# Digital SAT Math Practice Questions – Medium : Evaluating statistical claims: observational studies and Experiments

## SAT MAth Practice questions – all topics

• Problem-solving and Data Analysis Weightage: 15%  Questions: 5-7
• Ratios, rates, proportional relationships, and units
• Percentages
• One-variable data: distributions and measures of centre and spread
• Two-variable data: models and scatterplots
• Probability and conditional probability
• Inference from sample statistics and margin of error
• Evaluating statistical claims: observational studies and Experiments

## SAT MAth and English  – full syllabus practice tests

[Calc]  Question   medium

For the elements in the periodic table having atomic numbers 1 though 18, the bar graph summarizes the number of elements by their number of valence electrons. How many of these elements have at most 4 valence electrons?

Ans:10

To find the elements with at most 4 valence electrons, we count the elements with 1, 2, 3, or 4 valence electrons:

• 1 valence electron: Hydrogen (H), Lithium (Li), Sodium (Na)
• 2 valence electrons: Helium (He), Beryllium (Be), Magnesium (Mg)
• 3 valence electrons: Boron (B), Aluminum (Al)
• 4 valence electrons: Carbon (C), Silicon (Si)

Summing these:

• Elements with 1 valence electron: 3 (H, Li, Na)
• Elements with 2 valence electrons: 3 (He, Be, Mg)
• Elements with 3 valence electrons: 2 (B, Al)
• Elements with 4 valence electrons: 2 (C, Si)

Adding these counts together, we get:

$3+3+2+2=10$

Therefore, 10 elements have at most 4 valence electrons.

[Calc]  Question  medium

A state representative wants to increase the amount of driving practice time required before a student can earn a driver’s license. The representative surveyed a random sample of 50 from the 250 students taking the driver’s education class at Jefferson High School.

The survey reported that 62% of students taking the class agree that driving practice time should be increased, with an associated margin of error of 5%.

Which of the following populations can the results of the survey at Jefferson High School be extended to?
A. Students at any US high school
B. Students taking a driver’s education class at any US high school
C. Students at Jefferson High School
D. Students taking a driver’s education class at Jefferson High School

Ans: D

Given information:

• The survey was conducted on a random sample of 50 students from the population of 250 students taking the driver’s education class at Jefferson High School.

Based on this information, the results of the survey can only be reliably extended to the population from which the sample was drawn, which is:D.

The survey results cannot be generalized to other populations, such as:

A. Students at any US high school (because the sample was specific to Jefferson High School) B. Students taking a driver’s education class at any US high school (because the sample was specific to Jefferson High School) C. Students at Jefferson High School (because the sample was specific to the driver’s education class)

To extend the results to a larger population, such as students taking driver’s education classes at other high schools or all high school students, a different sampling method that ensures representativeness of the larger population would be required.

[Calc]  Question  medium

In 1980, the world population was 4.44 billion people. A model created at the time predicted that this population would grow at a rate of 0.076 billion people each year after 1980. In 2015, the world population was 7.35 billion people. What is the positive difference, in billions of people, between the actual world population in 2015 and the model’s predicted world population in 2015?

Ans: 1/4, .25

To find the model’s predicted world population in 2015, we need to use the model that predicts the population growth rate.

Given:
World population in 1980: $$4.44$$ billion people
Population growth rate: $$0.076$$ billion people each year after 1980

calculate the predicted population in 2015 using the model:

$\text{Predicted population in 2015} = 4.44 + (2015 – 1980) \times 0.076$

$= 4.44 + 35 \times 0.076$

$= 4.44 + 2.66$

$= 7.10 \text{ billion people}$

The actual world population in 2015 was $$7.35$$ billion people.

Now,  difference between the actual and predicted populations in 2015:

$\text{Difference} = \text{Actual population} – \text{Predicted population}$

$= 7.35 – 7.10$

$= 0.25 \text{ billion people}$

So, the positive difference between the actual world population in 2015 and the model’s predicted world population in 2015 is $$0.25$$ billion people.

