Idea Transcript
LESSON
23.3
Name
Sampling Distributions
Class
Date
23.3 Sampling Distributions Essential Question: How is the mean of a sampling distribution related to the corresponding population mean or population proportion?
Common Core Math Standards The student is expected to: COMMON CORE
Resource Locker
S-IC.B.4
Explore 1
Use data from a sample survey to estimate a population mean or proportion; develop a margin of error through the use of simulation models for random sampling.
Developing a Distribution of Sample Means
The tables provide the following data about the first 50 people to join a new gym: member ID number, age, and sex.
Mathematical Practices COMMON CORE
MP.6 Precision
Language Objective Work with a partner to compare and contrast the standard error of the mean and the standard error of the proportion.
Essential Question: How is the mean of a sampling distribution related to the corresponding population mean or population proportion? The mean of the sampling distribution of the sample mean is equal to the population mean. Similarly, the mean of the sampling distribution of the sample proportion is equal to the population proportion.
PREVIEW: LESSON PERFORMANCE TASK
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Enter the age data into a graphing calculator and find the mean age μ and standard deviation σ for the population of the gym’s first 50 members. Round each statistic to the nearest tenth.
µ ≈ 41.2; σ ≈ 15.3
Use a graphing calculator’s random number generator to choose a sample of 5 gym _ members. Find the mean age x for your sample. Round to the nearest tenth.
Answers will vary. Possible answer: x¯ = 39.5 Module 23
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Date
ling 23.3 Samp
ng correspondi d to the ution relate ing distrib Resource of a sampl proportion? Locker is the mean population pa ion: How mean or tion; develo population mean or propor population ng. estimate a random sampli survey to ns a sample models for Mea tion from ple data of simula COMMON S-IC.B.4 Use h the use CORE tion of Sam er ID number, age, and sex. error throug margin of a Distribu memb ing gym: join a new 1 Develop people to Explore the first 50 about data following provide the The tables
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rd deviation μ and standa t tenth. mean age 10 and find the c to the neares calculator each statisti a graphing ers. Round age data into gym’s first 50 memb Enter the ation of the for the popul of 5 gym a sample σ ≈ 15.3 to choose µ ≈ 41.2; t tenth. generator m number to the neares rando Round e. ator’s _ sampl ing calcul x for your Use a graph mean age Find the x¯ = 39.5 members. answer: Possible 1141 will vary. Answers
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Credits: ©Juice
View the Engage section online. Discuss the photo and note that the U.S. Census Bureau gathers data on the number of people in each household and the number of households that have each number of people, and that sampling could be used to find relevant probabilities. Then preview the Lesson Performance Task.
© Houghton Mifflin Harcourt Publishing Company • Image Credits: ©Juice Images/Shutterstock
ENGAGE
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2:55 PM
04/04/14 11:21 AM
Report your sample mean to your teacher. As other students report their sample means, create a class histogram. To do so, shade a square above the appropriate interval as each sample mean is reported. For sample means that lie on an interval boundary, shade a square on the interval to the right. For instance, if the sample mean is 39.5, shade a square on the interval from 39.5 to 40.5. Answers will vary. A possible Make your own copy of the class histogram using the grid shown. histogram is shown. 8
Frequency
C
EXPLORE 1 Developing a Distribution of Sample Means
6
INTEGRATE TECHNOLOGY
4
Students will use graphing calculators to first calculate the mean and standard deviation of a population, then randomly select samples from that population, and finally calculate statistics for the distribution of sample means. This simulation will enable them to see the effect of sample size on how closely the sample means approximate the population mean.
2 25.5 27.5 29.5 31.5 33.5 35.5 37.5 39.5 41.5 43.5 45.5 47.5 49.5 51.5 53.5 55.5 26.5 28.5 30.5 32.5 34.5 36.5 38.5 40.5 42.5 44.5 46.5 48.5 50.5 52.5 54.5
Sample Mean
D
Calculate the mean of the sample means, μ x_, and the standard deviation of the sample means, σ x_.
Answers will vary. Possible answer: μ x_ = 39.8; σ x_ = 7.9
E
Now use a graphing calculator’s random number generator to choose a sample of 15 gym members. Find the mean for your sample. Round to the nearest tenth.
QUESTIONING STRATEGIES
_ Answers will vary. Possible answer: x = 40.5 Report your sample mean to your teacher. As other students report their sample means, create a class histogram and make your own copy of it. Answers will vary. A possible histogram is shown. 8 6 4 2 25.5 27.5 29.5 31.5 33.5 35.5 37.5 39.5 41.5 43.5 45.5 47.5 49.5 51.5 53.5 55.5 26.5 28.5 30.5 32.5 34.5 36.5 38.5 40.5 42.5 44.5 46.5 48.5 50.5 52.5 54.5
Sample Mean
G
Calculate the mean of the sample means, μ x_, and the standard deviation of the sample means, σ x_.
© Houghton Mifflin Harcourt Publishing Company
Frequency
F
How can you change the way a sample is selected to make the sample mean better match the population mean? Increase the sample size. As the sample size increases, the sample mean approaches the population mean.
Answers will vary. Possible answer: μ x_ = 40.8; σ x_ = 3.4
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Learning Progressions In previous lessons, students constructed probability distributions, and they explored normal distributions in detail. In this lesson, students will construct distributions of data obtained from different samples of the same population. Students should understand that the mean and standard deviation for a sample usually will not match the statistics for the entire population or for a different sample of the same population. Students will learn how to use population statistics to determine the likelihood that a sample will have certain characteristics. In the next lesson they will build on this knowledge to make predictions about population parameters from sample statistics.
