Download Analyzing a Survey The Lesson Activities will help you meet these

Probability
and Statistics
Lesson Activities
Student Answer Sheet
Analyzing a Survey
The Lesson Activities will help you meet these educational goals:
 Mathematical Practices—You will make sense of problems and solve them, use
mathematics to model real-world situations, and use appropriate tools strategically.
 Inquiry—You will perform an investigation in which you will make observations, make
hypotheses, analyze results, communicate your results in graphs and written form, and
draw conclusions.
 STEM—You will use mathematical processes and analysis in scientific investigation to
analyze real-world situations.
 21stCentury Skills—You will employ online tools for analysis, use critical-thinking and
problem-solving skills, communicate effectively, assess and validate information, and
perform large-scale data analysis.
Directions
You will evaluate some of these activities yourself, and your teacher may evaluate others.
Please save this document before beginning the lesson and keep the document open for
reference during the lesson. Type your answers directly in this document for all activities.
_________________________________________________________________________
Self-Checked Activities
Read the instructions for the following activities and type in your responses. At the end of the
lesson, click the link to open the Student Answer Sheet. Use the answers or sample
responses to evaluate your own work.
1. Finding Sample Mean
As a part of your project studying tipping behavior, you ask people the question, How
much should you tip a grocery bagger for carrying five bags to your car?
The Grocery Bagger Surveys tab of this spreadsheet gives the responses to this question
from 10 independent surveys, each with 50 respondents. The respondents were asked to
round to the nearest quarter, since few people tip with pennies, nickels, and dimes.
In the spreadsheet, you can calculate the mean, or average, of data using the Average
function:
 Select the cell in which you want the mean to appear.
 Type =AVERAGE( in the formula bar.
 Select all the cells containing the values for which you want to find the average,
and then press Enter.
 The average of the values in the cells will be calculated and displayed in the cell in
which you entered the formula.
Copyright © 2012 PLATO Learning, Inc. All rights reserved.
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a. What is the sample mean of the responses in survey 1?
Sample answer:
The sample mean for survey 1 is $1.69.
b. Using the responses to survey 1, plot a histogram using the Histogram tool. If you
need help, follow these instructions for using the online probability tools. Recall that
you can copy a column of data from the spreadsheet and paste it into the histogram
tool. Set appropriate limits and interval for your plot, and be sure to label it properly.
Export an image of the plot, and paste it in the space below.
Sample answer:
c. The Histogram tool calculates and displays the mean of the data. What was the mean
value displayed in the Histogram tool? How does it compare with the mean obtained
using the spreadsheet?
Sample answer:
The mean displayed in the histogram tool is $1.68. This value is almost equal to the
mean obtained using the spreadsheet formula ($1.69). This is probably because of a
rounding difference in the two applications.
d. What can you observe about the distribution of the data? Does it resemble a normal
distribution?
Sample answer:
The histogram shows that most of the data is around the center and tapers off at both
ends. The distribution of the data does resemble a normal distribution.
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2. The Central Limit Theorem
The results for all 10 grocery surveys were actually generated using a mathematical
simulation, found on the Simulation tab of the spreadsheet. To use it, enter a population
mean and population standard deviation in the two yellow cells at the top of worksheet
and follow the instructions. Every time a new set of data is generated, the data changes,
providing a new sample. You can copy this data and paste it into the histogram tool to
analyze it. (You don’t have to do that in this activity, but it’s there for you to experiment
with, if you like.)
Any given set of sample data generated in this way would not typically have the same
mean, distribution, and standard deviation as the population values, but the simulated
sample data would be consistent with a randomly selected sample from such a
population.
Note that the simulated population mean is 1.73, and the simulated standard deviation is
0.657. Imagine that this random sample survey was simulated an infinite number of times.
In that case, the population mean (µ) of all the samples would be $1.73. All the sample
means would be normally distributed, according to the central limit theorem.
a. If the standard deviation of the population is $0.657 (as it is in the simulation), what is
the standard error of the mean? Round up your answer to the nearest tenth of a cent.
Sample answer:
The standard error of the mean is equal to
σ
n
,where σ is the standard deviation of
the population and n is the sample size.
Here, σ = 0.675 and n = 50.
The standard error of the mean is
σ
n

