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SAT Math: Extension Workshops / LEVEL 1 · DIFFICULTY 3/5

SAT: Statistical Evidence

Focused SAT extension practice.

2 stages · 16 practice problems · two 4-question assessment forms

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MINI QUEST SAT 1.1Estimates & Margin of ErrorRead the lesson
MINI QUEST SAT 1.2Sampling, Experiments & Causal ClaimsRead the lesson
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STAGE SAT 1.1 · 8 PRACTICE PROBLEMS · 4 NEW-CONTEXT APPLICATIONS

Estimates & Margin of Error

Useful preparation: Samples, Population Estimates & Limits

Goal: Build and interpret an estimate interval, and explain how sample size changes sampling uncertainty.

Before you begin: Sampling, percentages and data displays.

Same method. More observations.

A separate example: estimate $\displaystyle 50\%$ , starting margin 10 percentage points. Keep the confidence level and sampling method fixed.

Sample-size multiplier: 1×

$\displaystyle n = 100$ ; approximate margin=10 percentage points. Quadrupling n halves the margin. A biased sample can still give a narrow, misleading interval.

Understand the idea

A sample estimate is not the population value. An estimate p with margin m gives the interval $\displaystyle \left[p - m, p + m\right]$ . For a percentage, m is in percentage points, not a percent of p.

Choose and justify a method

At the same confidence level and with otherwise comparable random samples, the margin is approximately proportional to $\displaystyle \frac{1}{\sqrt{n}}$ . Four times as many observations roughly halves the margin. This approximation does not remove selection bias.

Check the conditions

$\displaystyle A\, 95\%$ confidence procedure captures the fixed population parameter in about $\displaystyle 95\%$ of repeated samples under its assumptions. It does not say that $\displaystyle 95\%$ of individuals lie inside this interval or guarantee that this particular interval contains the parameter.

WORKED EXAMPLE 1

A random-sample estimate $\displaystyle 52\%$ with a margin of error of 4 percentage points. What is the upper endpoint, as a percent? Enter the number without %.

Use estimate plus margin.
Both quantities are percentage points.
Upper endpoint = $\displaystyle 52 + 4$ .
This is $\displaystyle 56\%$ ; the interval is an estimate, not a guarantee.

WORKED EXAMPLE 2

A school wants to enlarge a random survey of 500 people. To reduce the margin to one third at the same confidence level, approximately how many people are needed?

A margin factor $\displaystyle \frac{1}{3}$ needs a square-root sample factor 3.
Square the factor to get the sample-size multiplier.
The sample size must be multiplied by $\displaystyle 3^{2} = 9$ .
$\displaystyle 9 \times 500 = 4500$ people.

4500

Common pitfalls

Possible mix-up: A margin of 3 means multiply the estimate by $\displaystyle 3\%$ .

Add and subtract 3 percentage points when the margin is stated that way.

Possible mix-up: A larger convenience sample eliminates bias.

Random sampling supports population inference; size alone does not repair selection bias.

Explain it to yourself

Explain why your method works in estimates & margin of error. State one condition it needs and describe what can go wrong without that condition.

Preview the eight practice prompts

Problem 1
A random-sample estimate $\displaystyle 50\%$ with a margin of error of 2 percentage points. What is the upper endpoint, as a percent? Enter the number without %.
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Problem 2
An estimate interval from $\displaystyle 30\%$ to $\displaystyle 40\%$ . What is its margin of error in percentage points?
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Problem 3
A study has a margin of 6 percentage points. With the same method and confidence level, it quadruples the random-sample size. Using the inverse-square-root approximation, what is the new margin?
Practise this exact problem →
Problem 4
A random sample contains 200 people. To reduce the margin to one third at the same confidence level, approximately how many people are needed?
Practise this exact problem →
Problem 5New context
A town survey estimates support at $\displaystyle 51\%$ with a margin of error of 3 percentage points. What is the upper endpoint, as a percent? Enter the number without %.
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Problem 6New context
A survey reports an interval for support from $\displaystyle 31\%$ to $\displaystyle 43\%$ . What is its margin of error in percentage points?
Practise this exact problem →
Problem 7New context
A polling team has a margin of 8 percentage points. With the same method and confidence level, it quadruples the random-sample size. Using the inverse-square-root approximation, what is the new margin?
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Problem 8New context
A school wants to enlarge a random survey of 300 people. To reduce the margin to one third at the same confidence level, approximately how many people are needed?
Practise this exact problem →

Open stage SAT 1.1 in the student workspace →

STAGE SAT 1.2 · 8 PRACTICE PROBLEMS · 4 NEW-CONTEXT APPLICATIONS

Sampling, Experiments & Causal Claims

Useful preparation: Estimates & Margin of Error

Goal: Distinguish who was sampled from how treatments were assigned, and identify the conclusions a design supports.