04/04/14 11:21 AM
Sampling Distributions 1142
v Reflect
EXPLORE 2
1.
In the class histograms, how does the mean of the sample means compare with the population mean?
The mean of the sample means is close to the population mean.
Developing a Distribution of Sample Proportions
2.
What happens to the standard deviation of the sample means as the sample size increases?
The standard deviation of the sample means decreases. 3.
QUESTIONING STRATEGIES
What happens to the shape of the histogram as the sample size increases?
The histogram gets closer to the shape of a normal distribution.
What is the difference between this sampling distribution and the one in the first Explore activity? The first one was a distribution of sample means, while this is a distribution of sample proportions.
Explore 2
Developing a Distribution of Sample Proportions
Use the tables of gym membership data from Explore 1. This time you will develop a sampling distribution based on a sample proportion rather than a sample mean.
A
Find the proportion p of female gym members in the population.
p = 0.6
INTEGRATE MATHEMATICAL PRACTICES Focus on Reasoning MP.2 When creating a histogram of sample
Use a graphing calculator’s random number generator to choose a sample of 5 gym members. Find the proportion p̂ of female gym members for your sample.
Answers will vary. All sample proportions will be 0, 0.2, 0.4, 0.6, 0.8, or 1.
C
Report your sample proportion to your teacher. As other students report their sample proportions, create a class
10 8
Answers will vary. A possible histogram is shown.
Frequency
histogram and make your own copy of it.
© Houghton Mifflin Harcourt Publishing Company
proportions for a sample size of 5, ask students why there are only a few different values for the sample proportion, even when many samples are selected. Students should recognize that the only possible values for the proportion are 0, 0.2, 0.4, 0.6, 0.8 and 1.
B
6 4 2 -0.05 0.15 0.35 0.55 0.75 0.95 0.05 0.25 0.45 0.65 0.85 1.05
Sample Proportion
D
Calculate the mean of the sample proportions, μ p̂, and the standard deviation of the sample proportions, σ p̂. Round to the nearest hundredth.
Answers will vary. Possible answer: µ p̂ = 0.68; σ p = 0.21
E
Now use your calculator’s random number generator to choose a sample of 10 gym members. Find the proportion of female members p̂ for your sample.
Answers will vary. All sample proportions will be 0, 0.1, 0.2, …, 0.9, or 1.
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Peer-to-Peer Activity Have students work in pairs. Give each pair the mean and standard deviation for a population. Have each student choose a sample size, determine the standard error of the mean for that sample size, and identify an interval that contains either 68%, 95%, or 99.7% of the sample means. Next, have students tell their partners the sample size they used and the upper and lower bounds of the interval they found. Have each student calculate the percent of sample means that fall within the partner’s interval. Have partners check each other’s results.
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Lesson 23.3
04/04/14 11:21 AM
Answers will vary. A possible histogram is shown.
G
10
Report your sample proportion to your teacher. As other students report their sample proportions, create a class histogram and make your own copy of it.
Calculate the mean of the sample proportions, μ p̂, and the standard deviation of the sample proportions, σ p̂. Round to the nearest hundredth.
EXPLAIN 1
8
Frequency
F
Answers will vary. Possible answer: µ p̂ = 0.55; σ p̂ = 0.16
6
Using the Sampling Distribution of the Sample Mean
4 2
QUESTIONING STRATEGIES
-0.05 0.15 0.35 0.55 0.75 0.95 0.0 0.25 0.45 0.65 0.85 1.05
If you create two sampling distributions by randomly selecting samples from the same population, with one distribution having samples of size 25 and the other having samples of size 100, how will the standard errors of the means of the two distributions compare? Explain. The standard error of the mean for samples of 100 will be half as large as the standard error for samples of 25. Standard error is inversely proportional to the square root of the sample size, so quadrupling the sample size reduces the standard error by a factor of 2.
Sample Proportion v Reflect In the class histograms, how does the mean of the sample proportions compare with the population proportion?
4.
The mean of the sample proportions is close to the population proportion. What happens to the standard deviation of the sample proportions as the sample size increases?
5.
The standard deviation of the sample proportions decreases.
Explain 1
Using the Sampling Distribution of the Sample Mean
The histograms that you made in the two Explores are sampling distributions. A sampling distribution shows how a particular statistic varies across all samples of n individuals from the same population. In Explore 1, you _ approximated sampling distributions of the sample mean, x, for samples of size 5 and 15. (The reason your sampling distributions are approximations is that you did not find all samples of a given size.) The mean of the sampling distribution of the sample mean is denoted µ x_. The standard deviation of the sampling distribution of the sample mean is denoted σ x_ and is also called the standard error of the mean.
Properties of the Sampling Distribution of the Sample Mean If a random sample of size n is selected from a population with mean μ and standard deviation σ, then
1. 2. 3.
µ x_ = µ, σ_ , and σ x_ = _ √n The sampling distribution of the sample mean is normal if the population is normal; for all other populations, the sampling distribution of the mean approaches a normal distribution as n increases.
Review how to use a graphing calculator to find the probability that a value in a normal distribution is between two specified values. Find the cumulative distribution function by going to the DISTR menu and selecting 2:normalcdf(, and entering the given information in the correct order (smaller value, larger value, mean, standard deviation) to find the probability. To find the probability that a quantity is less than a given value, enter −1 EE 99 for the lower value. To find the probability that a quantity is greater than a given value, enter 1 EE 99 for the upper value.
