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STATWAY™ INSTRUCTOR NOTES

Lesson 9.1.1 Sampling Distribution of Differences of Two Proportions

ESTIMATED TIME

50 minutes

MATERIALS REQUIRED

For each group of students, one black deck and one blue deck of cards is required. Before beginning the activity, remove 11 red cards from each black deck and 18 red cards from each blue deck. These cards are used to simulate a collection of differences between sample proportions in Part II of this lesson. If you feel that the group simulations may take too much time, you can do the simulations in front of the entire class.

BRIEF DESCRIPTION

In this activity, students begin to understand the distribution of differences between sample proportions from independent samples. The lesson begins with extraordinarily small populations, which allow students to generate an entire distribution of differences between sample proportions. From this small distribution of differences, students observe the bell-shaped nature of their distribution and quantify its mean and standard error.

Students learn that the mean of differences can be found by subtracting the means and the variance of differences can be found by adding the variances of the original populations.

The initial example is small but complete and should help students see exactly what is meant by a distribution of differences of sample proportions.

These ideas are broadened in Part II, where students simulate the sampling process from larger populations using two decks of cards. From these, informal inferences are made regarding the mean, standard error, and shape of the distribution of all differences of proportions.

© 2011 THE CARNEGIE FOUNDATION FOR THE ADVANCEMENT OF TEACHINGA PATHWAY THROUGH STATISTICS, VERSION 1.5, STATWAY™ - INSTRUCTOR NOTES

STATWAY INSTRUCTOR NOTES | 2


Take some time to review the basics of the sampling distributions of sample proportions. Remind students of the following important points through a question-and-answer interaction.

Distributions of sample proportions are approximately normal in their distributions, provided that nπ ≥ 10 and n(1 – π) ≥ 10.

The mean of sample proportions is the population proportion (µp = π). Thus, the sample proportion is an unbiased estimate of the population proportion.

The standard deviation (or error) of sample proportions is

σ p=√ π (1−π )n

The variance of sample proportions is the square of the standard deviation:

σ p2=π (1−π )n

LEARNING GOALS

Students will begin to understand that:

Sampling variability extends to statistics based on independent samples taken from two populations.

The mean of a distribution of sample differences is equal to the difference in the means for the populations.

The variance of a distribution of sample differences is equal to the sum of the variances for the populations.

Students will begin to be able to:

Describe the shape, center, and standard error of the sampling distribution of the difference in two proportions.

Quantify sampling variability resulting from the distribution of differences in two proportions, computed from random samples selected from different populations using the formula




sp1−p2=√ p1 (1−p1 )

n1+p2 (1−p2 )n2

In this lesson, students build upon their understanding of sampling distribution of sample proportions to include the distribution of the differences in two sample proportions, so that comparisons can be made between their populations.

Students have learned that under the condition of sufficiently large sample size, the distributions of sample proportions can be approximately normal. In this module, students compare sample proportions by computing their differences. In doing so, they learn that the approximate normality of the distributions of sample proportions is inherited by the distribution of their differences. With this assurance of approximate normality, students are immediately able to apply the tools gained previously—margin of error, confidence intervals, and hypothesis tests—to this new context of comparing differences of sample proportions.

INTRODUCTION

The GPS software company, TeleNav, recently commissioned a study on proportions of people who text while they drive. The study suggests that there are differences in the texting-while-driving habits of men and women.1

We will compare the proportions of male and female texters who text frequently while driving. Our primary tool for this comparison is subtraction. That sounds simple enough, but before we can really know when one of these differences between two proportions is significant, we must understand the nature of the distribution of such differences. This understanding allows us to apply the methods of inference we have learned—margins of error, confidence intervals, and hypothesis tests—to the comparisons between two population proportions.

For any two distributions of sample proportions, the distribution of differences between sample proportions can be very large and difficult to picture. To ease the comprehension of what a distribution of differences looks like, we will turn to two very small populations of texting drivers. The first population consists of two males who text while driving and the second population consists of three females who text while driving.

