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<\/p>\nExample 1: Describing a Population of Numerical Data<\/p>\n
The course project in a computer science class was to create 100 computer games of various levels of difficulty that had ratings on a scale from 1 (easy) to 20 (difficult). We will examine a representation of the data resulting from this project. Working in pairs, your teacher will give you a page that contains 100 rectangles of various sizes.<\/p>\n
a. What do you think the rectangles represent in the context of the 100 computer games?
\nAnswer:
\nEach rectangle represents a computer game.<\/p>\n
b. What do you think the sizes of the rectangles represent in the context of the 100 computer games?
\nAnswer:
\nThe size of the rectangle, or the number of squares that comprise it, represents the difficulty rating of the computer game. The minimum rating is 1; the maximum is 20.<\/p>\n
c. Why do you think the rectangles are numbered from 00 to 99 instead of from 1 to 100?
\nAnswer:
\nAnticipating that a random sample will be taken later in the lesson, It is easiest if all the labels have the same number of digits. So, 100 is conveniently designated as 00. The integers from 1 to 9 are represented by 01, 02, and so on.<\/p>\n
Exploratory Challenge 1\/Exercises 1 – 3: Estimate the Population Mean Rating<\/p>\n
Exercise 1.
\nWorking with your partner, discuss how you would calculate the mean rating of all 100 computer games (the population mean).
\nAnswer:
\nTo find the population mean, all 100 ratings would have to be added and then divided by 100. This is not hard, but it can be a tedious calculation to make. If students do not think adding 100 numbers is too bad, suggest that the number of computer games might have been 1,000.<\/p>\n
Exercise 2.
\nDiscuss how you might select a random sample to estimate the population mean rating of all 100 computer games.
\nAnswer:
\nA good answer would include stating a reasonable sample size, e.g., 10 or more. It should also state that a random number table or a calculator with a random number generator should be used to generate the 10 random two-digit numbers. The generated numbers identify the rectangles (computer games) that would be chosen for the sample.<\/p>\n
Exercise 3.
\nCalculate an estimate of the population means rating of all 100 computer games based on a random sample of size 10. Your estimate is called a sample mean, and It is denoted by. Use the following random numbers to select your sample:
\nAnswer:
\n34 86 80 58 04 43 96 29 44 51
\nThe respective ratings for the given random numbers are 12, 5, 2, 4, 1, 4, 18, 10, 1, and 16. Based on this sample,
\nthe estimate for the population mean rating is \\(\\frac{73}{10^{\\prime}}\\) or 7.3.<\/p>\n
Exploratory Challenge 2\/Exercises 4 – 6: Build a Distribution of Sample Means<\/p>\n
Exercise 4.
\nWork in pairs. Using a table of random digits or a calculator with a random number generator, generate four sets of ten random numbers. Use these sets of random numbers to identify four random samples of size 10. Calculate the sample mean rating for each of your four random samples.
\nAnswer:
\nAnswers will vary. To build a distribution, you should have 50 or more randomly generated sample mean estimates. So, if you have between 25 and 30 students in your class, then 12 to 15 pairs should generate about four sample mean estimates, which will provide the right number of estimates for the distribution.<\/p>\n
Exercise 5.
\nWrite your sample means on separate sticky notes, and post them on a number line that your teacher has prepared for your class.
\nAnswer:
\nBe sure that there is enough length on the number line so that the sticky notes do not overlap.<\/p>\n
Exercise 6.
\nThe actual population mean rating of all 100 computer games is 7.5. Does your class distribution of sample mean center at 7.5? Discuss why it does or does not.
\nAnswer:
\nA possible reason why their sample means do not center at 7. 5 is that they need more samples. They may suggest that they were very unlucky and got a distribution that centered well above or well below 7.5. That is possible, but it is highly unlikely.<\/p>\n
Note: There is a theoretical result that says, for random samples, the expected value of the sample mean is the mean value of the population from which the sample was taken. But that theory is not part of the curriculum. However, students may reason that if a population is divided into samples of equal size, then the mean of sample means is the same as the mean of the whole. They might give an example, such as the following:<\/p>\n
\u201cConsider four samples each of size three (e.g., 4, 1, 3; 2, 2, 7; 5, 9, 6; 3, 6, 5). The respective means are , \\(\\frac{8}{3}\\), \\(\\frac{11}{3}\\), \\(\\frac{20}{3}\\) and \\(\\frac{14}{3}\\) whose mean is their sum divided by 4, precisely the same as the sum of the 12 values divided by 12.\u201d<\/p>\n
Students might then argue that this would also apply to random samples and then go on to try to demonstrate conceptually that taking means produces values that tend to congregate or balance around the population mean.<\/p>\n
Example 2: Margin of Error<\/p>\n
Suppose that 50 random samples, each of size ten, produced the sample means displayed in the following dot plot.<\/p>\n
<\/p>\n
Note that almost all of the sample means are between 4 and 11. That is, almost all are roughly within 3.5 rating points of the population mean 7. 5. The value 3.5 is a visual estimate of the margin of error. It is not really an \u201cerror\u201d in the sense of a \u201cmistake.\u201d Rather, it is how far our estimate for the population mean is likely to be from the actual value of the population mean.<\/p>\n
Based on the class distribution of sample means, is the visual estimate of margin of error close to 3.5?