© Houghton Mifflin Harcourt Publishing Company
_ In Explore 1, you may have discovered that µ x_ is close to x regardless of the sample size and that σ x_ decreases as the sample size n increases. You based these observations on simulations. When you consider all possible samples of n individuals, you arrive at one of the major theorems of statistics.
INTEGRATE TECHNOLOGY
The third property stated above is known as the Central Limit Theorem.
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Sampling Distributions 1144
Example 2
AVOID COMMON ERRORS Remind students that when using a calculator to find the probability that the mean of a sample is between two given values (or above or below a given value), they must first calculate σ x_, the standard error of the mean, then use that value as the standard deviation that they enter on the calculator. If, instead, they use the standard deviation of the population, they will obtain an incorrect result.
Boxes of Cruncho cereal have a mean mass of 323 grams with a standard deviation of 20 grams.
For random samples of 36 boxes, what interval centered on the mean of the sampling distribution captures 95% of the sample means? Write the given information about the population and a sample. = 323
σ = 20
n = 36
Find the mean of the sampling distribution of the sample mean and the standard error of the mean. 20 20 σ_ = _ _ = _ ≈ 3.3 σ x_ = _ √n 6 √ 36 The sampling distribution of the sample mean is approximately normal. In a normal distribution, 95% of the data fall within 2 standard deviations of the mean. x_ = = 323
x_ - 2σ x_ = 323 - 2(3.3) = 316.4
x_ + 2σ x_ = 323 + 2(3.3) = 329.6
So, for random samples of 36 boxes, 95% of the sample means fall between 316.4 grams and 329.6 grams.
What is the probability that a random sample of 25 boxes has a mean mass of at most 325 grams? Write the given information about the population and the sample. = 323
σ = 20
n = 25
Find the mean of the sampling distribution of the sample mean and the standard error of the mean.
© Houghton Mifflin Harcourt Publishing Company
x_ = = 323
20 20 σ_ = _ _ = _= σ x_ = _ √n 5 25
√
4
The sampling distribution of the sample mean is approximately normal. Use a graphing calculator to find _ P(x ≤ 325). _ P(x ≤ 325) = normalcdf -1e99, 325 , 323 , 4 ≈ 0.69
(
)
So, the probability that the random sample has a mean mass of at most 325 grams is about
0.69 .
Your Turn
Boxes of Cruncho cereal have a mean mass of 323 grams with a standard deviation of 20 grams. 6.
For random samples of 50 boxes, what interval centered on the mean of the sampling distribution captures 99.7% of the sample means?
x_ = = 323
20 σ σ x_ = ___ ― ≈ 2.8 ― = ____ √n
√50
The sampling distribution of the sample mean is approximately normal. In a normal distribution, 99.7% of the data fall within 3 standard deviations of the mean. x_ -3σ x_ = 323 - 3(2.8) = 314.6
x_ +3σ x_ = 323 + 3(2.8) = 331.4
So, for random samples of 50 boxes, 99.7% of the sample means fall between 314.6 grams and 331.4 grams. Module 23
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What is the probability that a random sample of 100 boxes has a mean mass of at least 320 grams?
7.
20 20 σ σ x_ = ___ =2 ―― = __ ― = _____ 10
x_ = = 323
√n
EXPLAIN 2
√100
The sampling distribution of the sample mean is approximately normal. _ P (x ≥ 320) = normalcdf (320, 1E99, 323, 2) ≈ 0.93
Using the Sampling Distribution of the Sample Proportion
So, the probability that the random sample has a mean mass of at least 320 grams is about 0.93.
Explain 2
Using the Sampling Distribution of the Sample Proportion
QUESTIONING STRATEGIES
When you work with the sampling distribution of a sample proportion, p represents the proportion of individuals in the population that have a particular characteristic (that is, the proportion of “successes”) and p̂ is the proportion of successes in a sample. The mean of the sampling distribution of the sample proportion is denoted p̂. The standard deviation of the sampling distribution of the sample proportion is denoted σ p̂ and is also called the standard error of the proportion.
How are the mean and standard deviation of the sampling distribution of a sample proportion similar to and different from the mean and standard deviation of the sampling distribution of a sample mean? In both cases, the mean of the sampling distribution is equal to the corresponding value (mean or proportion) for the population. However, the standard deviation is calculated somewhat differently for sample means and sample proportions.
Properties of the Sampling Distribution of the Sample Proportion If a random sample of size n is selected from a population with proportion p of successes, then
1.
p̂ = p,
2.
σ p̂ =
3.
n if both np and n(1− p) are at least 10, then the sampling distribution of the sample proportion is approximately normal.
Example 2
――――
p(1 - p) _ , and
40% of the students at a university live off campus. When sampling from this population, consider “successes” to be students who live off campus.
INTEGRATE MATHEMATICAL PRACTICES Focus on Critical Thinking MP.3 Discuss with students the criteria for
For random samples of 50 students, what interval centered on the mean of the sampling distribution captures 95% of the sample proportions?
p = 0.4
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Write the given information about the population and a sample. n = 50
Find the mean of the sampling distribution of the sample proportion and the standard error of the proportion. p(1 - p) 0.4(1 - 0.4) µ p̂ = p = 0.4 σ p̂ = ______ = ________ ≈ 0.069 n 50
―――
――――
Check that np and n(1 - p) are both at least 10.
np = 50 • 0.4 = 20
n(1 - p) = 50 • 0.6 = 30
Since np and n(1 - p) are both greater than 10, the sampling distribution of the sample proportion is approximately normal. In a normal distribution, 95% of the data fall within 2 standard deviations of the mean. µ p̂ - 2σ p̂ = 0.4 - 2(0.069) = 0.262
determining whether a sampling distribution is approximately normal. If the population is normally distributed, the sampling distribution of the sample mean will also be normal. Also, the sampling distribution of a sample mean approaches a normal distribution as the sample size n increases. The distribution is approximately normal when n is very large, but we do not have precise definitions of “approximately normal” or “very large.”