0.657
50

0.657
 0.093.
7.071
The standard error of the mean is $0.093.
b. According to the central limit theorem, in which interval will 68% of the sample means
occur?
Sample answer:
According to the central limit theorem, since the survey is simulated an infinite number
of times, the sample means would be normally distributed. So, 68% of the sample
σ
means would fall within the interval μ 
.
n
µ (population mean) = $1.73
0.657
0.657
So, 68% of the sample means would lie between 1.73and 1.73 
.
50
50
1.73 – 0.093 and 1.73 + 0.093
$1.64 and $1.82 (rounded to the nearest cent)
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c. According to the central limit theorem, in which interval will 95% of the sample means
occur?
Sample answer:
According to the central limit theorem, since the survey is simulated an infinite number
of times, the sample means would be normally distributed. So, 95% of the sample
σ
means would fall within the interval μ  2 
.
n
µ (population mean) = $1.73
0.657
0.657
So, 95% of the sample means would fall between 1.73-2 
and 1.73  2 
.
50
50
1.73 – 2(0.093) and 1.73 + 2(0.093)
$1.54 and $1.92
d. According to the central limit theorem, in which interval will 99.7% of the sample
means occur?
Sample answer:
According to the central limit theorem, since the survey is simulated an infinite number
of times, the sample means would be normally distributed. So, 99.7% of the sample
σ
means would fall within the interval μ  3 
.
n
µ (population mean) = $1.73
So, 99.7% of the sample means would fall between
0.657
0.657
1.73-3 
and 1.73  3 
.
50
50
1.73 – 3(0.093) and 1.73 + 3(0.093)
$1.45 and $2.01
e. The spreadsheet has information about nine other surveys. Consider these surveys as
part of the infinite simulations carried out to find the population mean. Plot a histogram
of the responses to survey 4 with the Histogram tool. Export an image of the graph,
and paste it in the space below. Also copy and paste in your plot from survey 1 for
comparison. How do the two plots compare?
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Sample answer:
The spread of both plots is about the same. They both look roughly like normal
distributions, but they are clearly not identical.
f. Using the Average formula in the spreadsheet, find the sample mean for each of the
10 surveys and record it below.
Sample answer:
#1
#2
1.69
1.60
#3
1.78
#4
1.78
#5
1.74
#6
1.63
#7
1.76
#8
1.83
#9
1.79
#10
1.59
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g. Calculate the mean of the sample means and compare it with the population mean
($1.73). What do you observe with respect to the central limit theorem?
Sample answer:
The mean of the sample means is $1.72.This value is very close to the population
mean ($1.73). This observation conforms to the central limit theorem, which states that
for a random sample survey simulated an infinite number of times, the mean of the
sample means will be the same as the population mean.
h. Copy the data for all 10 surveys, one column at a time, into the Histogram tool and plot
a histogram for all 500 responses. Export an image of the graph, and paste it in the
space below.
Sample answer:
i.
Compare the histogram with the plots you created for surveys 1 and 4.
Sample answer:
The histogram of 500 responses resembles a normal distribution more closely than the
distribution of either of the individual sample surveys. This is because the number of
responses is 10 times greater.
j.
How does the mean of the data displayed in the histogram tool compare with the
population mean?
Sample answer:
The mean displayed by the histogram tool is 1.72, which is very close to the population
mean (1.73). The standard deviation for this larger sample is 0.64, which is very close
to the 0.657 modeled in the simulation.
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3. Margin of Error of a Sample Mean
In the previous activity, you calculated the intervals such that:
 There is a 68% chance that a given American will tip a grocery bagger between $1.64
and $1.82.
 There is a 95% chance that a given American will tip a grocery bagger between $1.54
and $1.92.
 There is a 99.7% chance that a given American will tip a grocery bagger between
$1.45 and $2.01.
a. You calculated the sample mean for survey 1 as $1.69, and the population standard
deviation is 0.657. Calculate the margin of error of the sample mean. Which interval
usually constitutes the margin of error?
Sample answer:
The 95% confidence interval constitutes the margin of error.
σ
The 95% confidence interval is 1.69  2 
.
n
Here, σ = 0.657 and n = 50.
0.657
 1.69  2(0.093)  1.69  0.186.
So the 95% confidence interval is 1.69  2 
50
The margin of error of the sample mean is $0.186. I can conclude that Americans
believe that a grocery bagger who carries five bags to the car should receive $1.69 as
a tip, with a margin of error of $0.186.
b. You calculated the mean values of the 10 surveys whose data is given in the
spreadsheet. Plot a histogram of these mean values using the Histogram tool. Set
appropriate limits and intervals, and label the plot properly. Export an image of the
graph, and paste it in the space below. Comment on the distribution of these 10
means.
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Sample answer:
The 10 means seem to be distributed roughly as a normal distribution. The mean from
the histogram tool is 1.72, which is very close to the 1.73 population mean entered in
the simulation.
c. How closely do these 10 results fit the theory on margin of error for the trial means?
That is, how many of the 95% ranges include the population mean?
Sample answer:
Sample
1
2
3
4
5
6
7
8
9
10
Mean
1.69
1.60
1.78
1.78
1.74
1.63
1.76
1.83
1.79
1.59
95% Range
1.504 - 1.876
1.414 - 1.786
1.594 - 1.966
1.594 - 1.966
1.554 - 1.926
1.444 - 1.816
1.574 - 1.946
1.644 - 2.016
1.604 - 1.976
1.404 - 1.776
95% Range
Includes 1.73?
yes
yes
yes
yes
yes
yes
yes
yes
yes
yes
All the 95% ranges include the population mean.
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4. Finding a Sample Proportion and Its Margin of Error
In your study on tipping behavior, you are asking the question, Should a customer at a
coffee shop leave a tip in the tip jar? In the spreadsheet, the Coffee Shop Tip Survey tab
gives the responses of 50 people to this question.
a. What is the sample proportion of the data?
Sample answer:
The sample proportion is the percentage of positive responses, in this case, the
percentage of yes responses.
Total number of yes responses: 9
Total number of responses: 50
9
× 100 = 0.18
50
Sample proportion of the data ( p̂ ) = 0.18
Percentage of positive responses:
b. What is the 68% confidence interval of the data?
Sample answer:
The 68% confidence interval is pˆ 
pˆ (1-pˆ )
n
0.18(1- 0.18)
 0.18  0.054
50
= 0.126 to 0.234.
 0.18 
c. What is the margin of error of this data?
Sample answer:
The 95% confidence interval constitutes the margin of error.
pˆ (1-pˆ )
The 95% confidence interval is pˆ  2
.
n
Here, p̂ = 0.18 and n = 50.
So, the 95% confidence interval is 0.18  2 
0.18(1- 0.18)
50
= 0.18 ± 2(0.054)
= 0.18 ± 0.108.
The margin of error of the sample proportion is 10.8%.
Thus, 18% of Americans believe that customers at a coffee shop should leave a tip,
and the margin of error is 10.8%.
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