1 TeleNav, Inc. (2010). TeleNav-commissioned survey suggests both genders have similar views on abiding by and breaking the rules of the road. Retrieved July 14, 2010, from www.telenav.com/about/driving-behavior.




Within these populations, frequent texting while driving is defined to be at least seven texts while driving per week. Rarely texting is defined as less than seven texts per week while driving. These are represented as follows:

F = frequently texting (at least seven texts while driving per week)R = rarely texting (less than seven texts while driving per week)

From these small populations, we will collect sample proportions of drivers who text frequently. We will then create a distribution of differences between male and female proportions. This small distribution will help us find a way to quantify the center and spread of such distributions of differences.

TRY THESE

As mentioned, the first population we will consider consists of two men who text while driving, where one of these men texts frequently. The observations are

{Frequent, Rare}.

This is a small population of one frequent texter out of two, with a population proportion of 1/2, which is represented as π = ½ = 0.5.

Now construct a sampling distribution of sample proportions from this population.

1 Treating this collection of two texting men as a small population, we will compute all sample proportions from samples of size n = 2. Sampling is done with replacement to preserve the independence of the trials (so some samples will include the same observation twice).

In the following table, all possible samples are listed. For the case the frequent texter is chosen twice, the proportion is 2/2 = 1.0. For the case where the first pick is a frequent texter but the second is not, the proportion is 1/2 = 0.5.

A Fill in the remaining sample proportions of men who text frequently, each of which are denoted as pm.




Sample Drawn With Replacement

Sample Proportion, pm=number of frequent texters

sample¿¿¿

{F , F } 22=1.0

{F ,R }

{R , F } 12=0.5

{R , R }

B We have previously given formulas for the mean, standard error, and variance of a sampling

distribution of sample proportions. These formulas require the population proportion of men who frequently text while driving, (π = ½ = 0.50) and the sample sizes (n = 2). Evaluate these formulas for the current sampling distribution.

πm=¿

σ pm=√ πm (1−πm )nm

=¿

variance=σ pm2 =¿

Answer: πm=0.500; σ pm=√ 0.5×0.52

=0.354; σ pm2 =0.125

2 Next, suppose we have a population of three women who text while driving and only one of these women texts frequently. These are represented as

{Frequent, Rare, Rare}

A What is the population proportion of frequent texters in the population of women?

πw=¿

Answer: πw=1/3=0.333




B Compute all sample proportions from samples of women texters of size n = 2. Denote the sample proportion of frequently texting women as pw.

Sample Drawn With Replacement

Sample Proportion,

pw=number of frequent texters

sample¿¿¿

{F , F } 2/2 = 1.0

{F ,R } 1/2 = 0.5

{F , R }

{R , F }

{R , R } 0/2 = 0.0

{R , R }

{R , F } 1/2 = 0.50

{R , R }

{R , R } 0/2 = 0.0

C For this collection of proportions from the women drivers whose population proportion is πw = 1/3 = 0.333, compute the mean, standard error, and variance. Remember that the samples are all of size n = 2.

πw=¿

σ pw=√ π w (1−πw )n

=¿

variance=σ pw2 =¿

Answer: πw = 0.333, σ pw=√ 0.3333×0.66672

=0.333, σ pw2 = 0.111

3 Our goal is to discover the nature of the distribution of all differences (pm – pw). To accomplish this, consider the difference of every value of pm minus every value of pw.




A In the top row of the following table, all men’s proportions (pm) found in Question 1b are listed. In the left column, all women’s proportions (pw) found in Question 2b are listed. Most of the differences (pm – pw) are listed as well, but seven are missing. In any cell where no difference is given, record the value of pm – pw, where pm is recorded at the top of the cell’s column and pw is recorded at the left of the cell’s row.

Proportions of Men Who Text Frequently

pm−pw pm=1.00 pm=0.50 pm=0.50 pm=0.00

Prop

ortio

ns o

f Wom

en

Who

Tex

t Frequ

ently

pw=1.00 0.00 –0.50

pw=0.50 0.50 0.00 –0.50

pw=0.50 0.50 0.00 0.00 –0.50

pw=0.50 0.50 0.00 0.00 –0.50

pw=0.00 0.50 0.50

pw=0.00 1.00 0.50 0.50 0.00

pw=0.50 0.00 0.00

pw=0.00 1.00 0.50 0.50 0.00

pw=0.00 1.00 0.50 0.50 0.00

The total number of differences (pm – pw) is 9 × 4 = 36.