µ p̂ + 2σ p̂ = 0.4 + 2(0.069)= 0.538
So, for random samples of 50 students, 95% of the sample proportions fall between 26.2% and 53.8%. Module 23
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LANGUAGE SUPPORT A2_MNLESE385900_U9M23L3.indd 1146
Communicate Math Have students work with a partner to compare and contrast the standard error of the mean and the standard error of the proportion. Monitor students’ use of terminology to learn what words and concepts are difficult for them to understand or to communicate.
10/16/14 8:00 AM
For the sampling distribution of a sample proportion, how close the distribution is to normal depends on both n and the proportion p. If p is very large or very small, n must be especially large for the distribution to be approximately normal. We can use the rule of thumb that says that if both np and n(1 - p) are at least 10, we can consider the distribution approximately normal.
Sampling Distributions 1146
B
What is the probability that a random sample of 25 students has a sample proportion of at most 37%? Write the given information about the population and the sample, where a success is a student who lives off campus. p = 0.4 n = 25 Find the mean of the sampling distribution of the sample proportion and the standard error of the proportion. 0.4 1 - 0.4 p(1 - p) _____________ = 0.098 μ p̂ = p = 0.4 σ p̂ = _______ = n 25
――――――
―――
(
Check that np and n(1 - p) are both at least 10.
np = 25
•
n(1 - p) = 25
0.4 = 10
•
)
0.6 = 15
Since np and n(1 - p) are greater than or equal to 10, the sampling distribution of the sample proportion is approximately normal. Use a graphing calculator to find P(p̂ ≤ 0.37).
(
)
P(p̂ ≤ 0.37) = normalcdf -1e99, 0.37 , 0.4 , 0.098 ≈ 0.38 So, the probability that the random sample has a sample proportion of at most 37% is about
0.38 .
YourTurn
40% of the students at a university live off campus. When sampling from this population, consider “successes” to be students who live off campus. 8.
For random samples of 80 students, what interval centered on the mean of the sampling distribution captures 68% of the sample proportions?
© Houghton Mifflin Harcourt Publishing Company
µ p̂ = p = 0.4
――
µ p̂ - σ p̂ = 0.4 - 0.055 = 0.345
µ p̂ + σ p̂ = 0.4 + 0.055 = 0.455
So, for random samples of 80 students, 68% of the sample proportions fall between 34.5% and 45.5%.
A2_MNLESE385900_U9M23L3 1147
Lesson 23.3
―――
p(1 - p) 0.4(1 - 0.4) = __________ ≈ 0.055 √_______ n 80
Since np = 32, n(1 - p) = 48, and both are greater than 10, the sampling distribution is approximately normal. In a normal distribution, 68% of the data fall within 1 standard deviation of the mean.
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σ p̂ =
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9.
What is the probability that a random sample of 60 students includes more than 18 students who live off campus?
µ p̂ = p = 0.4
σp =
―――
ELABORATE
――――
p(1 - p) 0.4(1 - 0.4) = √__________ ≈ 0.063 √_______ n 60
Since np = 24, n(1 - p) = 36, and both are greater than 10, the sampling distribution is approximately normal. p̂ min =
INTEGRATE MATHEMATICAL PRACTICES Focus on Communication MP.3 Ask students to describe different methods
18 _ = 0.3
60 P (p̂ > 0.3) = normalcdf (0.3, 1e99, 0.4, 0.063) ≈ 0.94 So, the probability that the random sample includes more than 18 students who live off campus is about 0.94.
for determining the probability that a sample mean or proportion falls between two given values. They should understand that if the given values are multiples of a standard deviation away from the mean of the sampling distribution, they can use the known percentages of data that fall within 0, 1, 2, or 3 standard deviations of the mean. If not, they can calculate z-scores for the given values and look up the corresponding probabilities on a standard normal table. Alternatively, they can use the cumulative distribution function on a graphing calculator.
Elaborate 10. What is a sampling distribution? A sampling distribution is a distribution that shows how a particular statistic varies across
all samples of n individuals from the same population. 11. What allows you to conclude that 95% of the sample means in a sampling distribution are within 2 standard deviations of the population mean? The Central Limit Theorem says that the sampling distribution of the sample mean is
normal or approximately normal, so 95% of the sample means will fall within 2 standard deviations of the mean of the sampling distribution, but the mean of the sampling distribution is equal to the population mean, so 95% of the sample means will fall within
SUMMARIZE THE LESSON
2 standard deviations of the population mean.
turn means that the sample mean or sample proportion will be more accurate because it is more likely to fall closer to the population mean or the population proportion. 13. Essential Question Check-In When you repeatedly take random samples of the same size from a population, what does the mean of the samples approximate? The mean of the samples approximates the population mean (if the data are numerical) or
the population proportion (if the data are categorical).
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Given the mean and standard deviation of a population, how can you determine the mean and standard deviation of a sampling distribution of sample means? The mean of the sampling distribution of the sample mean is equal to the mean of the population. To find the standard deviation, divide the population standard deviation by the square root of the sample size.