[Calc]  Question  Medium

The dot plot shows the estimated market values for 15 houses in a neighborhood. What is the maximum estimated market value, in thousands of dollars, for the data set?

Ans: 178

The dot plot shows the estimated market values for 15 houses in a neighborhood. To determine the maximum estimated market value, we need to identify the highest value on the horizontal axis that has at least one dot above it.

From the provided dot plot, the horizontal axis is labeled with market values in thousands of dollars, ranging from 168 to 178. The maximum estimated market value is represented by the rightmost dot on the plot.

The highest value on the plot where dots are present is at 178 thousand dollars.

Therefore, the maximum estimated market value for the data set is 178 thousand dollars.

[Calc]  Question   Medium

The bar graph shows the number of discovered near-Earth asteroids, by classification, as of April 2017 Of the near-Earth asteroids, how many more are classified as medium, large, or very large than are classified as very small or small?

Ans:1422

From the bar graph, the numbers are:
Very Small: 2911
Small: 4406
Medium: 4055
Large: 3808
Very Large: 876

First, sum the numbers for medium, large, and very large asteroids:
$4055 + 3808 + 876 = 8739$

Next, sum the numbers for very small and small asteroids:
$2911 + 4406 = 7317$

Now, find the difference between these two sums:
$8739 – 7317 = 1422$

Therefore, there are 1422 more near-Earth asteroids classified as medium, large, or very large than are classified as very small or small.

[Calc]  Question   medium

What is the median of the 25 data values represented in the dot plot above?

9

Based on the dot plot shown in the image, the median of the 25 data values appears to be 9. In a dot plot, each dot represents one data point, and the data values are arranged along the horizontal axis from lowest to highest. By counting the dots, We can see there are 25 total data points concentrated primarily around the values 6 through 12. Since the median is the middle value when the data is arranged in numerical order, and there is a prominent cluster of dots at the value 9, the median of this dataset is 9

$2,3,5,5,5,6,6,7,8,8,8,8,9,9,9,9,10,10,10,11,12,13,13,13,14$

This data set contains 25 values. Since 25 is an odd number, the median will be the value at the $$\left(\frac{25+1}{2}\right)$$ th position, which is the 13 th position.

Let’s identify the 13 th value in the ordered data set:
$2,3,5,5,5,6,6,7,8,8,8,8, \underline{9}, 9,9,9,10,10,10,11,12,13,13,13,14$

The 13 th value is 9 .

[Calc]  Question  Medium

A researcher estimates that there is a population of 618 gray wolves in the Upper Peninstula of Michigan, which covers an area of approximately 16,452 square miles. Which of the following is closest to the estimated population density, in gray wolves per square mile, in this area?
A) 0.04
B) 3.76
C) 26.62
D) 53.24

A

$\text{Population Density} = \frac{\text{Number of gray wolves}}{\text{Area in square miles}}$

Given:
Estimated population of gray wolves: $$618$$
Area of the Upper Peninsula: $$16,452$$ square miles

The population density calculation is:

$\text{Population Density} = \frac{618}{16,452}$

$\text{Population Density} = \frac{618}{16,452} \approx 0.0376$

$\text{Population Density} \approx 0.04$

Therefore, the closest estimated population density, in gray wolves per square mile, is $$\boxed{0.04}$$.

[Calc]  Question  medium

For a survey, students were assigned to either group R or group V. Combined, the students in both groups answered a total of 17 questions. Of these, a total of 9 questions were answered by the students in group V. The equation 4r + 9 = 17 describes this situation, where r represents the number of questions answered by each student in group R. Which of the following is the best interpretation of 4r in this context?

A)The number of students in group R

B)The number of students in group V

C)The total number of questions answered by students in group R

D)The total number of questions answered by students in group V

C)The total number of questions answered by students in group R

The equation $$4r + 9 = 17$$ describes the total number of questions answered by students in groups $$R$$ and $$V$$, where $$r$$ represents the number of questions answered by each student in group $$R$$.

