribution value to a standard normal distribution value to a standard normal distribution value is called a distribution value is called a population population zz-score-score..

• The population The population zz-score of a -score of a measurement, measurement, xx, is given by:, is given by:

xz

Question:Question:

If a data value in a normal If a data value in a normal distribution has a population z-score distribution has a population z-score of 0, we know that of 0, we know that ..

a. The data value is equal to the mean.a. The data value is equal to the mean.b. The data value is larger than the mean.b. The data value is larger than the mean.c. The data value is smaller than the c. The data value is smaller than the mean.mean.d. The data value is equal to the standard d. The data value is equal to the standard deviation.deviation.

Example 5Example 5

• Suppose a normal distribution has a Suppose a normal distribution has a mean of 4 and a standard deviation of mean of 4 and a standard deviation of 3. 3.

• Find the Find the zz-scores of the measurements -scores of the measurements -1, 2, 3, 5, and 9. -1, 2, 3, 5, and 9.



• Solution, cont’d: The relationship between Solution, cont’d: The relationship between the normal values and the standard normal the normal values and the standard normal values is illustrated. values is illustrated.

Example 6Example 6

• In 1996, the finishing times for the New York In 1996, the finishing times for the New York City Marathon were approximately normal, City Marathon were approximately normal, with a mean of 260 minutes and a standard with a mean of 260 minutes and a standard deviation of about 50 minutes.deviation of about 50 minutes.

• What percentage of the finishers that year What percentage of the finishers that year had times between 285 minutes and 335 had times between 285 minutes and 335 minutes.minutes.


• Solution: Find the Solution: Find the zz-scores.-scores.• For a time of 285 minutes,For a time of 285 minutes,

• For a time of 335 minutes,For a time of 335 minutes,

285 260 250.5

50 50z

335 260 751.5

50 50z


• Solution, cont’d: Find the areasSolution, cont’d: Find the areas• The area from 0 to 0.5 is 0.1915.The area from 0 to 0.5 is 0.1915.

• The area from 0 to 1.5 is 1.4332.The area from 0 to 1.5 is 1.4332.


• Solution, cont’d: Subtract the areas to find Solution, cont’d: Subtract the areas to find 0.4332 – 0.1915 = 0.2417.0.4332 – 0.1915 = 0.2417.• The conclusion is that 24.17% of the finishing The conclusion is that 24.17% of the finishing

times were between 285 and 335 minutes. times were between 285 and 335 minutes.

Example 7Example 7

• Recall the distribution of HDL cholesterol Recall the distribution of HDL cholesterol levels from the previous example, with a levels from the previous example, with a mean of 64 mg/dL and a standard deviation mean of 64 mg/dL and a standard deviation of 15 mg/dL.of 15 mg/dL.

• If an HDL level of 40 mg/dL signals an If an HDL level of 40 mg/dL signals an increased risk for coronary heart disease, increased risk for coronary heart disease, what percentage of the women studied are what percentage of the women studied are at increased risk? at increased risk?


• Solution: Find the Solution: Find the z-z-score for an HDL score for an HDL level of 40 mg/dL: level of 40 mg/dL:

• The area between 0 and -1.6 is The area between 0 and -1.6 is 0.4452.0.4452.

40 641.6

15z


• Solution: The area to the left of -1.6 is Solution: The area to the left of -1.6 is 0.5 – 0.4452 = 0.0548.0.5 – 0.4452 = 0.0548.• In this group of women, 5.48% of them In this group of women, 5.48% of them

are at increased risk for coronary heart are at increased risk for coronary heart disease because of low HDL levels.disease because of low HDL levels.

11.2 Initial Problem Solution11.2 Initial Problem Solution• Two suppliers make an engine part.Two suppliers make an engine part.

• Supplier A charges $120 for 100 parts which Supplier A charges $120 for 100 parts which have a standard deviation of 0.004 mm from have a standard deviation of 0.004 mm from the mean size.the mean size.

• Supplier B charges $90 for 100 parts which Supplier B charges $90 for 100 parts which have a standard deviation of 0.012 mm from have a standard deviation of 0.012 mm from the mean size.the mean size.

