Quiz & test questions Grade 11 Math
Pellow River Nitrate: A Grade 11 Statistics Unit Test
A 30-point grade 11 statistics test on a fictional river's nitrate data, with items on spread, margin of error, and least-squares regression.
Objectives
- Students will compute and compare measures of center and spread for a data set.
- Students will build and interpret an interval estimate from a sample mean and margin of error.
- Students will use a regression line to predict a value, find a residual, and judge the fit.
Vocabulary
Key words: mean, range, margin of error, residual, regression line
Teacher overview
This 40-minute unit test, worth 30 points, checks whether students can compare data sets, work with a margin of error, and use a regression line. It draws on nitrate monitoring for a fictional utility, the Marchmont water district on the Pellow River. The test mixes two multiple-choice items, three short-answer items, and five computation items, spanning recall and application. Point values are printed with each item.
Assessment
The Marchmont water district (a fictional utility) measures nitrate in the Pellow River. Its eight monthly readings, in milligrams per liter, are 4.2, 5.1, 4.8, 5.5, 4.9, 6.0, 5.2, and 4.6. For this test, the district's action level is 10 milligrams per liter.
- (2 points) Two rivers have the same mean nitrate, but River X has a much larger standard deviation than River Y. Which statement is best supported?
- A) River X is more consistent.
- B) River Y is more consistent.
- C) They are equally consistent.
- D) River Y has a higher mean.
- (4 points) Compute the mean of the eight monthly nitrate readings. Round to the nearest hundredth.
- (3 points) Compute the range of the eight monthly readings.
- (3 points) How far below the district's 10 milligrams per liter action level is the mean from item 2?
- (4 points) A citywide sampling program estimates the mean nitrate from a random sample of taps. The sample mean is 5.0 milligrams per liter, with a simulated margin of error of 0.6 milligrams per liter. Give the interval estimate for the true mean.
- (3 points) Explain in one or two sentences what the margin of error of 0.6 milligrams per liter tells you about the estimate.
- (4 points) A scatter plot pairs monthly rainfall in inches with river nitrate. The least-squares line is y equals 0.8x plus 2.0, where x is rainfall in inches and y is nitrate in milligrams per liter. Predict the nitrate when rainfall is 5 inches.
- (3 points) One month had 5 inches of rain and a measured nitrate of 6.8 milligrams per liter. Using the line from item 7, compute the residual.
- (2 points) A residual plot shows points scattered randomly around zero with no pattern. Which conclusion is best supported?
- A) The linear model fits the data well.
- B) The linear model is a poor fit.
- C) There is no correlation at all.
- D) The data are categorical.
- (2 points) State one reason the district uses a random sample of taps instead of testing every tap in the city.
From Water for Schools (waterforschools.org). This material is in the public domain (CC0) and may be copied, edited, and shared freely.
Answer Key
- Answer: B) River Y is more consistent. A smaller standard deviation means the readings are packed more tightly around the mean, so River Y is the more consistent of the two.
- Answer: 5.04 mg/L. Sum = 4.2 + 5.1 + 4.8 + 5.5 + 4.9 + 6.0 + 5.2 + 4.6 = 40.3; mean = 40.3 ÷ 8 = 5.0375, about 5.04 milligrams per liter.
- Answer: 1.8 mg/L. Range = 6.0 − 4.2 = 1.8 milligrams per liter.
- Answer: 4.96 mg/L. The mean of 5.04 is 10 − 5.04 = 4.96 milligrams per liter below the action level.
- Answer: 4.4 to 5.6 mg/L. The interval is 5.0 minus 0.6 to 5.0 plus 0.6, which is 4.4 to 5.6 milligrams per liter.
- Answer: The margin of error says the true citywide mean is likely within 0.6 milligrams per liter of the sample mean, so the estimate could reasonably be as low as 4.4 or as high as 5.6 rather than being exact.
- Answer: 6.0 mg/L. y = 0.8(5) + 2.0 = 4.0 + 2.0 = 6.0 milligrams per liter.
- Answer: 0.8 mg/L. Residual = observed − predicted = 6.8 − 6.0 = 0.8 milligrams per liter.
- Answer: A) The linear model fits the data well. Randomly scattered residuals with no pattern are the sign that a straight line is an appropriate model.
- Answer: Testing every tap would be too costly and slow, so a random sample lets the district infer the citywide level from a manageable number of measurements.
Differentiation
Support
Provide the sum of the eight readings already computed for item 2, and give a printed reminder that a residual is observed minus predicted. Allow a calculator for all computation items.
Extension
Ask students to predict the nitrate at 8 inches of rainfall using the same line, then explain why extending the line far beyond the observed rainfall range could give an unreliable prediction.
ELL supports
Provide the interval-estimate template "sample mean minus margin to sample mean plus margin" with blanks for item 5, and pair "residual" with a small picture showing the gap between a data point and the line.