Super-Hard SAT Math Question 102 | Free SAT Prep | Aniko

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Question 102·200 Super-Hard SAT Math Questions·Problem Solving and Data Analysis

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Sample	Percent in favor	Margin of error
A	64.2%	5.1%
B	60.8%	3.4%

Two random samples were selected from the same population, and the margins of error were calculated using the same method.

Suppose the margin of error (MOE) for a random sample is proportional to

\sqrt{\frac{p(1-p)}{n}},

where $p$ is the sample proportion in favor (as a decimal) and $n$ is the sample size.

Approximately how many times as large was the sample size of sample B as the sample size of sample A? A‍nik⁠ο ‍Quеst‍iοn Вanк

When a margin of error is given as a proportional relationship, first solve symbolically for how $n$ scales (here, $n \propto \frac{p(1-p)}{\text{MOE}^2}$ ). Then take a ratio between the two samples so the constant cancels, and compute the ratio carefully—watch for traps like forgetting to square the MOE ratio or ignoring other factors like $p(1-p)$ when they are part of the stated relationship. Aniкo - Frеe Ѕ⁠AT⁠ Рre‍р

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Start by introducing a constant of proportionality

Write $\text{MOE}=k\sqrt{\frac{p(1-p)}{n}}$ and remember $k$ is the same for both samples.

Solve for how $n$ depends on $p$ and MOE

Square the equation and rearrange to express $n$ in terms of $p(1-p)$ and $\text{MOE}^2$ .

Use a ratio to cancel the constant

Form $\frac{n_B}{n_A}$ so that the constant $k$ disappears, leaving an expression involving $p_A,p_B,\text{MOE}_A,\text{MOE}_B$ only. a⁠nikо.аі

Desmos Guide

Compute each $p(1-p)$ term

In Desmos, compute 0.642*(1-0.642) and 0.608*(1-0.608). Anі⁠ko АI Tutor

Build the ratio expression

Enter (0.608*(1-0.608))/(0.642*(1-0.642))*(5.1/3.4)^2 to represent $\frac{n_B}{n_A}$ under the given proportionality.

Interpret the output and match a choice

Interpret the value as “sample B is about that many times as large as sample A,” and choose the closest option.

Step-by-step Explanation

Rewrite the proportionality and solve for how $n$ scales

If $\text{MOE} \propto \sqrt{\frac{p(1-p)}{n}}$ , we can write

\text{MOE} = k\sqrt{\frac{p(1-p)}{n}}

for some constant $k$ (the same method means the same $k$ ). © an‌ikο.аi

Square both sides:

\text{MOE}^2 = k^2\frac{p(1-p)}{n}.

Rearrange to see how $n$ depends on $p$ and MOE:

n = k^2\frac{p(1-p)}{\text{MOE}^2} \quad\Rightarrow\quad n \propto \frac{p(1-p)}{\text{MOE}^2}.

Set up the ratio $\frac{n_B}{n_A}$ so $k$ cancels

Using $n \propto \frac{p(1-p)}{\text{MOE}^2}$ ,

\frac{n_B}{n_A} = \frac{\frac{p_B(1-p_B)}{\text{MOE}_B^2}}{\frac{p_A(1-p_A)}{\text{MOE}_A^2}} = \frac{p_B(1-p_B)}{p_A(1-p_A)}\left(\frac{\text{MOE}_A}{\text{MOE}_B}\right)^2.

Compute the two factors

Convert the percents to decimals:

$p_A=0.642\Rightarrow p_A(1-p_A)=0.642(0.358)\approx 0.229836$
$p_B=0.608\Rightarrow p_B(1-p_B)=0.608(0.392)\approx 0.238336$

\frac{p_B(1-p_B)}{p_A(1-p_A)}\approx \frac{0.238336}{0.229836}\approx 1.037.

For the MOE factor (using the given percent values consistently):

\left(\frac{\text{MOE}_A}{\text{MOE}_B}\right)^2 = \left(\frac{5.1}{3.4}\right)^2 = (1.5)^2.

Multiply and select the closest choice

Now multiply the factors:

\frac{n_B}{n_A}\approx (1.037)(2.25)\approx 2.33.

So sample B’s sample size was about $2.33$ times sample A’s sample size.

Therefore, the correct choice is About 2.33 times as large.

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Step-by-step Explanation

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Step-by-step Explanation