• If parts must be within 0.012 mm to be If parts must be within 0.012 mm to be acceptable, which supplier is a better acceptable, which supplier is a better choice?choice?


• Determine the cost for each acceptable Determine the cost for each acceptable part from each supplier.part from each supplier.• Supplier A: Since Supplier A: Since σσ = 0.004 mm, all parts = 0.004 mm, all parts

within 3 standard deviations will be within 3 standard deviations will be acceptable.acceptable.

• We know that 99.7% of the parts are within 3 We know that 99.7% of the parts are within 3 standard deviations of the mean.standard deviations of the mean.

• Each acceptable part costs Each acceptable part costs $120 $1.2099.7


• Determine the cost for each acceptable Determine the cost for each acceptable part from each supplier.part from each supplier.• Supplier B: Since Supplier B: Since σσ = 0.012 mm, all parts = 0.012 mm, all parts

within 1 standard deviation will be within 1 standard deviation will be acceptable.acceptable.

• We know that 68% of the parts are within 1 We know that 68% of the parts are within 1 standard deviation of the mean.standard deviation of the mean.

• Each acceptable part costs Each acceptable part costs $90 $1.3268


• Overall each part from supplier B costs Overall each part from supplier B costs less than each part from supplier A, but less than each part from supplier A, but more parts from B will have to be more parts from B will have to be thrown away.thrown away.

• Each acceptable part from supplier B Each acceptable part from supplier B costs more than each acceptable part costs more than each acceptable part from supplier A.from supplier A.

• They should choose supplier A. They should choose supplier A.


Confidence IntervalsConfidence Intervals• GoalsGoals

• Study proportionsStudy proportions• Study population proportionsStudy population proportions• Study sample proportionsStudy sample proportions

• Study confidence intervalsStudy confidence intervals• Study margin of errorStudy margin of error

11.3 Initial Problem11.3 Initial Problem• A candy company prints prize tickets A candy company prints prize tickets

inside the wrappers of some of their inside the wrappers of some of their candy bars.candy bars.

• Suppose you buy 400 candy bars and Suppose you buy 400 candy bars and find that 25 of them have prizes. If you find that 25 of them have prizes. If you buy 1000 more, how many prizes would buy 1000 more, how many prizes would you expect to win? you expect to win? • The solution will be given at the end of the section.The solution will be given at the end of the section.

ProportionsProportions• A fraction of the population under A fraction of the population under

consideration is called a consideration is called a population population proportionproportion..• The notation for a population proportion is The notation for a population proportion is pp..

• For example, if 65,000,000 of 130,000,000 For example, if 65,000,000 of 130,000,000 people support the President’s budget, the people support the President’s budget, the population proportion of people who support population proportion of people who support the budget isthe budget is 65,000,000

50%130,000,000

p

Proportions, cont’dProportions, cont’d

• A fraction of the sample being A fraction of the sample being measured is called a measured is called a sample proportionsample proportion..• The notation for a sample proportion is .The notation for a sample proportion is .

• For example, if 198 of 413 people polled For example, if 198 of 413 people polled support the President’s budget, the support the President’s budget, the sample proportion of people who sample proportion of people who support the budget issupport the budget is 198

ˆ 48%413

p

p̂

Example 1Example 1

• A college has 3520 freshman, of which 1056 A college has 3520 freshman, of which 1056 have consumed an alcoholic beverage in the have consumed an alcoholic beverage in the last 30 days.last 30 days.

• Of the 50 students surveyed in a health Of the 50 students surveyed in a health class, 11 say they have had an alcoholic class, 11 say they have had an alcoholic beverage in the last 30 days.beverage in the last 30 days.

• What are the population proportion and the What are the population proportion and the sample proportion?sample proportion?


• Solution: The population is the 3520 Solution: The population is the 3520 freshmen at the college.freshmen at the college.

• The population proportion is The population proportion is

105630%

3520p


• Solution, cont’d: The sample is the 50 Solution, cont’d: The sample is the 50 students who were surveyed.students who were surveyed.

• The sample proportion is The sample proportion is

11ˆ 22%50

p


• Solution, cont’d: Notice that the Solution, cont’d: Notice that the population proportion and the sample population proportion and the sample proportion were not identical.proportion were not identical.