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12. When finding a sample mean or a sample proportion, why is using the greatest sample size possible (given constraints on the cost and time of sampling) a desirable thing to do? Increasing the sample size decreases the variation in the sampling distribution, which in
Lesson 3
9/1/14 7:35 PM
Sampling Distributions 1148
Evaluate: Homework and Practice
EVALUATE 1.
The general manager of a multiplex theater took random samples of size 10 from the audiences attending the opening weekend of a new movie. From each sample, the manager obtained the mean age and the proportion of those who said they liked the movie. The sample means and sample proportions are listed in the tables.
ASSIGNMENT GUIDE
Sample number
Sample mean (age)
Sample proportion (liked the movie)
Sample number
Sample mean (age)
Sample proportion (liked the movie)
Concepts and Skills
Practice
1
22.1
0.6
11
24.2
0.5
Explore 1 Developing a Distribution of Sample Means
Exercise 1
2
25.7
0.9
12
26.4
0.6
3
24.8
0.7
13
25.9
0.7
Explore 2 Developing a Distribution of Sample Proportions
Exercise 18
Example 1 Using the Sampling Distribution of the Sample Mean
Exercises 2–9
Example 2 Using the Sampling Distribution of the Sample Proportion
Exercises 10–17
When finding the probability that a sample has a value within a given range, remind students to pay attention to whether they are working with sample means or sample proportions, so that they use the correct formula for the standard deviation of the sample distribution.
24.3
0.7
14
23.8
0.8
23.9
0.8
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a. Based on the 20 samples, what is the best estimate for the mean age of all the people who saw the movie? Explain.
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AVOID COMMON ERRORS
4 5
The best estimate of the population mean is the mean of the sample means: µ x_ = 24.1. b. Based on this sample, what is the best estimate for the proportion of all the people who saw the movie and liked it? Explain.
The best estimate of the population proportion is the mean of the sample proportions: µ p̂ = 0.69. c.
What could the manager have done to improve the accuracy of both estimates?
The manager could have improved the accuracy of the estimates by using a sample size greater than 10.
Module 23
Exercise
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Lesson 23.3
Lesson 3
1149
Depth of Knowledge (D.O.K.)
COMMON CORE
Mathematical Practices
1 Recall of Information
MP.5 Using Tools
2–5
2 Skills/Concepts
MP.5 Using Tools
6–9
2 Skills/Concepts
MP.6 Precision
10–13
2 Skills/Concepts
MP.5 Using Tools
1
1149
• Online Homework • Hints and Help • Extra Practice
04/04/14 11:21 AM
On a standardized science test, the seniors at Fillmore High School have a mean score of 425 with a standard deviation of 80. 2.
VISUAL CUES
For random samples of 30 seniors, what interval centered on the mean of the sampling distribution captures 95% of the mean scores?
When finding an interval that captures a given percentage of the sample means or proportions, encourage students to look back at the figure in the previous lesson that divides the normal curve into parts and indicates what percent of the data in a normal distribution are contained in each part. This visual representation may make it easier for visual learners to identify the required interval.
80 σ _____ μ x_ = μ = 425; σ x_ = ____ ― = ― ≈ 14.6 √n
√30
μ x_ - 2σ x_ = 425 - 2(14.6) = 395.8
μ x_ + 2σ x_ = 425 + 2(14.6) = 454.2 So, for random samples of 30 seniors, 95% of the sample means fall between 395.8 and 454.2.
3.
For random samples of 100 seniors, what interval centered on the mean of the sampling distribution captures 68% of the mean scores? σ μ x_ = μ = 425; σ x_ = ____ ―= √n
μ x_ - σ x_ = 425 - 8 = 417
80 ______ =8 √―― 100
μ x_ + σ x_ = 425 + 8 = 433 So, for random samples of 100 seniors, 68% of the sample means fall between 417 and 433. 4.
What is the probability that a random sample of 50 seniors has a mean score of at most 415? μ x_ = μ = 425; σ x_ = σ― = 80― ≈ 11.3 √n √50 _ P(x ≤ 415) = normalcdf (-1e99, 415, 425, 11.3) ≈ 0.19 So, the probability that a random sample of 50 seniors has a mean score of at most 415 is about 0.19.
____ _____
5.
What is the probability that a random sample of 25 seniors has a mean score of at least 430? σ = 80― = 16 μ x― = μ = 425; σ x― = √25 √n _ P(x ≥ 430) = normalcdf (430, 1e99, 425, 16) ≈ 0.38 So, the probability that a random sample of 25 seniors has a mean score of at least 430 is about 0.38.
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_ _____ ―
For Exercises 6–9, use the following information: The safety placard on an elevator states that up to 8 people (1200 kilograms) can ride the elevator at one time. Suppose the people who work in the office building where the elevator is located have a mean mass of 80 kilograms with a standard deviation of 25 kilograms. 6.
For random samples of 8 people who work in the office building, what interval centered on the mean of the sampling distribution captures 95% of the mean masses? 25 σ = ≈ 8.8 μ x― = μ = 80; √n √8 μ x― - 2σ x― = 80 - 2(8.8) = 62.4 μ x― + 2σ x― = 80 + 2(8.8) = 97.6 So, for random samples of 8 people who work in the office building, 95% of the sample means fall between 62.4 kilograms and 97.6 kilograms.
