## Engage NY Eureka Math 7th Grade Module 5 Lesson 22 Answer Key

### Eureka Math Grade 7 Module 5 Lesson 22 Example Answer Key

Examples 1–3

Tamika’s mathematics project is to see whether boys or girls are faster in solving a KenKen-type puzzle. She creates a puzzle and records the following times that it took to solve the puzzle (in seconds) for a random sample of 10 boys from her school and a random sample of 11 girls from her school:

Example 1.

On the same scale, draw dot plots for the boys’ data and for the girls’ data. Comment on the amount of overlap between the two dot plots. How are the dot plots the same, and how are they different?

Answer:

The dot plots appear to have a considerable amount of overlap. The boys’ data may be slightly skewed to the left, whereas the girls’ are relatively symmetric.

Example 2.

Compare the variability in the two data sets using the MAD (mean absolute deviation). Is the variability in each sample about the same? Interpret the MAD in the context of the problem.

Answer:

The variability in each data set is about the same as measured by the mean absolute deviation (around 4 sec.)

For boys and girls, a typical deviation from their respective mean times (35 for boys and 39 for girls) is about 4 sec.

Example 3.

In the previous lesson, you learned that a difference between two sample means is considered to be meaningful if the difference is more than what you would expect to see just based on sampling variability. The difference in the sample means of the boys’ times and the girls’ times is 4.1 seconds (39.4 seconds – 35.3 seconds). This difference is approximately 1 MAD.

a. If4 sec. is used to approximate the value of 1 MAD for both boys and for girls, what is the interval of times that are within 1 MAD of the sample mean for boys?

Answer:

35.3 sec.+ 4 sec.=9.3 sec. , and 35.3 sec.- 4 sec.=31.4 sec. The interval of times that are within 1 MAD of the boys’ mean time is approximately 31.4 sec. to 39.3 sec.

b. Of the 10 sample means for boys, how many of them are within that interval?

Answer:

Six of the sample means for boys are within the interval.

c. Of the 11 sample means for girls, how many of them are within the interval you calculated in part (a)?

Answer:

Seven of the sample means for girls are within the interval.

d. Based on the dot plots, do you think that the difference between the two sample means is a meaningful difference? That is, are you convinced that the mean time for all girls at the school (not just this sample of girls) is different from the mean time for all boys at the school? Explain your choice based on the dot plots.

Answer:

Answers will vary. Sample answer: I don’t think that the difference is meaningful. The dot plots overlap a lot, and there is a lot of variability in the times for boys and the times for girls.

Examples 4–7

How good are you at estimating a minute? Work in pairs. Flip a coin to determine which person in the pair will go first. One of you puts your head down and raises your hand. When your partner says “Start,” keep your head down and your hand raised. When you think a minute is up, put your hand down. Your partner will record how much time has passed. Note that the room needs to be quiet. Switch roles, except this time you talk with your partner during the period when the person with his head down is indicating when he thinks a minute is up. Note that the room will not be quiet.

Answer:

Use your class data to complete the following.

Example 4.

Calculate the mean minute time for each group. Then, find the difference between the quiet mean and the talking mean.

Answer:

The mean of the quiet estimates is 58.8 sec.

The mean of the talking estimates is 64.8 sec.

64.8 – 58.8 = 6

The difference between the two means is 6 sec.

Example 5.

On the same scale, draw dot plots of the two data distributions, and discuss the similarities and differences in the two distributions.

Answer:

The dot plots have quite a bit of overlap. The quiet group distribution is fairly symmetric; the talking group distribution is skewed somewhat to the right. The variability in each is about the same. The quiet group appears to be centered around 60 sec., and the talking group appears to be centered around 65 sec.

Example 6.

Calculate the mean absolute deviation (MAD) for each data set. Based on the MADs, compare the variability in each sample. Is the variability about the same? Interpret the MADs in the context of the problem.

Answer:

The MAD for the quiet distribution is 2.68 sec.

The MAD for the talking distribution is 2.73 sec.

The MAD measurements are about the same, indicating that the variability in each data set is similar. In both groups, a typical deviation of students’ minute estimates from their respective means is about 2.7 sec.

Example 7.

Based on your calculations, is the difference in mean time estimates meaningful? Part of your reasoning should involve the number of MADs that separate the two sample means. Note that if the MADs differ, use the larger one in determining how many MADs separate the two means.

Answer:

The number of MADs that separate the two sample means is \(\frac{6}{2.73}\), or 2.2. There is a meaningful difference between the means.

### Eureka Math Grade 7 Module 5 Lesson 22 Problem Set Answer Key

Question 1.

A school is trying to decide which reading program to purchase.

a. How many MADs separate the mean reading comprehension score for a standard program (mean = 67.8, MAD = 4.6, n = 24) and an activity-based program (mean = 70.3, MAD = 4.5, n = 27)?

