Question 17·50 Super-Hard SAT Problem Solving and Data Analysis Questions·Problem Solving and Data Analysis

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A school district has 6,000 high school students and 3,000 middle school students. A random sample of high school students indicates that $18\%$ of high school students participate in an after-school arts program, with a margin of error of $4$ percentage points. Independently, a random sample of middle school students indicates that $10\%$ of middle school students participate in the program, with a margin of error of $4$ percentage points.

Which choice lists all and only the appropriate conclusions?

I. It is plausible that $22\%$ of all high school students in the district participate in the program.

II. Based on the two samples, it is appropriate to estimate that $14\%$ of all students in the district participate in the program.

III. It is plausible that $19\%$ of all students in the district participate in the program.

Separate samples with a margin of error give a plausible range for each group's percentage. To estimate a percentage for everyone, multiply each group's rate by its size, add the resulting counts, and divide by the total; Desmos can evaluate that expression. With unequal subgroup sizes, averaging the two reported percentages gives the smaller group too much weight.

Hints

Hint 1
Percentage points are added directly to a percentage estimate. A margin of error gives a range from the estimate minus the margin to the estimate plus the margin. Where does the high-school claim fall?
Hint 2
For a combined percentage, each group's rate applies to that group's student count. Estimate the number of participants in each group, add those counts, and divide by the total number of students.
Hint 3
To test whether the district-wide percentage is plausible, combine the upper ends of the two groups' ranges. Can a percentage between the district estimate and that upper end occur?

Step-by-step

Weight the group percentages
Step 1
Check the high-school claim
A margin of error of $4$ percentage points means adding or subtracting $4$ points, not taking $4\%$ of the estimate. Subtract for the low end:
$18\%-4\%=14\%$
Add for the high end:
$18\%+4\%=22\%$
So I is plausible: $22\%$ is the upper end of the high-school range.
Step 2
Find the middle-school range
Apply the same margin to the middle-school estimate. Subtract for the low end:
$10\%-4\%=6\%$
Add for the high end:
$10\%+4\%=14\%$
So a middle-school rate from $6\%$ to $14\%$ is plausible.
Step 3
Build the district estimate
The groups have different sizes, so estimate participants in each group before dividing by all students:
$\frac{(18\%)(6000)+(10\%)(3000)}{6000+3000}$
Each product estimates one group's participants. Weight each group's percentage by its own student count. Averaging $18\%$ and $10\%$ would treat the two groups as equally large.
Step 4
Evaluate the district estimate
Type $\frac{18\%\operatorname{of}6000+10\%\operatorname{of}3000}{6000+3000}$ in Desmos. It prints $0.15333\ldots$ , or $15\frac{1}{3}\%$ . So $14\%$ is not the district estimate in II. It could be plausible, but II asks for the estimate.
Step 5
Build the highest plausible district percentage
For III, use both upper endpoints: a higher rate in either group raises the district percentage. That gives
$\frac{(22\%)(6000)+(14\%)(3000)}{6000+3000}$
Both endpoints come from the separate samples' margins of error.
Step 6
Check the proposed district percentage
Add $\frac{22\%\operatorname{of}6000+14\%\operatorname{of}3000}{6000+3000}$ in Desmos. It prints $0.19333\ldots$ , or $19\frac{1}{3}\%$ . The proposed $19\%$ lies between the district estimate of $15\frac{1}{3}\%$ and this upper end. Rates within the groups' plausible ranges can produce it. So I and III are appropriate, but II is not. Choice B.

Question 17·50 Super-Hard SAT Problem Solving and Data Analysis Questions·Problem Solving and Data Analysis

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