Unit 7, Lesson 9
Variability in Samples
Integrated Math 3
9.1
Warm-up
Standards Alignment
Building On
Addressing
Building Toward
HSS-IC.B.4
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.
Instructional Routines
Poll the ClassActivity Narrative
The mathematical goal of this activity is for students to calculate sample means and sample proportions to estimate the population mean and proportion in a future activity. Students should recognize that different samples can result in different (but close) estimates for means and proportions. The data collected in this Warm-up is used in a later activity focused on margin of error.
Launch
Give each student 1 standard number cube and 1 copy of the table from the blackline master. Demonstrate that students should select a value for their sample by rolling the number cube once for the row of the table and once for the column of the table. Tell students that a class histogram will be created using their data. Collect the sample means and proportions from the class to create histograms using the class data in the discussion.
Activity
NoneCoins are usually stamped with the year and location of the mint where they were made. D represents the mint in Denver, Colorado, and a blank or P represents the mint in Philadelphia, Pennsylvania.
Diego has a jar containing 36 coins. Select a sample of 5 coins by rolling your number cube once to represent the row and then rolling again to find the column. For example, rolling a 3 and then a 5 would represent selecting the coin marked “2000 P.” Repeat this process to collect a sample of 5 coins. The samples are drawn with replacement, which means we allow for the opportunity to draw the same coin more than once into our sample.
| coin 1 | coin 2 | coin 3 | coin 4 | coin 5 | sample mean year | sample proportion minted in Denver | |
|---|---|---|---|---|---|---|---|
| sample 1 | |||||||
| sample 2 | |||||||
| sample 3 |
- Find the mean of the years for the sample of 5 coins.
- Find the proportion of the sample of 5 coins that were minted in Denver.
- Repeat the process to find 2 more samples of 5 coins, and compute the mean year and the proportion that were minted in Denver for each sample.
Activity Synthesis
The goal of this discussion is to make sure students understand the difference between a sample mean and a sample proportion, and for students to examine the variability present in the sample characteristics collected.
Collect the sample means and proportions from the class, and create two histograms, one for the sample means and one for the proportions, to be used in a later activity.
Here are some questions for discussion:
- “What is the lowest sample mean? The highest?” (2003.6 and 2013.8)
- “What is the lowest sample proportion? The highest?” (0.4 and 0.8)
- “When making a histogram of the sample means for the year and another histogram for the sample proportions minted in Denver, what interval size makes sense for each histogram?” (It makes sense to have an interval of 0.5 or 1 for the years and 0.2 for the sample proportions.)
- “What information does the sample mean tell us?” (It tells us the mean of the years the 5 randomly selected coins were minted.)
- “What information does the sample proportion tell us?” (It tells us the proportion of the 5 randomly selected coins that were minted in Denver.)
Display histograms of sample means and proportions from the class.
- “Not all the samples have the same mean and proportion. Why not?” (Because each sample has a chance of getting different coins, the values won’t always match.)
- “For a distribution of sample means like this histogram, which, if any, of these years would be surprising to you if they were the mean year of all 36 coins: 2011, 2017, or 1999?” (2011 is not surprising because it is close to the middle, but the others would be surprising because there are no sample means near those values.)
- “The actual mean year of all 36 coins is 2010. Does that make sense based on the histogram of sample means?” (Yes, that makes sense because the year 2010 is the middle of distribution shown in the histogram.)
- “The actual proportion of these 36 coins that were minted in Denver is about 0.42. Does that make sense based on the histogram of sample proportions?” (Yes, 0.4 is the most frequent value shown in the histogram.)
9.2
Activity
Instructional Routines
NoneMaterials
Activity Narrative
In this activity, students estimate a margin of error to be used with an estimate for a population proportion based on a sample. The proportion found in the sample is used to estimate the proportion expected for the population. A margin of error is given along with the point estimate to provide a range of values that would not be surprising for the population proportion.
To determine a margin of error, students run a simulation to create a sampling distribution consisting of proportions of samples drawn from the original sample data. The standard deviation from the sampling distribution is used to find a margin of error.
Launch
Lesson Synthesis
Here are some questions for discussion:
- “Why do we see variability between different random samples from the same population?” (If there is variation in the population, a sample will usually include some of this variation. With a random sample, we would expect differences in the sample to be reflective of variation in the population.)
- “Think about flipping a coin 8 times. How many heads would you expect to get? Explain your reasoning.” (I think I would get between 3 and 5 or between 2 and 6. It seems unlikely to get 0, 1,7, or 8 heads.)
- “Why does it make sense to report the number of heads with more than one value?” (It makes sense because the number of heads when flipping a coin 8 times is not predictable, but we can say that we expect it to happen in a certain interval most of the time or almost all of the time.)
- “If everyone in the class flipped a coin 8 times, how could we use the results to find the margin of error?” (We could find the mean and standard deviation of the number of heads flipped. The margin of error would be 2 times the standard deviation.)
- “Why do we double the standard deviation to find the margin of error?” (For normally distributed data, we expect 95% of the data to be within 2 standard deviations of the mean. So it is likely that this range of estimates will include the actual population characteristic.)
Student Lesson Summary
In many cases, it is difficult to collect data from an entire population, so using data from a small subset of the larger group, called a sample, is needed. The trade-off is that the incomplete information from samples can provide only estimates of characteristics for the population.
For example, an ecologist may wonder what proportion of trees in a particular large forest contains bird nests. It would be hard to look at every tree in the forest, so the researcher might take a random sample of 1,000 trees to find the proportion that have nests. Let’s say the ecologist found that a proportion of 0.04 trees in the sample have nests. That is a good number to give as an estimate for the percentage of trees in the forest that have nests, but because the sample was randomly selected, it would probably not be surprising if 0.028 or 0.05 of the trees have nests.
To give a sense of the variability and confidence in estimates, a margin of error is usually given along with the point estimate. A margin of error is the maximum expected difference between a point estimate for a population characteristic and the actual value of the population characteristic. For means and proportions, the distribution of point estimates derived from samples, called the sampling distribution, tends to be approximately normal and centered around the actual population characteristic, so it is reasonable to expect that about 95% of the point estimates are within 2 standard deviations of the actual population characteristic. In this unit, we will use 2 standard deviations of the sampling distribution as the margin of error.
The ecologist could take the data from their sample and use a computer to simulate drawing, with replacement, thousands more samples of the same size from the original sample to create a sampling distribution. If the standard deviation for the sampling distribution is 0.006, then they can use a margin of error of 0.012 along with the estimate of 0.04 for the population proportion and report an estimate of for the proportion of trees in the forest that have nests. This means that any value in the range of 0.028 to 0.052 would not be surprising for the proportion of trees in the forest that have nests.