_ _ ― ―
Module 23
Exercise
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Lesson 3
1150
Depth of Knowledge (D.O.K.)
COMMON CORE
Mathematical Practices
14–17
2 Skills/Concepts
MP.6 Precision
18
2 Skills/Concepts
MP.3 Logic
19
2 Skills/Concepts
MP.3 Logic
20
3 Strategic Thinking
MP.6 Precision
21
3 Strategic Thinking
MP.3 Logic
4/3/14 2:51 AM
Sampling Distributions 1150
7.
AVOID COMMON ERRORS
For random samples of 8 people who work in the office building, what interval centered on the mean of the sampling distribution captures 99.7% of the mean masses? 25 σ ≈ 8.8 = μ x― = μ = 80; σ x― = √n √8 μ x― - 3σ x― = 80 - 3(8.8) = 53.6
_ _ ― ―
When using a graphing calculator to determine the probability that a sample mean or proportion is above or below a given value, students will obtain incorrect results if they enter the given information in the wrong order. Remind them to read each problem carefully. If the problem asks about means or proportions that are “more than” a certain value, that value should be the first number entered in the calculator’s cumulative distribution function. For questions about means or proportions that are “at most” a certain value, that value should be the second number entered.
μ x― + 3σ x― = 80 + 3(8.8) = 106.4 So, for random samples of 8 people who work in the office building, 99.7% of the sample means fall between 53.6 kilograms and 106.4 kilograms.
8.
What is the probability that a random sample of 8 people who work in the office building has a mean mass of at most 90 kilograms? 25 σ ≈ 8.8 = μ x― = μ = 80; σ x― = √n √8 ― P(x ≤ 90) = normalcdf (-1e99, 90, 80, 8.8) ≈ 0.87 So, the probability that a random sample of 8 people who work in the office building has a mean mass of at most 90 kilograms is about 0.87.
_ _ ― ―
9.
Based on the elevator’s safety placard, what is the maximum mean mass of 8 people who can ride the elevator at one time? What is the probability that a random sample of 8 people who work in the office building exceeds this maximum mean mass?
The maximum mean mass of 8 people who can ride the elevator at one time 1200 = 150 kilograms. is 8 25 σ μ x― = μ = 80; σ x― = ≈ 8.8 = √n √8 ― P(x > 150) = normalcdf (150, 1e99, 80, 8.8) ≈ 0.00000000000000091 So, the probability that a random sample of 8 people who work in the office building has a mean mass that exceeds 150 kilograms is less than one quadrillionth
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_
A popcorn manufacturer puts a prize in 25% of its bags of popcorn. When sampling from this population, consider “successes” to be bags of popcorn containing a prize. 10. For random samples of 100 bags of popcorn, what interval centered on the mean of the sampling distribution captures 95% of the sample proportions?
μ p̂ = p = 0.25; σ p̂ =
―――――
μ p̂ - 2σ p̂ = 0.25 - 2(0.043) = 0.164
μ p̂ + 2σ p̂ = 0.25 + 2(0.043) = 0.336 So, for random samples of 100 bags of popcorn, 95% of the sample proportions fall between 16.4% and 33.6%.
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Lesson 23.3
―――
p(1 - p) 0.25(1 - 0.25) = √__ ≈ 0.043 √_ n 100
Since np = 25, n(1 - p) = 75, and both are greater than 10, the sampling distribution is approximately normal.
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11. For random samples of 80 bags of popcorn, what interval centered on the mean of the sampling distribution captures 68% of the sample proportions?
μ p̂ = p = 0.25; σ p̂ =
―――
―――――
p(1 - p) 0.25(1 - 0.25) = √__ ≈ 0.048 √_ n 80
Since np = 20, n(1 - p) = 60, and both are greater than 10, the sampling distribution is approximately normal. μ p̂ - σ p̂ = 0.25 - 0.048 = 0.202 μ p̂ + σ p̂ = 0.25 + 0.048 = 0.298 So, for random samples of 80 bags of popcorn, 68% of the sample proportions fall between 20.2% and 29.8%. 12. What is the probability that a random sample of 120 bags of popcorn has prizes in at most 30% of the bags?
μ p̂ = p = 0.25; σ p̂ =
―――
―――――
p(1 - p) 0.25(1 - 0.25) = √__ ≈ 0.040 √_ n 120
Since np = 30, n(1 - p) = 90, and both are greater than 10, the sampling distribution is approximately normal. P(p̂ ≤ 0.3) = normalcdf (-1e99, 0.3, 0.25, 0.04) ≈ 0.89
So, the probability that a random sample of 120 bags of popcorn has prizes in at most 30% of the bags is about 0.89.
13. What is the probability that a random sample of 60 bags has prizes in more than 12 bags?
μ p̂ = p = 0.25; σ p̂ =
―――
―――――
p(1 - p) 0.25(1 - 0.25) = √__ ≈ 0.056 √_ n 60
Since np = 15, n(1 - p) = 45, and both are greater than 10, the sampling distribution is approximately normal. p̂̂ min =
12 _ = 0.2
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60 P(p̂ > 0.2) = normalcdf (0.2, 1e99, 0.25, 0.056) ≈ 0.81
So, the probability that a random sample of 60 bags of popcorn has prizes in more than 12 bags is about 0.81.
Module 23
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Lesson 3
9/1/14 10:02 AM
Sampling Distributions 1152
About 28% of students at a large school play varsity sports. When sampling from this population, consider “successes” to be students who play varsity sports. 14. For random samples of 75 students, what interval centered on the mean of the sampling distribution captures 95% of the sample proportions?