First, we solve for $$r$$:
$4r + 9 = 17$
$4r = 17 – 9$
$4r = 8$
$r = 2$

Now, we interpret $$4r$$. Given $$r$$ represents the number of questions answered by each student in group $$R$$, and there are $$4r$$ total questions answered by all students in group $$R$$, $$4r$$ is the total number of questions answered by students in group $$R$$.

Therefore, the best interpretation of $$4r$$ is:

C) The total number of questions answered by students in group $$R$$

[Calc]  Question   Medium

The effectiveness of a mineral supplement in the soil on the growth of a particular species of plant is being studied. A botanist planted 1,000 seeds in a greenhouse so that the growing conditions for all seeds would be as identical as possible. The seeds were obtained from two 500 -seed packages. The seeds from one package were planted in soil that had the supplement added, and the seeds from the second package were planted in soil that did not have the supplement added. How should the experiment be changed to allow the researcher to conclude whether the supplement has an effect on plant growth?
A) One of the packages of seeds should be planted outdoors rather than in a greenhouse.
B) Half of the seeds from each package should be randomly assigned to each soil type.
C) All 1,000 seeds should receive the supplement.
D) No changes to the experiment are needed.

B

To conclude whether the mineral supplement has an effect on plant growth, the experiment should be designed to compare the growth of seeds planted with the supplement to those planted without it, while controlling for other variables.

Option C) All 1,000 seeds should receive the supplement, is not ideal because it does not provide a comparison group. Instead, the experiment should involve a controlled group (seeds without supplement) and a treatment group (seeds with supplement).

Option D) No changes to the experiment are needed, is incorrect because the current setup lacks a proper control group.

Therefore, the most appropriate change is:
$\boxed{B) \, \text{Half of the seeds from each package should be randomly assigned to each soil type.}}$

[Calc]  Question  medium

The box plots shown summarize the data in each of four data sets. Which of the four data sets has a range of 6 ?

A)Data set A

B)Data set B

C)Data set C

D)Data set D

D)Data set D

To determine which data set has a range of 6, We need to look at the minimum and maximum values represented by the box plots for each data set.

Data set A: The box spans from around 3 to around 5, with the whiskers extending a bit further in each direction. The range appears to be less than 6.

Data set B: The box spans from around 4 to around 8, with short whiskers on each end. The range is likely around 8 or more, greater than 6.

Data set C: The box spans from around 2 to around 8, with longer whiskers extending the range. Based on the spacing, the full range greater than 6.

Data set D: The box spans from around 4 to around 6, with whiskers extending a couple units further in each direction. The range appears to be 4 or 6.

Based on the visual representation, Data set C seems most likely to have an exact range of 6 across its data points.

Therefore, the answer is D) Data set D.

[No- Calc]  Question  Medium

The function $$A(t)=12(2)^{\frac{t}{6}}$$ models the number of water hyacinths in a population over time, where A(t) is the number of water hyacinths and t is the time, in days, since the population was first measured. Which is the best interpretation of $$(2)^{\frac{t}{6}}$$ in this context?

A) The number of water hyacinths doubled t times.

B) The number of water hyacinths doubled every 6 days.

C) The number of water hyacinths increased by 2 every  t/6 days.

D) The number of water hyacinths increased by 2 every t days.

Ans: B

In the function $$A(t) = 12(2)^{\frac{t}{6}}$$, the expression $$(2)^{\frac{t}{6}}$$ represents the growth factor for the number of water hyacinths over time. Let’s interpret it in the context of the problem:

A) The number of water hyacinths doubled $$t$$ times. – This interpretation suggests exponential growth where the number of water hyacinths doubles repeatedly over time. However, this doesn’t match the structure of the function $$A(t)$$ because the base of the exponential function is $$2$$, not $$2t$$.

