AP Stats Chapter 3 Practice Test

A school guidance counselor examines how many extracurricular activities students participate in and their grade point average. The guidance counselor says, “The evidence indicates that the correlation between the number of extracurricular activities a student participates in and his or her grade point average is close to 0.” Which of the following is the most appropriate conclusion?

• Students with good grades tend to be students who are not involved in many extracurricular activities.

• Students involved in many extracurricular activities are just as likely to get good grades as bad grades.

• No conclusion should be made based on the correlation without looking at a scatterplot of the data.

• Students with good grades tend to be students who are involved in many extracurricular activities.

• Students involved in many extracurricular activities tend to be students with poor grades.

Answer explanation

Depending on whether the correlation is zero due to random scatter or due to a strong curved relationship. It is possible to have a zero correlation where there is a strong relationship, just not a linear one.

An AP®® Statistics student designs an experiment to see whether today’s high school students are becoming too calculator-dependent. She prepares two quizzes, both of which contain 40 questions that are best done using paper-and-pencil methods. A random sample of 30 students participates in the experiment. Each student takes both quizzes—one with a calculator and one without—in a random order. To analyze the data, the student constructs a scatterplot that displays a linear association between the number of correct answers with and without a calculator for the 30 students. A least-squares regression yields the equation𝐶𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑜𝑟ˆ=−1.2+0.865(𝑃𝑒𝑛𝑐𝑖𝑙)Calculator^=−1.2+0.865(Pencil) 𝑟=0.79.

• 1: If the student had used Calculator as the explanatory variable, the correlation would remain the same.

• 2: If the student had used Calculator as the explanatory variable, the slope of the least-squares line would remain the same.

• 3: The standard deviation of the number of correct answers on the paper-and-pencil quizzes was smaller than the standard deviation on the calculator quizzes.

• 1 and 3 only

• 1,2, and 3

Answer explanation

The slope of the least-squares line depends on which variable is the explanatory variable and which is the response.

Scientists examined the activity level of 7 fish at different temperatures. Fish activity was rated on a scale of 0 (no activity) to 100 (maximal activity). The temperature was measured in degrees Celsius. A computer regression printout and a residual plot are provided. Notice that the horizontal axis on the residual plot is labeled “Fitted value,” which means the same thing as “predicted value.” What is the correlation between temperature and fish activity?

• –0.95

• 0.91

• 0.45

• 0.95

• –0.91

Scientists examined the activity level of 7 fish at different temperatures. Fish activity was rated on a scale of 0 (no activity) to 100 (maximal activity). The temperature was measured in degrees Celsius. A computer regression printout and a residual plot are provided. Notice that the horizontal axis on the residual plot is labeled “Fitted value,” which means the same thing as “predicted value.” What was the actual activity level rating for the fish at a temperature of 20°C?

• 66

• 81

• 87

• 84

• 3

Answer explanation

The line predicts that a fish would have activity level 𝑦̂ =148.62−3.2167(20) = 84.286 Looking at the residual plot, the fish with a predicted activity level of about 84 has a residual of approximately +3.  

Scientists examined the activity level of 7 fish at different temperatures. Fish activity was rated on a scale of 0 (no activity) to 100 (maximal activity). The temperature was measured in degrees Celsius. A computer regression printout and a residual plot are provided. Notice that the horizontal axis on the residual plot is labeled “Fitted value,” which means the same thing as “predicted value.”Which of the following gives a correct interpretation of 𝑠s in this setting?

• The typical distance of the temperature readings from their mean is about 4.785°C.

• The typical distance of the activity level readings from their mean is about 4.785 units.

• At a temperature of 0°C, this model predicts an activity level of 4.785 units.

• For every 1°C increase in temperature, fish activity is predicted to increase by 4.785 units.

• The typical distance of the activity level ratings from the least-squares line is about 4.785 units.

Answer explanation

This is another way of saying the typical error between the actual values and the predicted values using the linear model.

Which of the following statements is not true of the correlation 𝑟 between the lengths (in inches) and weights (in pounds) of a sample of brook trout?

• 𝑟 is measured in inches

• 𝑟 would not change if we measured the weights of the trout in kilograms instead of pounds.

• 𝑟 would not change if we measured the lengths of the trout in centimeters instead of inches.

• 𝑟 must be a value between -1 and 1.

• If longer trout tend to also be heavier, then

  𝑟 > 0.

Answer explanation

Correlation does not have units attached to it.

When we standardize the values of a variable, the distribution of standardized values has mean 0 and standard deviation 1. Suppose we measure two variables 𝑋 and 𝑌 on each of several subjects. We standardize both variables and then compute the least-squares regression line. Suppose the slope of the least-squares regression line is –0.44.

• the correlation will also be –0.44.

• the intercept will be 1.0.

• The correlation will be 1/–0.44.

• the intercept will also be –0.44.

• the correlation will be 1.0.

There is a linear relationship between the number of chirps made by the striped ground cricket and the air temperature. A least-squares fit of some data collected by a biologist gives the model 𝑦̂ =25.2+3.3𝑥, where 𝑥 is the number of chirps per minute and 𝑦̂ is the estimated temperature in degrees Fahrenheit. What is the predicted increase in temperature for an increase of 5 chirps per minute?

• 3.3°F

• 25.2°F

• 41.7°F

• 28.5°F

• 16.5°F

Answer explanation

The slope gives the predicted increase in 𝑦 for each 1 unit increase in 𝑥. So,if 𝑥 increases 5 units, then the predicted value of 𝑦 increases 5 times the slope, or 5(3.3)=16.5°F.

The scatterplot shows the relationship between the number of people per television set and the number of people per physician for 40 countries, along with the least-squares regression line. In Ethiopia, there were 503 people per TV and 36,660 people per doctor. Which of the following is correct?

• Ethiopia has more people per doctor than expected, based on how many people it has per TV.

• Increasing the number of TVs in a country will attract more doctors.

• The correlation is greater than 1.

• The slope of the least-squares regression line is less than 1.

• The point for Ethiopia is decreasing the slope of the least-squares regression line.

Answer explanation

Ethiopia's residual is very large, indicating that there are many more people per doctor than is predicted by the least-squares regression line.

The first scatterplot shows the lean body mass and metabolic rate for a sample of 5 adults. For each person, the lean body mass is the subject’s total weight in kilograms less any weight due to fat. The metabolic rate is the number of calories burned in a 24-hour period. Because a person with no lean body mass should burn no calories, it makes sense to model the relationship with a direct variation function in the form 𝑦=𝑘𝑥. Models were tried using different values of 𝑘(𝑘=25,𝑘=26), and the sum of squared residuals (SSR) was calculated for each value of 𝑘. Given is a second scatterplot, this one showing the relationship between SSR and 𝑘: According to the scatterplot, what is the ideal value of 𝑘 to use for predicting metabolic rate?

• 24

• 26

• 25

• 31

• 36

Answer explanation

When 𝑘=31, the sum of the squared residuals is minimized.