End-of-Unit Assessment

1.

Noah gathered data at his school among 7th and 8th graders to see if there was an association between grade level and handedness. The table and graph show his data, but the number of right-handed 8th graders is missing.

	left-handed	right-handed
7th grade	11	72
8th grade	24

Noah found there was no evidence of an association between grade level and handedness. Which of these could be the number of right-handed 8th graders?

A.

33

B.

85

C.

107

D.

157

Answer:

157

Teaching Notes

Students who select A reverse the proportion between the columns ( $72 \cdot \frac{11}{24}$ ). Students who select B preserve the absolute difference in each column $(|11 - 24| = |72-85|)$ . Students who select C add the numbers in the problem: $(11 + 24 + 72)$ .

2.

Here is a scatter plot:

Scatterplot, x, y. Points start at point 2 comma 2 point 1 and trend right and down toward 5 point 5 comma point 2.

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

A.

$y=\text-\frac13x+2$

B.

$y=\text-\frac13x+6$

C.

$y = \frac 1 3 x + 2$

D.

$y = \frac 1 3 x + 6$

Answer:

$y=\text-\frac13x+2$

Teaching Notes

Students who select B choose the incorrect intercept, using $(6,0)$ instead of $(0,2)$ . Students who select C have the intercept right but switch the sign of the slope. Students who select D do both of these things.

3.

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

A.

Outside temperature and cost to heat a home

B.

Length of time walked and distance traveled

C.

Pounds of cherries bought and amount of money spent on cherries

D.

Speed of a train and the amount of time it takes for the train to get to its destination

E.

Number of people in a grocery check-out line and amount of time waiting in line

Answer: B, C, E

Teaching Notes

A student who selects A may not be thinking about the way outside temperature is interpreted. As temperature increases, the cost to heat a home generally decreases. A student who selects choice D may be looking for any association, without regard to whether it is positive or negative. A student who does not select E may be thinking about a required linear relationship, instead of a generally positive association.

4.

Draw a scatter plot that shows a positive, linear association and has 1 clear outlier. Circle the outlier.
Draw a scatter plot that shows a negative 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 positive slope, and 1 circled point not near the line.
Plot shows points that are not nearly in the same line, with a generally negative 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.

Jada surveyed all 7th and 8th graders at her school about whether they have pets. Complete the missing entries in this two-way table.

	has pet	has no pet	total
7th grade	102		150
8th grade		68	175
total

Answer:

	has pet	has no pet	total
7th grade	102	48	150
8th grade	107	68	175
total	209	116	325

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 a school social, children attend with family members. Everyone had a choice between a sweet snack and a salty snack. Here is a two-way table showing the number of adults and children who made each choice of snack.

	sweet snack	salty snack	total
adult	57	88	145
child	77	31	108
total	134	119	253

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

sweet snack salty snack 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.

	sweet snack	salty snack	total
adult			100%
child			100%

Answer:

sweet snack salty snack total

adult 39% 61% 100%

child 71% 29% 100%

	sweet snack	salty snack	total
adult	39%	61%	100%
child	71%	29%	100%

<p>Two bar graphs one for adult and one for child.</p>

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 completely 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.

Lin opened a lemonade stand during the summer. She noticed that she sold more lemonade on warmer days. For each day she sold lemonade, she plotted the point $(t,c)$ , where $t$ represents daily high temperature in degrees Fahrenheit and $c$ represents cups of lemonade sold.

Scatterplot, temperature, degrees Fahrenheit, cups sold. Points begin at 70 comma 50 and trend up and right toward 87 comma 97.

On the same axes, draw a line that you think is a good fit for the data.
A computer program found that the line $c = 2t - 89$ is a good fit for the data. Use this equation to predict how many cups of lemonade Lin might sell on a day when the daily high temperature is 74 degrees Fahrenheit.
The daily high temperature this Sunday is expected to be 5 degrees warmer than the daily high temperature this Saturday. Using the line $c = 2t - 89$ , how many more cups of lemonade should Lin expect to sell on Sunday than Saturday? Explain or show your reasoning.

Answer:

Sample response:
59 cups. $2 \boldcdot 74 - 89 = 59$ .
10 more cups. The slope of the line is 2. This means that for each one-degree increase in temperature, Lin can expect to sell about two more cups of lemonade. If the temperature increases by 5 degrees, she can expect to sell about 10 more cups of lemonade.

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.
59
10, because the slope is 2 and $2 \boldcdot 5 = 10$ .

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 cups for 74 degrees; explanation missing for why 10 more cups are expected on Sunday.

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 $c = 2t - 89$ .

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 70 degrees, then using the equation $c = 2t - 89$ to calculate the expected number of cups sold for 70 degrees and for 75 degrees, then comparing the answers.