Math problem set Grade 8 Science

Thistlebrook Creek: Water Temperature and Oxygen Scatter Plot

A grade 8 science problem set where students scatter-plot creek temperature and dissolved oxygen, fit a trend line, and predict when oxygen drops below the fish limit.

Objectives

  1. Students will construct a scatter plot and describe the association between two water-quality variables.
  2. Students will fit a trend line, estimate its slope, and use it to predict a value beyond the data.

Vocabulary

Key words: scatter plot, association, trend line, outlier

Teacher overview

A volunteer monitoring the fictional Thistlebrook Creek records water temperature and dissolved oxygen on eight summer days, because warmer water holds less oxygen and low oxygen stresses fish. This problem set has students plot the data, describe the pattern, fit a trend line by eye, and use it to predict when oxygen would fall below the 5 mg/L that most fish need. A common misconception the set is built to correct is the belief that a best-fit line must pass through every data point; a good trend line comes close to the points overall while ignoring a single outlier.

Answer Key

  1. Answer: Negative, roughly linear. The points fall from upper left to lower right, so as temperature rises, dissolved oxygen drops in a nearly straight-line pattern.
  2. Answer: Decrease. Higher water temperatures line up with lower dissolved-oxygen readings throughout the table.
  3. Answer: The point at 16 °C, 9.6 mg/L. Its oxygen is higher than the downward trend, which near 16 °C would sit closer to 8.9 mg/L.
  4. Answer: -0.3 mg/L per °C. From (12, 10.1) to (26, 5.9), the oxygen changes by 5.9 − 10.1 = −4.2 over 26 − 12 = 14 degrees, and −4.2 ÷ 14 = −0.3 mg/L per °C.
  5. Answer: Each 1 °C rise in water temperature lowers dissolved oxygen by about 0.3 mg/L, so warmer water carries steadily less oxygen for fish.
  6. Answer: 4.7 mg/L. Using DO = −0.3 × temperature + 13.7, DO = −0.3 × 30 + 13.7 = −9 + 13.7 = 4.7 mg/L (accept roughly 4.5 to 5).
  7. Answer: 29 °C. Setting −0.3 × temperature + 13.7 = 5 gives 0.3 × temperature = 8.7, so temperature = 29 °C.
  8. Answer: Hot days would raise the water temperature, which lowers dissolved oxygen along the trend, so above about 29 °C the oxygen would drop under 5 mg/L and the fish would be stressed or begin to die.

Differentiation

Support

Provide pre-labeled and scaled axes so students plot points without setting up the grid, and mark the 5 mg/L line on the graph in advance. Give the trend-line equation DO = −0.3 × temperature + 13.7 for students who need it to attempt problems 6 and 7.

Extension

Ask students to compute how far each of the eight readings sits above or below the line DO = −0.3 × temperature + 13.7, add up the sizes of those gaps, and use the total to argue how well the line fits.

ELL supports

Pre-teach "association," "trend line," "slope," and "outlier" with a simple sketch showing points that rise, points that fall, and one stray point. Allow students to answer problems 5 and 8 with a labeled arrow on the graph plus a short phrase.