B Fill in the frequency table with the frequencies and their corresponding probabilities as relative frequencies for the various values of pm – pw. Record the probabilities as unreduced fractions whose denominators are each 36. The distribution of probabilities is the sampling distribution of differences.


-1.0 -0.5 0.0 0.5 1.0

2/36

4/36

6/36

8/36

10/36

12/36

The Sampling Distribution of Differences,



pm−pw Frequency Probability

–1.0

–0.5

0.0

0.5

1.0

Total 36 36/36 = 1

C Sketch a histogram of the probability distribution of pm – pw on the graph below.

This distribution gives probabilities for every possible difference between male proportions of frequent texters and female proportions of frequent texters.

Is the distribution approximately symmetric?

Is the distribution approximately bell-shaped?

Recalling that the mean is a similar to a balancing point for a probability distribution, give a visual estimate of the mean.




Answer: The distribution is roughly symmetric and bell-shaped. The exact value of the mean = 1/6 = 0.1667.

4 Use technology to compute the mean and standard deviation of all differences summarized by the frequency table in Question 3 (or your instructor can provide these).

A What is the mean of differences? Is this value close to the estimate made in Question 3c?

Answer: Mean of differences = 6/36 = 1/6 = 0.167

B Compute the difference of the means of the men and women from Questions 1b and 2c.

Answer: 1/2 – 1/3 = 0.500 – 0.333 = 0.167

C Comparing your answers in Questions 4a and 4b, complete the statement below.

The mean of differences in Question 4a is the __________________ of the means from Questions 1b and 2c.

Answer: Difference.

D Compute the variance of all differences (by squaring the standard deviation) summarized by the frequency table in Question 3b (or your instructor can provide this).

Answer: Variance of differences = 17/72 = 0.236

E Compute the sum of the variances of men and women texters from Questions 1b and 2c.


-1.0 -0.5 0.0 0.5 1.0

2/36

4/36

6/36

8/36

10/36

12/36



Answer: Sum of variances = 0.125 + 0.111 = 0.236

F Comparing your answers in Questions 4d and 4e, complete the following statement.

The variance of differences in Question 4d is the _______________ of the variances from Questions 1b and 2c.

Answer: Sum.

G Compute the standard deviation (or error) of all differences of proportions by taking the square root of the variance from Question 4d. Round this answer to the nearest tenth.

Answer: Standard error = √0.236 = 0.486 ≈ 0.5

Do you consider a difference of pm – pw = 0.5 to be unusually high? Why?

Answer: No it is less than one standard error above the mean.

WRAP-UP/TRANSITION

In Part 1 of this activity, students learned to quantify the mean and standard error of differences between sample proportions. They also learned that the distribution of such differences could be bell-shaped. The following principles were demonstrated and should be emphasized:

The mean of differences was the difference of the means.

The variance of differences was the sum of variances.

The standard error of differences is the square root of the variance.

The distribution of differences between sample proportions can be bell-shaped.

Knowing the mean and standard error of differences allows you to make inferences regarding them.

These principles are explored on a larger scale in Part II of this activity where students simulate the process of sampling women texters and men texters by drawing cards from two different decks.




If the simulation will take too long for students to do, you can do this simulation for the entire group.

NEXT STEPS

For this part of the lesson, we will expand our simulated populations of men and women using decks of cards. Your instructor will provide a deck of black cards and a deck of blue cards. These decks represent the following.

The black deck is the population of males who text while driving.

The blue deck is the population of females who text while driving.

Do not count the cards in the decks! Just as the exact truth is always unknown in large populations, the truth must remain unknown with the decks of cards. In this way, you can experience the uncertainty that occurs when working with samples.

Within each deck, there are red cards and black cards. The color of each card’s front side represents the following:

A card with a red front is a person who sends texts frequently while driving.