• The sample proportion can vary The sample proportion can vary depending on what random sample of depending on what random sample of students is chosen. students is chosen.

Example 1, cont’dExample 1, cont’d• Solution, cont’d: A Solution, cont’d: A

distribution of the distribution of the sample sample proportions for proportions for various possible various possible samples of this samples of this population is population is shown at right.shown at right.

Sample Proportions DistributionSample Proportions Distribution

• If samples of size If samples of size nn are taken from a are taken from a population having a population population having a population proportion proportion pp, then the set of all sample , then the set of all sample proportions has a mean and standard proportions has a mean and standard deviation of:deviation of:•

•

p

1p p

n

Sample Proportions, cont’dSample Proportions, cont’d• If two conditions are met, then If two conditions are met, then nn is large is large

enough and the distribution of sample enough and the distribution of sample proportions is approximately normal. proportions is approximately normal.

• The conditions are: The conditions are:

•

•

13 0p p

pn

13 1p p

pn

Example 2Example 2

• Suppose the population proportion of a Suppose the population proportion of a group is 0.4, and we choose a simple group is 0.4, and we choose a simple random sample of size 30.random sample of size 30.

• Find the mean and standard deviation Find the mean and standard deviation of the set of all sample proportions.of the set of all sample proportions.


• Solution: In this case, Solution: In this case, pp = 0.4 and = 0.4 and nn = = 30.30.• The mean is The mean is

• The standard deviation is The standard deviation is

0.4p

1 0.4 1 0.40.09

30

p p

n

Example 2, cont’dExample 2, cont’d• Solution, cont’d: The sample proportion Solution, cont’d: The sample proportion

distribution is graphed below.distribution is graphed below.

Question:Question:

If a population proportion is known to If a population proportion is known to be 0.25, is a sample size of 20 large be 0.25, is a sample size of 20 large enough to guarantee that the enough to guarantee that the distribution of sample proportions is distribution of sample proportions is approximately normal? approximately normal?

a. yesa. yes b. nob. no

Example 3Example 3

• Fox News asked 900 registered voters Fox News asked 900 registered voters whether or not they would take a smallpox whether or not they would take a smallpox vaccine.vaccine.

• Suppose it is known that 60% of all Suppose it is known that 60% of all Americans would take the vaccine. What is Americans would take the vaccine. What is the approximate percentage of samples for the approximate percentage of samples for which between 58% and 62% of voters in the which between 58% and 62% of voters in the sample would take the shot? sample would take the shot?


• Solution: We know that Solution: We know that pp = 0.6, so the mean = 0.6, so the mean of the sample proportion distribution is 0.6.of the sample proportion distribution is 0.6.

• The sample size is The sample size is nn = 900, so the standard = 900, so the standard deviation is deviation is

1 0.6 1 0.60.02

900

p p

n

Example 3, cont’dExample 3, cont’d• Solution, cont’d: A normal curve is shown, Solution, cont’d: A normal curve is shown,

labeled with sample proportion values as well labeled with sample proportion values as well as their as their zz-scores.-scores.


• Solution, cont’d: Approximately 68% of the Solution, cont’d: Approximately 68% of the samples would show a sample proportion of samples would show a sample proportion of between 58% and 62%.between 58% and 62%.

Standard ErrorStandard Error• In most situations, we do not know the In most situations, we do not know the

population proportion.population proportion.• The point of measuring the sample is to The point of measuring the sample is to

estimate the population proportion.estimate the population proportion.

• The The standard errorstandard error is the standard deviation is the standard deviation of the set of all sample proportions:of the set of all sample proportions:

ˆ ˆ1ˆ

p ps

n

Example 4Example 4

• What is the standard error in a sample What is the standard error in a sample of size 400 if the sample proportion in of size 400 if the sample proportion in one sample is 35%?one sample is 35%?