B) The number of water hyacinths doubled every $$6$$ days. – This interpretation fits with the exponent $$\frac{t}{6}$$ in the context of time measured in days. The base of the exponential function is $$2$$, indicating that the population doubles every $$6$$ days.

C) The number of water hyacinths increased by $$2$$ every $$\frac{t}{6}$$ days. – This interpretation doesn’t match the structure of the function. The base of the exponential function is $$2$$, not $$2 + \frac{t}{6}$$.

D) The number of water hyacinths increased by $$2$$ every $$t$$ days. – This interpretation doesn’t match the structure of the function. The base of the exponential function is $$2$$, not $$2t$$.

So, the best interpretation of $$(2)^{\frac{t}{6}}$$ in this context is B) The number of water hyacinths doubled every $$6$$ days.

[Calc]  Question  Medium

The Metropolitan Museum of Art has plates on display from the Roman Empire and ancient Greece. The box plots shown summarize the distributions of the diameters, in centimeters, of all the museum’s
plates from each region. How does the median diameter of the plates from the Roman Empire, r, compare to the median diameter of the plates from ancient Greece, g ?
A) r < g
B) r> g
C) r = g
D) There is not enough information to compare the medians.

Ans: A

To compare the median diameters of the plates from the Roman Empire (r) and ancient Greece (g) using the box plots:

1. Identify the median:
• The median is represented by the line inside the box of each box plot.
• For the Roman Empire, the median appears to be at around 20 cm.
• For ancient Greece, the median appears to be at around 25 cm.

By comparing these medians:

• The median diameter of the plates from the Roman Empire (r) is less than the median diameter of the plates from ancient Greece (g).

Therefore, the correct answer is:

A) r < g

Question

The box plots above show the distributions of two data sets. Which of the following statements
must be true? 3.9

1. The minimum value of numbers in data set A is greater than the minimum value of numbers in data set B.
2. The range of data set A is greater than the range of data set B.
3. Data set A has more data values than data set B has.
4. The median of data set A is greater than the median of data set B.

D

Question

The line graph shows the wind speed recorded every 2 hours in a town, where $$x$$ is the number of hours after midnight and $$y$$ is the wind speed, in miles per hour (mph). What is the greatest average change in wind speed, in miles per hour per hour, between two readings shown in the graph?

5

Question

.

Data sets A and B are summarized in the graphs above. Each data set consists of 12 whole numbers. Which of the following statements must be true?

1. Data sets A and B have the same mean, but the standard deviation of data set A is greater than the standard deviation of data set B.
2. Data sets A and B have the same mean, but the standard deviation of data set B is greater than the standard deviation of data set A.
3. Data sets A and B have the same standard deviation, but the mean of data set A is greater than the mean of data set B.
4. Data sets A and B have the same standard deviation, but the mean of data set B is greater than the mean of data set A.

B

Questions

WA sample of 40 fourth-grade students was selected at random from a certain school. The 40 students completed a survey about the morning announcements, and 32 thought the announcements were helpful. Which of the following is the largest population to which the results of the survey can be applied?
A. The 40 students who were surveyed
B. All fourth-grade students at the school
C. All students at the school
D. All fourth-grade students in the county in which the school is located

Ans: B

Questions

$2,000-61 k=48$

In 1962, the population of a bird species was 2,000. The population $k$ years after 1962 was 48 , and $k$ satisfies the equation above. Which of the following is the best interpretation of the number 61 in this context?
A. The population $k$ years after 1962
B. The value of $k$ when the population was 48
C. The difference between the population in 1962 and the population $k$ years after 1962
D. The average decrease in the population per year from 1962 to $k$ years after 1962

Ans: D

Questions

A sample of 600 ninth graders was selected at random and asked how much time they spend on homework each day. Of the ninth graders selected, 220 spend less than 2 hours on homework each day. If the conclusion was drawn that “approximately 1.35 million ninth graders spend less than 2 hours on homework each day,” which of the following is closest to the population, in millions, of ninth graders?
A. 0.495
B. 1.35
C. 3.68
D. 5.84