Before you begin: Sampling, percentages and data displays.

Understand the idea

Random sampling selects participants from a target population. It helps justify generalizing to that population, provided nonresponse and measurement problems are controlled.

Choose and justify a method

Random assignment divides participants into treatment groups. It helps balance confounders and supports a causal comparison under a well-conducted experiment. Volunteers can be randomly assigned without being a random population sample.

Check the conditions

An observational association alone does not establish cause. Ask two separate questions: Were people randomly sampled? Were treatments randomly assigned? Neither a large sample nor a small margin of error replaces these design requirements.

WORKED EXAMPLE 1

A survey randomly samples 140 city residents and records reading time and grades without assigning treatments. Which conclusion type is supported? Enter 1 for population association only, 2 for causation only, 3 for both, 4 for neither.

Separate sampling from treatment assignment.
The residents are randomly sampled but no treatment is assigned.
Random sampling supports estimating an association for city residents.
Without an experiment, that association alone cannot establish a causal effect. Choose 1.

WORKED EXAMPLE 2

A school website poll receives 360 voluntary responses about study time and scores, with no treatment assigned. Enter 1 for justified population generalization only, 2 for causation only, 3 for both, 4 for neither.

Voluntary response is not random sampling.
There is also no random treatment assignment.
Respondents may differ from people who did not respond.
Neither broad population generalization nor a causal conclusion is justified by this design alone: 4.

Common pitfalls

Possible mix-up: Random assignment makes volunteers representative of everyone.

Assignment supports a treatment comparison; sampling determines the population to which it can be generalized.

Possible mix-up: An association proves one variable caused the other.

Observational associations may have confounders and do not by themselves establish causation.

Explain it to yourself

Explain why your method works in sampling, experiments & causal claims. State one condition it needs and describe what can go wrong without that condition.

Preview the eight practice prompts

Problem 1
A survey randomly samples 120 city residents and records reading time and grades without assigning treatments. Which conclusion type is supported? Enter 1 for population association only, 2 for causation only, 3 for both, 4 for neither.
Practise this exact problem →
Problem 2
An experiment randomly assigns 60 volunteers to two study methods. Volunteers were not randomly sampled. Enter 1 for population generalization only, 2 for causal comparison among participants only, 3 for both, 4 for neither.
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Problem 3
A study randomly samples 200 students from its district and randomly assigns a tutoring method, with complete follow-up. Enter 1 for generalization only, 2 for causal comparison only, 3 for both, 4 for neither.
Practise this exact problem →
Problem 4
A poll receives 300 voluntary responses about study time and scores, with no treatment assigned. Enter 1 for justified population generalization only, 2 for causation only, 3 for both, 4 for neither.
Practise this exact problem →
Problem 5New context
A library survey randomly samples 130 city residents and records reading time and grades without assigning treatments. Which conclusion type is supported? Enter 1 for population association only, 2 for causation only, 3 for both, 4 for neither.
Practise this exact problem →
Problem 6New context
A learning-app trial randomly assigns 64 volunteers to two study methods. Volunteers were not randomly sampled. Enter 1 for population generalization only, 2 for causal comparison among participants only, 3 for both, 4 for neither.
Practise this exact problem →
Problem 7New context
A district study randomly samples 210 students from its district and randomly assigns a tutoring method, with complete follow-up. Enter 1 for generalization only, 2 for causal comparison only, 3 for both, 4 for neither.
Practise this exact problem →
Problem 8New context
A school website poll receives 320 voluntary responses about study time and scores, with no treatment assigned. Enter 1 for justified population generalization only, 2 for causation only, 3 for both, 4 for neither.
Practise this exact problem →

Open stage SAT 1.2 in the student workspace →

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SAT: Statistical Evidence

Estimates & Margin of Error

Understand the idea

Choose and justify a method

Check the conditions

A random-sample estimate 52%\displaystyle 52\%52% with a margin of error of 4 percentage points. What is the upper endpoint, as a percent? Enter the number without %.

A school wants to enlarge a random survey of 500 people. To reduce the margin to one third at the same confidence level, approximately how many people are needed?

Common pitfalls

Explain it to yourself

Sampling, Experiments & Causal Claims

Understand the idea

Choose and justify a method

Check the conditions

A survey randomly samples 140 city residents and records reading time and grades without assigning treatments. Which conclusion type is supported? Enter 1 for population association only, 2 for causation only, 3 for both, 4 for neither.

A school website poll receives 360 voluntary responses about study time and scores, with no treatment assigned. Enter 1 for justified population generalization only, 2 for causation only, 3 for both, 4 for neither.

Common pitfalls

Explain it to yourself

A random-sample estimate $\displaystyle 52\%$ with a margin of error of 4 percentage points. What is the upper endpoint, as a percent? Enter the number without %.