# Digital SAT Math : Inference from sample statistics and margin of error -Practice Questions

## 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   Easy

The tables give the heights, in feet, of 5 peaks in the Rocky Mountains and 5 peaks in the Appalachian Mountains.

What is the height, in meters, of Blanca Peak?
(Use 1 meter $$=3.28$$ feet.)
A. 437.5
B. 4,375
C. 47,045
D. 47,068

Ans:B

To find the height of Blanca Peak in meters, we need to convert its height from feet to meters. Since $$1$$ meter is equal to $$3.28$$ feet, we can use this conversion factor.

Blanca Peak’s height in feet is $$14,350$$. To convert this to meters, we divide by $$3.28$$:

$14,350 \text{ feet} \div 3.28 = 4,375 \text{ meters}$

Rounding to the nearest tenth, the height of Blanca Peak in meters is $$4,375$$ meters.

[Calc]  Question   Easy

The tables give the heights, in feet, of 5 peaks in the Rocky Mountains and 5 peaks in the Appalachian Mountains.

For the given Appalachian Mountain peaks, the height of the highest peak is approximately what percent greater than the height of the lowest peak?
A. $$1.1 \%$$
B. $$9.9 \%$$
C. $$73.0 \%$$
D. $$101.1 \%$$

Ans:A

To find the percentage difference between the highest peak (Mount Mitchell) and the lowest peak (Balsam Cone) in the Appalachian Mountains, we can use the formula:

$\text{Percentage difference} = \frac{{\text{Difference in heights}}}{{\text{Height of lowest peak}}} \times 100\%$

The difference in heights is $$6,684 – 6,611 = 73$$ feet.

So, the percentage difference is:

$\frac{{73}}{{6,611}} \times 100\%$

$\approx 1.1\%$

Therefore, the correct answer is option A: $$1.1\%$$.

[Calc]  Question  Easy

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%.

Based on the results of the survey, how many of the students surveyed agree with the proposed change?

A. 19
B. 31
C. 124
D. 155

Ans: B

Given information:

• The survey was conducted on a random sample of 50 students from a total of 250 students taking the driver’s education class.
• The survey reported that 62% of the surveyed students agree with increasing the driving practice time.
• The associated margin of error is 5%.

To find the number of students who agree with the proposed change, we need to calculate 62% of the sample size (50 students).

62% of 50 students $= 0.62 × 50 = 31$ students

Therefore, the number of students surveyed who agree with the proposed change is 31.

Among the given options, the correct answer is B. 31.

Questions

A scientist conducted an experiment and selected a random sample of runners from a list of all high school track participants from a certain city. The scientist randomly assigned each runner to one of two treatment groups, and the results of the experiment were found to be statistically significant. To which of the following populations can the results of the experiment be safely generalized?
A. All high school athletes
B. All high school track participants from the city
C. All high school track participants from the country
D. All runners

Ans: B

Questions

. A city with 120,000 residents is voting on a proposal that would eliminate overnight parking of vehicles on the city’s streets. An independent company randomly surveys 1,200 residents to see whether or not residents would support this proposal. The outcome of the survey shows that $60 \%$ of the residents surveyed approve of the proposal with a margin of error of $2 \%$. Which of the following statements is a plausible conclusion from the outcome of the survey?
A. Exactly $60 \%$ of city residents approve eliminating overnight parking.
B. There are 72,000 city residents who approve eliminating overnight parking.
C. About $2 \%$ of the city residents do not approve eliminating overnight parking.
D. Between $58 \%$ and $62 \%$ of the city residents approve eliminating overnight parking.

Ans: D

Question

An analysis of a random sample of a type of laptop computer battery estimated that the mean working time was 4.7 hours with a margin of error of 0.7 hours. Which of the following is the most appropriate conclusion based on this analysis?
A. This type of laptop computer battery has a mean working time of at least 4.7 hours.
B. This type of laptop computer battery has a mean working time of at least 5.7 hours.
C. This type of laptop computer battery has a mean working time of between 4.0 and 5.4 hours.
D. This type of laptop computer battery has a mean working time of between 0.0 and 0.7 hours.

Ans: C

Questions

One hundred park-district members will be selected to participate in a survey about selecting a new park-district coordinator. Which of the following methods of choosing the 100 members would result in a random sample of members of the park district?

A. Obtain a numbered list of all park-district members. Use a random number generator to select 100 members from the list. Give the survey to those 100 members.
B. Obtain a list of all park-district members sorted alphabetically. Give the survey to the first 100 members on the list.
C. Tell all park-district members that volunteers are needed to take the survey. Give the survey to the first 100 members who volunteer.
D. Obtain a list of all park-district members who are attending an upcoming event. Give the survey to the first 100 members on the list.