End-of-Unit Assessment

1.

Jada gathered data at her school among 7th and 8th graders to see if there was an association between grade level and playing sports. The table shows her data, but the number of 8th graders who play sports is missing.

	play sports	don't play sports
7th grade	112	52
8th grade		84

Jada found there was no evidence of an association between grade level and playing sports. Which of these could be the number of 8th graders who play sports?

A.

248

B.

181

C.

144

D.

69?

Answer:

B

Teaching Notes

Students who select A add the numbers in the problem: ( $112+52+84 = 248$ ). Students who select C first subtract ( $84-52=32$ ), then add ( $32+112=144$ ). Students who select D reverse the proportion between the columns ( $112\boldcdot\frac{52}{84}\approx69$ ).

2.

Here is a scatter plot:

The graph of what linear equation is a good fit for this data?

A.

$y=\frac12x$

B.

$y=\text-\frac12x+1$

C.

$y=\text-\frac12x$

D.

$y=\frac12x+1$

Answer:

D

Teaching Notes

Students who select A do not choose the correct the intercept, using (0,0) instead of (0,1). Students who select B have the intercept right but switch the sign of the slope. Students who select C do both of these things.

3.

Select all the relationships that demonstrate a negative association between variables.

A.

Number of absences from school and final grades

B.

Outside temperature and ice cream sales

C.

Price of houses and house sales

D.

Number of rainy days and car accidents

E.

Number of hours playing video games and grades

Answer:

A, C, E

Teaching Notes

Students who do not select A, C, and E may be thinking about a required linear relationship, instead of a generally negative association. Students who select B and D may be looking for any association, without regard to whether it is positive or negative.

4.

Draw a scatter plot that shows a negative, linear association and has one clear outlier. Circle the outlier.
Draw a scatter plot that shows a positive association that is not linear.

Answer:

Answers vary.

Minimal Tier 1 response:

Work is complete and correct.
Sample:

Plot shows points nearly in the same line with a negative slope, and 1 circled point not near the line.
Plot shows points that are not nearly in the same line, with a generally positive trend.

Tier 2 response:

Work shows general conceptual understanding and mastery, with some errors.
Sample errors: One plot shows a significant error, but the other is correct; outlier not circled; plots have reversed negative and positive associations, but otherwise all work is correct.

Tier 3 response:

Significant errors in work demonstrate lack of conceptual understanding or mastery.
Sample errors: Both plots show errors beyond the ones listed in Tier 2.

Teaching Notes

This problem asks students to demonstrate their understanding of the terms “positive association,” “negative association,” and “outlier” by drawing scatter plots. The points need not be arranged close to a line to show an association: the important thing is that there is a general upward (or downward) trend.

5.

Andre surveyed all 7th and 8th grade students at his school about whether they prefer to walk or ride a bike. Complete the missing entries in this two-way table.

	prefer walk	prefer bike	total
7th grade students		219	293
8th grade students	168		261
total

Answer:

	prefer walk	prefer bike	total
7th grade students	74	219	293
8th grade students	168	93	261
total	242	312	554

Teaching Notes

This problem requires students to understand the structure of a two-way table. Successful students will subtract across the rows to find the missing entries in the first 2 rows, then add down the columns for the bottom row.

6.

At the school carnival, adults attended with children. Everyone had a choice between a piece of pizza and a hot dog. Here is a two-way table showing the number of adults and children who made each choice.

	pizza	hot dog	total
adult	48	105	153
child	92	73	165
total	140	178	318

Complete the table with relative frequency by row. Round to the nearest percent.

pizza hot dog total

adult 100%

child 100%
Make a segmented bar graph to represent the data in your table. Use one bar for each row of the table.

	pizza	hot dog	total
adult			100%
child			100%

Answer:

pizza hot dog total

adult 31% 69% 100%

child 56% 44% 100%

	pizza	hot dog	total
adult	31%	69%	100%
child	56%	44%	100%

2.

Minimal Tier 1 response:

Work is complete and correct.
Sample:
The table is completely correct.
The segmented bar graphs are accurate.

Tier 2 response:

Work shows general conceptual understanding and mastery, with some errors.
Sample errors: Percentages in table are nearly correct but wrong due to a calculation or rounding error; segmented bar graphs are inaccurate.
Acceptable errors: Segmented bar graph is correct based on incorrect percentages.

Tier 3 response:

Significant errors in work demonstrate lack of conceptual understanding or mastery.
Sample errors: A regular bar graph is drawn instead of a segmented bar graph; percentages in table are very inaccurate, notably if outside 0–100% range or if using an incorrect total as the basis for the percentages.

Teaching Notes

This problem assesses understanding of relative frequency, both by having students calculate relative frequency from a two-way table and by constructing segmented bar graphs to represent the data.

7.

A deli owner noticed that as the outside temperature increased, they sold less soup. For each day soup was sold, they plotted the point $(t,s)$ , where $t$ represents high temperature and $s$ represents bowls of soup sold.

On the same axes, draw a line that you think is a good fit for the data.
The deli owner found that the line $s = \text-t +120$ is a good fit for the data. Use this equation to predict how many bowls of soup they might sell on a day when the high temperature is 32 degrees.
The high temperature this Saturday is expected to be 10 degrees colder than the high temperature this Friday. Using the line $s = \text-t + 120$ , how many more bowls of soup should the deli expect to sell on Saturday than Friday? Explain or show your reasoning.

Answer:

Sample response:
88 bowls. $\text-32 + 120 = 88$
10 more bowls. The slope of the line is -1. This means that for each one-degree decrease in temperature, the deli can expect to sell about one more bowl of soup. If the temperature decreases by 10 degrees, they can expect to sell about 10 more bowls of soup.

Minimal Tier 1 response:

Work is complete and correct, with complete explanation or justification.
Sample:

The line reasonably represents a fitting line for the data.
88
10, because the slope is -1 and $\text–1 \boldcdot \text–10 = 10$

Tier 2 response:

Work shows good conceptual understanding and mastery, with either minor errors or correct work with insufficient explanation or justification.
Sample errors: Line of fit is not close but still generally follows direction of data; minor visible calculation error gives incorrect number of bowls for 32 degrees; explanation missing for why 10 more bowls are expected on Saturday.

Tier 3 response:

Work shows a developing but incomplete conceptual understanding, with significant errors.
Sample errors: Line of fit is missing or very inaccurate; incorrect answer given for part b with no work shown or explanation; incorrect application of rule $s = \text- t + 120$ .

Tier 4 response:

Work includes major errors or omissions that demonstrate a lack of conceptual understanding and mastery.
Sample errors: Two or more error types from Tier 3 response.

Teaching Notes

In part c, students could also solve this problem by choosing a temperature for Saturday, such as 40 degrees, then using the equation $s = \text-t + 120$ to calculate the expected number of bowls sold for 40 degrees and for 30 degrees, then comparing the answers.