μ p̂ = p = 0.28 σ p̂ =
―――
―――――
p(1 - p) 0.28(1 - 0.28) = √__ ≈ 0.052 √_ n 75
Since np = 21, n(1 - p) = 54, and both are greater than 10, the sampling distribution is approximately normal. μ p̂ - 2σ p̂ = 0.28 - 2(0.052) = 0.176 μ p̂ + 2σ p̂ = 0.28 + 2(0.052) = 0.384
So, for random samples of 75 students, 95% of the sample proportions fall between 17.6% and 38.4%. 15. For random samples of 100 students, what interval centered on the mean of the sampling distribution captures 99.7% of the sample proportions?
μ p̂ = p = 0.28; σ p̂ =
――― ――――― p(1 - p) 0.28(1 - 0.28) = √__ ≈ 0.045 √_
n 100 Since np = 28, n(1 - p) = 72, and both are greater than 10, the sampling distribution is approximately normal. μ p̂ - 3σ p̂ = 0.28 - 3(0.045) = 0.145
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μ p̂ + 3σ p̂ = 0.28 + 3(0.045) = 0.415 So, for random samples of 100 students, 99.7% of the sample proportions fall between 14.5% and 41.5%. 16. What is the probability that a random sample of 45 students includes more than 18 students who play varsity sports?
μ p̂ = p = 0.28; σ p̂ =
n 45 Since np = 12.6, n(1 - p) = 32.4, and both are greater than 10, the sampling distribution is approximately normal. p̂̂ min =
So, the probability that a random sample of 45 students includes more than 18 students who play varsity sports is about 0.037.
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Lesson 23.3
18 _ = 0.4
45 P(p̂ > 0.4) = normalcdf (0.4, 1e99, 0.28, 0.067) ≈ 0.037
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――― ――――― p(1 - p) 0.28(1 - 0.28) = √__ ≈ 0.067 √_
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Lesson 3
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17. What is the probability that a random sample of 60 students includes from 12 to 24 students who play varsity sports?
μ p̂ = p = 0.28; σ p̂
――― ――――― p(1 - p) 0.28(1 - 0.28) = √_ = √__ ≈ 0.058
INTEGRATE MATHEMATICAL PRACTICES Focus on Reasoning MP.2 To help students see that the formula for σ p‸
n 60 Since np = 16.8, n(1 - p) = 43.2, and both are greater than 10, the sampling distribution is approximately normal. 12 _ = 0.2; p̂̂
_
24 = 0.4 max = 60 60 P(0.2 ≤ p̂ ≤ 0.4) = normalcdf (0.2, 0.4, 0.28, 0.058) ≈ 0.90 p̂ min =
(the standard error of the proportion) is actually a special case of the formula for σ x_ (the standard error of the mean), you can show them the derivation of a formula for the standard deviation of a population consisting of categorical data that either meet or do not meet a criterion:
So, the probability that a random sample of 60 students includes from 12 to 24 students who play varsity sports is about 0.90.
18. Among the 450 seniors in a large high school, 306 plan to be in college in the fall following high school graduation. Suppose random samples of 100 seniors are taken from this population in order to obtain sample proportions of seniors who plan to be college in the fall. Which of the following are true statements? Select all that apply.
Let each data value that meets the criterion be 1, and let each data value that doesn’t meet the criterion be 0. This effectively converts the categorical data into numerical data. Assume that a proportion p of the population has a value of 1, while a proportion 1 - p has a value of 0. Then the mean value of the population is p(1) + (1 – p)(0) = p. The deviation of each 1 from the mean is 1 - p, while the deviation of each 0 from the mean is 0 - p, or -p. The standard deviation can be calculated by multiplying each fraction of the population by its squared deviation from the mean, then taking the square root:
a. Every sample proportion is 0.68. b. The mean of the sampling distribution of the sample proportion is 0.68. c.
The standard error of the proportion is about 0.047.
d. The standard error of the proportion is about 0.0047. e. The sampling distribution of the sample proportion is skewed. f.
The sampling distribution of the sample proportion is approximately normal.
The population proportion is p =
306 _ = 0.68. This is also the mean of the sampling
450 distribution of the sample proportion. However, sample proportions obtained from individual samples will vary. So, A is false, and B is true.
―――――
For random samples of size 100, the standard error of the proportion is σ p̂ =
Because np = 100(0.68) = 68, n(1 - p) = 100(0.32) = 32, and both are greater than 10, the sampling distribution of the sample proportion is approximately normal. So, E is false, and F is true. Answers: B, C, F
Module 23
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© Houghton Mifflin Harcourt Publishing Company
0.68(1 - 0.68) ≈ 0.047. So, C is true, and D is false. √____________ 100
σ=
――――――――― + (1 - p)(-p) ,
√p(1 - p)
2
2
―――
which can be simplified to σ = √p(1 - p) . The standard deviation of the sampling distribution of the mean is, therefore,
―――
√p(1 - p) σ = _______ = σ x_ = _ √― n √― n
――― p(1 - p) √______ , which exactly n
matches the formula for σ p‸.
Lesson 3
4/3/14 2:54 AM
Sampling Distributions 1154
JOURNAL
H.O.T. Focus on Higher Order Thinking
Have students create a flow chart showing how to determine the interval that contains 95% of the means of random samples of size n taken from a population whose mean and standard deviation are known.