Answer:

The number of MADs that separate the sample mean reading comprehension score for a standard program and an activity-based program is \(\frac{70.3-67.8}{4.6}\), or 0.54, about half a MAD.

b. What recommendation would you make based on this result?

Answer:

The number of MADs that separate the programs is not large enough to indicate that one program is better than the other program based on mean scores. There is no noticeable difference in the two programs.

Question 2.

Does a football filled with helium go farther than one filled with air? Two identical footballs were used: one filled with helium and one filled with air to the same pressure. Matt was chosen from the team to do the kicking. Matt did not know which ball he was kicking. The data (in yards) follow.

Answer:

a. Calculate the difference between the sample mean distance for the football filled with air and for the one filled with helium.

Answer:

The 17 air-filled balls had a mean of 27 yd. compared to 23.8 yd. for the 17 helium-filled balls, a difference of 3.2 yd.

b. On the same scale, draw dot plots of the two distributions, and discuss the variability in each distribution.

Answer:

Based on the dot plots, it looks like the variability in the two distributions is about the same.

c. Calculate the MAD for each distribution. Based on the MADs, compare the variability in each distribution. Is the variability about the same? Interpret the MADs in the context of the problem.

Answer:

The MAD is 2.59 yd. for the air-filled balls and 2.07 yd. for the helium-filled balls. The typical deviation from the mean of 27.0 is about 2.59 yd. for the air-filled balls. The typical deviation from the mean of 23.8 is about 2.07 yd. for the helium-filled balls. There is a slight difference in variability.

d. Based on your calculations, is the difference in mean distance meaningful? Part of your reasoning should involve the number of MADs that separate the sample means. Note that if the MADs differ, use the larger one in determining how many MADs separate the two means.

Answer:

\(\frac{3.2}{2.59}\) = 1.2

There is a separation of 1.2 MADs. There is no meaningful distance between the means.

Question 3.

Suppose that your classmates were debating about whether going to college is really worth it. Based on the following data of annual salaries (rounded to the nearest thousand dollars) for college graduates and high school graduates with no college experience, does it appear that going to college is indeed worth the effort? The data are from people in their second year of employment.

a. Calculate the difference between the sample mean salary for college graduates and for high school graduates.

Answer:

The 15 college graduates had a mean salary of $52,400, compared to $32,800 for the 15 high school graduates, a difference of $19,600.

b. On the same scale, draw dot plots of the two distributions, and discuss the variability in each distribution.

Answer:

Based on the dot plots, the variability of the two distributions appears to be about the same.

c. Calculate the MAD for each distribution. Based on the MADs, compare the variability in each distribution. Is the variability about the same? Interpret the MADs in the context of the problem.

Answer:

The MAD is 5.15 for college graduates and 5.17 for high school graduates. The typical deviation from the mean of 52.4 is about 5.15 (or $5,150) for college graduates. The typical deviation from the mean of 32.8 is about 5.17 ($5,170) for high school graduates. The variability in the two distributions is nearly the same.

d. Based on your calculations, is going to college worth the effort? Part of your reasoning should involve the number of MADs that separate the sample means.

Answer:

\(\frac{19.6}{5.17}\) = 3.79

There is a separation of 3.79 MADs. There is a meaningful difference between the population means. Going to college is worth the effort.

### Eureka Math Grade 7 Module 5 Lesson 22 Exit Ticket Answer Key

Suppose that Brett randomly sampled 12 tenth-grade girls and boys in his school district and asked them for the number of minutes per day that they text. The data and summary measures follow.

Question 1.

Draw dot plots for the two data sets using the same numerical scales. Discuss the amount of overlap between the two dot plots that you drew and what it may mean in the context of the problem.

Answer:

There is no overlap between the two data sets. This indicates that the sample means probably differ, with girls texting more than boys on average. The girls’ data set is a little more compact than the boys, indicating that their measure of variability is smaller.

Question 2.

Compare the variability in the two data sets using the MAD. Interpret the result in the context of the problem.

Answer:

The MAD for the boys’ number of minutes spent texting is 7.9 min., which is higher than that for the girls, which is 5.3 min. This is not surprising, as seen in the dot plots. The typical deviation from the mean of 70.9 is about 7.9 min. for boys. The typical deviation from the mean of 97.3 is about 5.3 min. for girls.

Question 3.

From 1 and 2, does the difference in the two means appear to be meaningful?

Answer:

97.3 – 70.9 = 26.4

The difference in means is 26.4 min.

\(\frac{26.4}{7.9}\) = 3.3

Using the larger MAD of 7.9 min., the means are separated by 3.3 MADs. Looking at the dot plots, it certainly seems as though a separation of more than 3 MADs is meaningful.