Penny Smeltzer began teaching AP Statistics when the course began and quickly created one of the largest programs in the nation in Austin, Texas. Her passion for the subject continues with decades of experience as an AP Statistics exam grader, grading leader, rubric writer, Test Development Committee member, and Question Leader. Her passion is creating meaningful interdisciplinary lessons with science using topics that matter in the lives of the human population and the planet, which she share on her website Lessons That Matter.

Penny was the question leader for 2023 #5.

Wildlife biologists are interested in the health of tule elk, a species of deer found in California. An important measurement of tule elk health is their weight. The weight of a tule elk is difficult to measure in the wild. However, chest circumference, which is believed to be related to the weight of a tule elk, can easily be measured from a safe distance using a harmless laser. A study was done to investigate whether chest circumference, in centimeters (cm), could be used to accurately estimate the weight, in kilograms (kg), of male tule elk. For the study, wildlife biologists captured 30 male tule elk, measured their chest circumference and weight and then released the elk. The data for the 30 male tule elk are shown in the scatterplot.

### Part (a)

Describe the relationship between chest circumference and weight of male tule elk in context.

## WOULD THIS GET CREDIT?

The rubric for this part of the question requires students to discuss at least 3 of 4 components (direction, strength, form, unusual features) AND must be in context.

The first partial credit response does contain adequate context and the concept of a positive direction, but there is no mention of form or strength. The second partial credit response does not contain any context. The full credit response does not address unusual features, but still satisfies 3 of the 4 components (direction, strength, form) plus context.

### Teaching Tips:

Consider using an acronym to help student to remember all of the ideas that must be discussed when describing the relationship between two quantitative variables. Many teachers use DUFS (direction, unusual features, form, strength) + context.

Always discuss context and include it in every response.

### Part (b)

Following is the equation of the least-squares regression line relating chest circumference and weight for male tule elk.

Predicted weight = -350.3 + 3.7455(chest circumference)

(b) The weight of one male tule elk with a chest circumference of 145.9 cm is 204.3 kg.

(i) Using the equation of the least-squares regression line, calculate the predicted weight for the male tule elk. Show your work.

(ii) Calculate the residual for this male tule elk. Show your work.

## WOULD THIS GET CREDIT?

Response 1:

(i) x = -350.3 + 3.7455(145.9) = 196.17 kg

(ii) residual = actual – predicted = 204.3 - 196.17 = 8.13 kg

Response 2:

(i) ŷ = -350.3 + 3.7455(145.9) = 196.17 kg

(ii) 196.17 – 204.3 = -8.13 kg

Response 3:

(i) 196.2 kg

(ii) 8.1 kg

Response 1 correctly calculates the predicted value but incorrectly labels the work as “x = ” (this should be y = ), which receives no credit for part (i). The residual value is correct with adequate work shown. Overall, this part (b) earns a score of P.

Response 2 correctly calculates the predicted value in part (i) with work shown but reverses the subtraction when calculating the residual, losing credit for part (ii). Overall, this part (b) earns a score of P.

Response 3 contains the correct values for both the predicted weight and the residual but does not include any work. Part (b) would score an I.

### Teaching Tips:

Always show work when making calculations.

Be mindful of variables used in your solutions. Practice defining variables any time they are included in a response.

Always include units in your final answers.

### Part (c)

The equation of the least-squares regression line relating chest circumference and weight for male tule elk is repeated here.

Predicted weight = -350.3 + 3.7455(chest circumference)

(c) Interpret the slope of the least-squares regression line in context.

## WOULD THIS GET CREDIT?

The response that received no credit has three mistakes, any of which would lower the score from an E to a P. First, there is no mention of an increase in the chest circumference. Second the increase in weight is written definitively rather than being described as a prediction or an average. And finally, there are no units, kg, for the weight.

### Teaching Tips:

Include the direction of change for both variables when interpreting the slope.

Discuss the least squares regression equation as a model of the points rather than the equation of the points. As a model, any predictions approximate the actual values.

Always include units of measurement for quantitative variables.

### Part (d)

The sambar, another species of deer, is similar in size to the tule elk. The slope of the population regression line relating chest circumference and weight for all male sambars is 4.5 kilograms per centimeter. A wildlife biologist wants to determine whether the slope of the population regression line for male tule elk is different than that for male sambars. Let β represent the slope of the population regression line for male tule elk. The wildlife biologist conducted a test of the following hypotheses using the sample of 30 tule elk.

Ho: β = 4.5

Ha: β ≠ 4.5

The test statistic was calculated to be 3.408. Assume all conditions for inference were met.

(i) Determine the p-value of the test.

(ii) At the significance level of α=0.05, what conclusion should the wildlife biologist make regarding the slope of the population regression line for male tule elk. Justify your response.

Response 1:

(i) p-value = 0.002

(ii) We have sufficient evidence that the slope of the lines for tule elk and sambar deer are different since 0.002 < 0.05.

Response 2:

(i) p-value = 0.04

(ii) Because the p-value < 0.05, we can conclude that the population slope for male tule elk is different than the slope for sambar deer.

Response 3:

(i) p-value = 0.001

(ii) Reject Ho. The data provide enough evidence to conclude that the slope of the equation relating chest circumference with weight for male tule elk is different than the slope for sambar deer.

Response 1 would score an essentially correct (E) due to the correct p-value, a correct decision based on a comparison of the p-value to the given alpha, and a non-definitive conclusion about Ha and in context.

Response 2 simply guesses a p-value which is incorrect. The conclusion is definitive (“we can conclude”) rather than allowing for the possibility of an error (“we have convincing evidence”) and so is incorrect. This response would be incorrect (I).

Response 3 fails to double the p-value for the 2-tailed test and does not support the conclusion with a comparison of the p-value and the given alpha. This response would be incorrect (I).

### Teaching Tips:

Practice determining a p-value given the test statistic for both one-tailed and two-tailed tests.

Practice writing a complete conclusion including an explicit comparison of the p-value to alpha, a non-definitive conclusion statement about the alternative hypothesis, and sufficient context.

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