A card with a black front is a person who sends texts rarely while driving.

Your instructor has manipulated the decks of cards so that the proportions of red cards in each deck roughly match the proportions of men and women drivers who text frequently in the real world.




TRY THESE

5 From each deck of cards, you and a partner will sample 10 cards, randomly replacing each card after it is drawn before drawing the next card to preserve the independence of the trials.

A Shuffle the black deck (representing the men who text while driving) and draw 10 cards with replacement. Count the red front cards drawn.

What is the sample proportion of frequent texters for the men (the number of red front cards divided by the sample size n = 10)?

pm=¿

The dotplot below includes a simulation of 30 additional proportions (pm) generated by drawing cards from a deck, just as you have done. Add your data point to this dotplot.

0.70.60.50.40.30.20.10.0Men Proportions

Men who Text FrequentlyProportions of Men who Text Frequently While Driving

(Seven or More Texts Per Week)

Population includes only men who admit to texting while driving.

B Shuffle the blue deck (representing the women who text while driving) and draw 10 cards with replacement. Count the red front cards drawn. What is the sample proportion of frequent texters for the women (the number of red front cards divided by the sample size n = 10)?

pw=¿

Once again, a dotplot is provided below which includes 30 proportions (pw) generated in the same way. Add your proportion to this dotplot.




0.70.60.50.40.30.20.10.0Women Proportions

Women who Text FrequentlyProportions of Women who Text Frequently While Driving


Population includes only women who admit to texting while driving.

C Compute the difference between the proportions found in Questions 5a and 5b.

pm−pw=¿

The dotplot below is generated by taking differences between the sample proportions pm and pw in the previous dotplots. Add your difference to the dotplot of differences below.

0.60.50.40.30.20.10.0-0.1-0.2-0.3Differences

Differences Between Men and Women ProportionsFor Frequent Texters


Populations include only people who admit to texting while driving.




Add any additional differences (pm – pw) generated by others in your class to this dotplot.

D From the dotplot of texting men, estimate the mean by picking one proportion that is representative of the group’s center.

Answer: μpm ≈ 0.4

E From the dotplot of texting women, estimate the mean by picking one proportion that is representative of that group’s center.

Answer: μpw ≈ 0.2

F The difference of means is the mean of the differences pm – pw. Thus, an estimate of the mean difference can be found by subtracting the estimated means from Questions 5d and 5e. Subtract these values.

Answer: pm−pw≈0.2

Does this value make a good representation of the center of the dotplot of differences?

Answer: Yes!

G Using complete sentences, refer to the dotplot in Question 5c to describe the distribution of differences pm – pw in terms of shape, symmetry, and center.




6 The variances of sample proportions (pm and pw) are each computed as the square of the standard error,

σ 2=π (1−π )n

When the population proportions (π) are unknown, estimate them using sample proportions pm and pw). The variance of the differences (pm – pw) is the sum of the individual variances. This variance is therefore estimated as a sum,

pm (1−pm )nm

+pw (1−pw )nw

Applying a square root gives an estimate of the standard error of differences (pm – pw).

s❑pm−pw=√ pm (1−pm )nm

+pw (1−pw )nw

A Use your representative values for pm and pw (from Questions 5d and 5e) to estimate the standard error of differences between all such proportions.

s❑pm−pw=√ pm (1−pm )nm

+pw (1−pw )nw

=¿

Answer: spm−p w=0.2.

B Using the estimated mean of pm – pw values (from Question 5f) and the standard error (from Question 6a), do you consider a difference of pm – pw = 0.6 to be unusually high?

Answer: 0.6 is around 2 standard errors above the estimated mean difference of 0.2. This is unusual.

C Using the estimated mean of the representative value of pm – pw (from Question 5f) and the standard error (from Question 6a), do you consider a difference of pm – pw = 0 to be unusually low?


0.060.050.040.030.020.010.00-0.01Male Proportion Minus Female Proportion

Differences of Left Handed ProportionsFor Men and Women

Twenty-five samples of men gathered with 575 members in each.Twenty Samples of women gathered with 815 members each.