• Solution: Use the formula from the Solution: Use the formula from the previous slide:previous slide:

ˆ ˆ1 0.35 1 0.35ˆ 0.024

400

p ps

n

Confidence IntervalsConfidence Intervals• According to the 68-95-99.7 Rule, 95% of the According to the 68-95-99.7 Rule, 95% of the

time the sample proportion will be within 2 time the sample proportion will be within 2 standard deviations of the population standard deviations of the population

proportion.proportion. • A A 95% confidence interval95% confidence interval is the interval is the interval

ˆ ˆ ˆ ˆ2 , 2p s p s

Confidence Intervals, cont’dConfidence Intervals, cont’d

• For a 95% confidence interval, the For a 95% confidence interval, the margin of errormargin of error is is

• Any value in the confidence interval is a Any value in the confidence interval is a reasonable estimate for the population reasonable estimate for the population proportion.proportion.

ˆ2s

Confidence Intervals, cont’dConfidence Intervals, cont’d

• For example, (a) and (b) below show For example, (a) and (b) below show good estimates while (c) shows an good estimates while (c) shows an unlikely estimate. unlikely estimate.

Example 5Example 5

• Determine the 95% confidence interval Determine the 95% confidence interval and the margin of error for a sample and the margin of error for a sample size of 400 with a sample proportion of size of 400 with a sample proportion of 35%.35%.


• Solution: In a previous example we found Solution: In a previous example we found the standard error in this case to be 2.4%.the standard error in this case to be 2.4%.

• Calculate the confidence interval bounds:Calculate the confidence interval bounds:•

•

• The margin of error is The margin of error is

ˆ ˆ2 35% 2 2.4% 30.2%p s

ˆ ˆ2 35% 2 2.4% 39.8%p s

ˆ2 2 2.4% 4.8%s

Question:Question:Find the 95% confidence interval for a Find the 95% confidence interval for a sample size of 100 with a sample sample size of 100 with a sample proportion of 25%. Round your answer proportion of 25%. Round your answer to the nearest hundredth of a percent. to the nearest hundredth of a percent.

a. (20.67%, 29.33%)a. (20.67%, 29.33%)

b. (24.63%, 25.38%)b. (24.63%, 25.38%)

c. (16.34%, 33.66%)c. (16.34%, 33.66%)

d. (12.01%, 37.99%)d. (12.01%, 37.99%)

Question:Question:

Find the margin of error for the 95% Find the margin of error for the 95% confidence interval in the previous confidence interval in the previous question.question.

Recall, the sample size was 100 and Recall, the sample size was 100 and the sample proportion was 25%. the sample proportion was 25%. Round to the nearest hundredth of a Round to the nearest hundredth of a percent.percent.

a. ± 4.33%a. ± 4.33% b. ± 17.32%b. ± 17.32%c. ± 2.17%c. ± 2.17% d. ± 8.66% d. ± 8.66%

Example 6Example 6

• In a sample of 600 U.S. citizens, 362 In a sample of 600 U.S. citizens, 362 people say they drive an American-built people say they drive an American-built car.car.

• Find the 95% confidence interval and Find the 95% confidence interval and the margin of error for the proportion of the margin of error for the proportion of the population that drive an American-the population that drive an American-built car. built car.


• Solution: The sample proportion is: Solution: The sample proportion is:

• The standard error is:The standard error is:

362ˆ 0.603600

p

ˆ ˆ1 0.603 1 0.603ˆ 0.020

600

p ps

n


• Solution, cont’d: Calculate the Solution, cont’d: Calculate the confidence interval bounds:confidence interval bounds:•

•

• The margin of error is The margin of error is

ˆ ˆ2 60.3% 2 2% 56.3%p s

ˆ2 2 2% 4%s

ˆ ˆ2 60.3% 2 2% 64.3%p s


• Solution, cont’d: With a confidence Solution, cont’d: With a confidence level of 95% we can say that 60.3% of level of 95% we can say that 60.3% of Americans drive American-built cars, Americans drive American-built cars, with a margin of error of with a margin of error of ± 4%.± 4%.

Example 7Example 7

• In a survey of 1000 adults, 44% said they In a survey of 1000 adults, 44% said they were satisfied with the quality of health care were satisfied with the quality of health care in the U.S.in the U.S.

• The margin of error was reported as The margin of error was reported as ± 3%.± 3%.• Assuming a 95% confidence interval was Assuming a 95% confidence interval was

used, verify that the margin of error is used, verify that the margin of error is correct and explain what it means. correct and explain what it means.