19. Explain the Error A student was told that a population has a mean of 400 and a standard deviation of 25. The student was asked to find the probability that a random sample of size 45 taken from the population has a mean of at most 401. The student entered normalcdf (−1e99, 401, 400, 25) on a graphing calculator and got a probability of about 0.516. What did the student do wrong? Show how to find the correct answer.
The student used the standard deviation of the population rather than the standard deviation of the sampling distribution.
_ ―
25 σ = ≈ 3.73 σ x― = ____ √― n √45 ― P(x ≤ 401) = normalcdf (-1e99, 401, 400, 3.73) ≈ 0.606
20. Draw Conclusions Amanda plans to use random sampling to estimate the percent of people who are truly ambidextrous (that is, they do not have a dominant right or left hand). She suspects that the percent is quite low, perhaps as low as 1%. If she wants the sampling distribution of the sample proportion to be approximately normal, what minimum sample size should she use? Explain.
The sampling distribution of the sample proportion is approximately normal provided np = 10. Substituting 0.01 for p and solving for n gives n = 1000. So, the minimum sample size should be 1000.
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21. Check for Reasonableness Given that about 90% of people are right-handed, you are interested in knowing what percent of people put their right thumb on top when they clasp their hands. Having no other information to go on, you assume that people who put their right thumb on top when they clasp their hands are those who are also right-handed. You then take a random sample of 100 people and find that 60 put their right thumb on top when they clasp their hands. Does this result lead you to question your assumption? Explain why or why not.
The assumption that people who put their right thumb on top when they clasp their hands are those who are also right-handed means that the population proportion p for right thumb on top should equal 0.9. For random samples of size 100, np = 100(0.9) = 90 and n(1 - p) = 100(0.1) = 10, so the sampling distribution of the sample proportion should be approximately normal with mean μ p̂ = 0.9 and standard deviation σ p̂ =
sample has produced a sample proportion that is 10 standard deviations below the mean. This result is so unlikely that you should reject your assumption.
Module 23
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Lesson 23.3
――――
0.9(1 - 0.9) 0.6 - 0.9 = 0.03. Since ________ = -10, your random √__________ 100 0.03
1155
Lesson 3
9/1/14 10:05 AM
Lesson Performance Task
AVOID COMMON ERRORS Recommend that students organize the information they need for the Lesson Performance Task in a table.
Among the data that the U.S. Census Bureau collects are the sizes of households, as shown in the table.
Number of people in household
Number of households
1
31,886,794
31,886,794
2
38,635,170
77,270,340
3
18,044,529
54,133,587
4
15,030,350
60,121,400
5
6,940,508
34,702,540
6
2,704,873
16,229,238
7 or more
1,749,501
12,246,507
Lower Upper Population bound bound mean
Number of people
Standard deviation
This will help students enter and track numbers in their correct order.
INTEGRATE MATHEMATICAL PRACTICES Focus on Critical Thinking MP.3 When using the normalcdf(function
a. In the table above, assume that you can simply use 7 as the number of people in households with 7 or more people. Complete the third column of the table. Then use that column to approximate the population mean μ (that is, the mean number of people in a household). Explain your reasoning.
Divide the total number of people by the total number of households.
on a graphing calculator, under what circumstances would you enter the desired mean of the sample first? When it is the least number in the range of data under consideration, enter the desired mean of the sample population first.
_________
31,886,794 + 77,270,340 + 54,133,587 + 60,121,400 + 34,702,540 + 16,229,238 + 12,246,507 286,590,406 _____________________________________________________________ = 114,991,725 31,886,794 + 38,635,170 + 18,044,529 + 15,030,350 + 6,940,508 + 2,704,873 + 1,749,501
≈ 2.49
So, the approximate population mean is about 2.49. b. Is the actual population mean greater than or less than the mean that you calculated? Explain.
with 7 people, 8 people, 9 people, and so on. So, the actual number of people living in those 286,590,406 households is greater than 12,246,507, which makes the numerator of greater, 114,991,725 which in turn makes the mean greater.
__
c. Given that μ ≈ 2.49 and σ ≈ 1.42, find the probability that a random sample of 100 households in the United States has a mean size of 2.3 people or less. σ 1.42 μ x¯ = μ ≈ 2.49; σ x― = ≈ = 0.142 √n √100 ― P(x ≤ 2.3) = normalcdf (-1e99, 2.3, 2.49, 0.142) ≈ 0.09
_ _ ― ――
© Houghton Mifflin Harcourt Publishing Company
The actual population mean is greater because the “7 or more” category includes households
d. Given that μ ≈ 2.49 and σ ≈ 1.42, find the probability that a random sample of 100 households in the United States has a mean size of 2.6 people or more. σ 1.42 = 0.142 = μ x¯ = μ ≈ 2.49; σ x― = √n √100 ― P(x ≥ 2.6) = normalcdf (2.6, 1e99, 2.49, 0.142) ≈ 0.22
_ _ ― ――
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Lesson 3
EXTENSION ACTIVITY A2_MNLESE385900_U9M23L3.indd 1156
Have different students choose a sample of either 200, 400, or 600 households; different ranges for the number of people in the household (for example 1 to 3); and then find the probability that, out of a random sample of that many households in the United States, the mean household size will be within that range. Have students compare their results and how the different sample sizes affected their results.
10/16/14 8:00 AM
Scoring Rubric 2 points: Student correctly solves the problem and explains his/her reasoning. 1 point: Student shows good understanding of the problem but does not fully solve or explain his/her reasoning. 0 points: Student does not demonstrate understanding of the problem.
Sampling Distributions 1156