Answer: 0.0 is around 1 standard error below the estimated mean difference of 0.2. This is not unusual.

D Suppose that the mean value of pm is the population proportion, πm, of men who text successively. Suppose also that the mean value of pw is the population proportion, πw, of women who text successively. What is implied by the approximate difference between πm and πw given in your answer from 6c above?

Answer: 0.0 is around 1 standard error below the estimated mean difference of 0.2, it is a plausible value for the difference in population proportions. Therefore there maybe no difference in the proportions of men and women who text frequently while driving.

TAKE IT HOME

To compare the proportion of men who are left-handed to the proportion of women who are left-handed, twenty-five samples, each containing 575 men each, were gathered. From each of these samples of men, a sample proportion, pm, of those who are left-handed was computed. Additionally, twenty samples of 815 women each were gathered, and from each, a sample proportion, pw, of women who are left-handed was computed.

Thus, twenty-five sample proportions, pm, of left-handed men, and twenty sample proportions, pw, of left-handed women were gathered. From these, the collection of all possible differences, pm – pw, was constructed. The distribution of differences is plotted below.




1 Do you consider an 8% difference between the proportions of men and women who are left- handed likely? If unlikely, is the difference too high or too low?

Answer: An 8% difference is unlikely, and too high.

2 If you assume an 8% difference, is the proportion of left-handedness higher for men or women?

Answer: It is higher for the men.

3 Do you consider a –1% difference likely? If unlikely, is the difference too high or too low?

Answer: A –1% difference seems unlikely as well. It is too low.

4 If you assume a –1% difference, is the proportion of left-handedness higher for men or women?

Answer: It is higher for the women.

5 Pick a difference that you consider a representative value of the differences on the dotplot.

Answer: 0.02 = 2%

6 Does your chosen difference allow for pm and pw to be equal? If not, then which is greater?

Answer: No. The proportion for men is higher.

7 Suppose that a random proportion of left-handedness for men, pm = 0.10, is chosen from a sample of size nm = 575. Suppose also that a random proportion of left-handedness for women, pw = 0.08, is chosen from a sample of size nw = 815. Estimate the standard error in the differences of sample proportions, pm – pw, rounded to two places after the decimal.

s❑pm− pw=√ pm (1−pm )nm

+pw (1−pw )nw

=¿




Answer: s❑pm−pw=√ 0.10 (1−0.10 )575

+0.08 (1−0.08 )

815≈0.016

8 Estimate, roughly, the number of standard errors that the difference, pm – pw = 0.08, lies from your representative difference in Question 5.

Answer: The difference is almost 4 standard errors away.

Does the estimated value support your answer from Question 1?

Answer: Yes

9 Estimate, roughly, the number of standard errors that the difference, pm – pw = – 0.01, lies from your representative difference in Question 5.

Answer: This difference is nearly two standard errors away.

Does the estimated value support your answer from Question 3?

Answer: Nearly.

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This lesson is part of STATWAY™, A Pathway Through College Statistics, which is a product of a Carnegie Networked Improvement Community that seeks to advance student success. Version 1.0, A Pathway Through Statistics, Statway™ was created by the Charles A. Dana Center at the University of Texas at Austin under sponsorship of the Carnegie Foundation for the Advancement of Teaching. This version 1.5 and all subsequent versions, result from the continuous improvement efforts of the Carnegie Networked Improvement Community. The network brings together community college faculty and staff, designers, researchers and developers. It is an open-resource research and development community that seeks to harvest the wisdom of its diverse participants in systematic and disciplined inquiries to improve developmental mathematics instruction. For more information on the Statway Networked Improvement Community, please visit carnegiefoundation.org. For the most recent version of instructional materials, visit Statway.org/kernel.

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STATWAY™ and the Carnegie Foundation logo are trademarks of the Carnegie Foundation for the Advancement of Teaching. A Pathway Through College Statistics may be used as provided in the CC BY license, but neither the Statway trademark nor the Carnegie Foundation logo may be used without the prior written consent of the Carnegie Foundation.


http://www.carnegiefoundation.org/

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