• Solution: We know Solution: We know nn = 1000 and = 1000 and • The standard error isThe standard error is

ˆ 0.44p

ˆ ˆ1 0.44 1 0.44ˆ 0.0157

1000

p ps

n


• Solution, cont’d: The margin of error is Solution, cont’d: The margin of error is

• The margin of error of approximately 3% The margin of error of approximately 3% indicates that the researchers are 95% indicates that the researchers are 95% confident that the true percentage of adults confident that the true percentage of adults satisfied with health care in the U.S. is satisfied with health care in the U.S. is between 41% and 47%.between 41% and 47%.

ˆ2 2 0.0157 0.0314s

Example 8Example 8

• A manufacturer tests 1000 computer A manufacturer tests 1000 computer chips and finds 216 defective ones. chips and finds 216 defective ones. Find a 95% confidence interval for the Find a 95% confidence interval for the population proportion of defective population proportion of defective chips.chips.


• Solution: Solution: We know We know nn = 1000 and = 1000 and

• The standard error isThe standard error is

ˆ 0.216p

ˆ ˆ1 0.216 1 0.216ˆ 0.013

1000

p ps

n



•

• The 95% confidence interval is The 95% confidence interval is (19.0%, 24.2%).(19.0%, 24.2%).

ˆ ˆ2 21.6% 2 1.3% 19.0%p s

ˆ ˆ2 21.6% 2 1.3% 24.2%p s

Example 9Example 9

• A manufacturer tests 10,000 computer A manufacturer tests 10,000 computer chips and finds 2160 defective ones. chips and finds 2160 defective ones. Find a 95% confidence interval for the Find a 95% confidence interval for the population proportion of defective population proportion of defective chips.chips.

• Does choosing a larger sample give Does choosing a larger sample give significantly better results?significantly better results?


• Solution: Solution: We know We know nn = 10,000 and = 10,000 and


ˆ 0.216p

ˆ ˆ1 0.216 1 0.216ˆ 0.004

10,000

p ps

n



•

• The 95% confidence interval is The 95% confidence interval is (20.8%, 22.4%).(20.8%, 22.4%).

ˆ ˆ2 21.6% 2 0.4% 20.8%p s

ˆ ˆ2 21.6% 2 0.4% 22.4%p s

Example 9, cont’dExample 9, cont’d• Solution, cont’d: Compare the results for Solution, cont’d: Compare the results for

sample sizes of 1000 and 10,000.sample sizes of 1000 and 10,000.• For For nn = 1000, the 95% confidence interval is = 1000, the 95% confidence interval is

(19.0%, 24.2%).(19.0%, 24.2%).

• For For nn = 10,000, the 95% confidence interval is = 10,000, the 95% confidence interval is (20.8%, 22.4%). (20.8%, 22.4%).

• Increasing the sample size to 10,000 does Increasing the sample size to 10,000 does give a significantly better estimate of the give a significantly better estimate of the population proportion. population proportion.

11.3 Initial Problem Solution11.3 Initial Problem Solution

• A candy company prints prize tickets A candy company prints prize tickets inside the wrappers of some of their inside the wrappers of some of their candy bars.candy bars.

• Suppose you buy 400 candy bars Suppose you buy 400 candy bars and find that 25 of them have prizes. and find that 25 of them have prizes. If you buy 1000 more, how many If you buy 1000 more, how many prizes would you expect to win?prizes would you expect to win?


• For the first 400 candy bars you For the first 400 candy bars you bought, the sample proportion of bought, the sample proportion of winning bars is winning bars is


25ˆ 0.0625400

p

ˆ ˆ1 0.0625 1 0.0625ˆ 0.0121

400

p ps

n


• Calculate the confidence interval bounds:Calculate the confidence interval bounds:

•

•

• Out of 1000 new candy bars, you should Out of 1000 new candy bars, you should expect between 3.83% and 8.67%, or expect between 3.83% and 8.67%, or between 38 and 87 bars, to be winners. between 38 and 87 bars, to be winners.

ˆ ˆ2 6.25% 2 1.21% 3.83%p s

ˆ ˆ2 6.25% 2 1.21% 8.67%p s

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A Mathematical View of Our World