\nAnswer:<\/p>\n
Example 3: Standard Deviation as a Refinement of Margin of Error<\/p>\n
Note that the margin of error is measuring how spread out the sample means are relative to the value of the actual population mean. From previous lessons, you know that the standard deviation is a good measure of spread. So, rather than producing a visual estimate for the margin of error from the distribution of sample means, another approach is to use the standard deviation of the sample means as a measure of spread.<\/p>\n
For example, the standard deviation of the 50 sample means in the example above is 1.7. Note that if you double 1.7, you get a value for margin of error close to the visual estimate of 3.5.<\/p>\n
Another way to estimate margin of error is to use two times the standard deviation of a distribution of sample means. For the above example, because 2(1.7) = 3.4, the refined margin of error (based on the standard deviation of sample means) is 3.4 rating points.<\/p>\n
An interpretation of the margin of error is that plausible values for the population mean rating are from 7.5 – 3.4 to 7. 5 + 3.4 (i.e., 4. 1 to 10.9 rating points).<\/p>\n
Exploratory Challenge 3\/Exercise 7:<\/p>\n
Exercise 7.
\nCalculate and interpret the margin of error for your estimate of the population mean rating of 100 computer games based on the standard deviation of your class distribution of sample means.
\nAnswer:
\nAnswers will vary based on the class distribution of sample means. Sample response: Suppose the sample mean is 7.2 and the standard deviation is 1.9. Because 2(1.9) = 3.8, the margin of error is 3.8 rating points. This means that the plausible values for the population mean rating is from 7.2 – 3.8 to 7.2 + 3.8, or from 3.4 to 11 rating points.<\/p>\n
Question 1.
\nSuppose you are interested in knowing how many text messages eleventh graders send daily.<\/p>\n
Describe the steps that you would take to estimate the mean number of text messages per day sent by all eleventh graders at a school.
\nAnswer:
\nIf I could not get responses from all eleventh graders, I would base my estimate on the responses from a random sample of students. I would need to find a record or list of all eleventh graders. If I had this list, I would number all of the students on it and use random numbers to generate a random selection of students.<\/p>\n
For example, if there are 450 students, I would number all of the students on the list from 1 to 450 and generate a selection of students using the random number generator on my calculator. If my sample Is 10 students, I would generate 10 random numbers from 1 to 450, identify the 10 students based on the random numbers, and ask the 10 students how many text messages they send during a school day.<\/p>\n
I would then find the mean of those 10 responses. The mean from this sample of students would be my estimate of the mean number of text messages sent by eleventh graders.<\/p>\n
Question 2.
\nSuppose that 62 random samples based on ten student responses to the question, \u201cHow many text messages do you send per day?\u201d resulted in the 62 sample means (rounded) shown below.<\/p>\n
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<\/p>\n
a. Draw a dot plot for the distribution of sample means.
\nAnswer:<\/p>\n
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<\/p>\n
b. Based on your dot plot, would you be surprised if the actual mean number of text messages sent per day for all eleventh graders in the school is 91.7? Why or why not?
\nAnswer:
\nNo. The distribution appears to be balanced around 92, so 91.7 is plausible.<\/p>\n
Question 3.
\nDetermine a visual estimate of the margin of error when a random sample of size 10 is used to estimate then population mean number of text messages sent per day.
\nAnswer:
\nAlmost all sample means are roughly within 10 text messages of the population mean 91.7. So, visually, the margin of error is 10 text messages on average.<\/p>\n
Question 4.
\nThe standard deviation of the above distribution of sample mean a number of text messages sent per day is 7. 5. Use this to calculate and interpret the margin of error for an estimate of the population mean number of text messages sent daily by eleventh graders (based on a random sample of size 10 from this population).
\nAnswer:
\nUsing the standard deviation of the sampling distribution, since 2(7.5) = 15, the margin of error is 15 text messages. Note that the visual estimate is quite a bit smaller than the one using the standard deviation. However, they are the same if the visual estimate were to include all of the sample means from 77 to 107.<\/p>\n
Question 1.
\nAt the beginning of the school year, school districts implemented a new physical fitness program. A student project involves monitoring how long it takes eleventh graders to run a mile. The following data were taken midyear.<\/p>\n
a. What is the estimate of the population mean time it currently takes eleventh graders to run a mile based on the following data (minutes) from a random sample of ten students?
\n6.5, 8.4, 8.1, 6.8, 8.4, 7.7, 9.1, 7.1, 9.4, 7.5
\nAnswer:
\nThe mean of the ten times is 7.9 minutes.<\/p>\n
b. The students doing the project collected 50 random samples of 10 students each and calculated the sample means. The standard deviation of their distribution of 50 sample means was 0.6 minutes. Based on this standard deviation, what is the margin of error for their sample mean estimate? Explain your answer.
\nAnswer:
\nThe margin of error is twice the standard deviation of the sampling distribution.
\nSince 2(0.6) = 1.2, the margin of error is 1.2 minutes.<\/p>\n
c. Interpret the margin of error you found in part (b) in the context of this problem.
\nAnswer:
\nBecause 7.9 – 1.2 = 6.7 and 7.9 + 1. 2 = 9. 1, plausible values for the population mean time it takes eleventh graders to run the mile midyear are between 6.7 to 9. 1 minutes.<\/p>\n","protected":false},"excerpt":{"rendered":"
Engage NY Eureka Math Algebra 2 Module 4 Lesson 20 Answer Key Eureka Math Algebra 2 Module 4 Lesson 20 Exercise Answer Key Example 1: Describing a Population of Numerical Data The course project in a computer science class was to create 100 computer games of various levels of difficulty that had ratings on a … Read more<\/a><\/p>\n","protected":false},"author":3,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[6],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/posts\/92638"}],"collection":[{"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/users\/3"}],"replies":[{"embeddable":true,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/comments?post=92638"}],"version-history":[{"count":1,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/posts\/92638\/revisions"}],"predecessor-version":[{"id":136844,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/posts\/92638\/revisions\/136844"}],"wp:attachment":[{"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/media?parent=92638"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/categories?post=92638"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/tags?post=92638"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}