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Digital SAT Math Practice Questions – Advanced : Inference from sample statistics and margin of error

Digital SAT Math Practice Questions - Advanced : Inference from sample statistics and margin of error - New Syllabus

DSAT 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

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DSAT MAth and English  – full syllabus practice tests

Question Hard

Micha and Rana each selected a random sample of students at their school and asked how many soft drink servings each student had consumed the previous week. Micha estimated that the mean number of soft drink servings was 7.1, with an associated margin of error of 1.2. Rana estimated that the mean number of soft drink servings was 8.3, with an associated margin of error of 0.8. Assuming the margins of error were calculated in the same way, which of the following best explains why Rana obtained a smaller margin of error than Micha?

A) Rana’s sample contained more students than Micha’s sample contained.

B) Rana’s sample contained more students who drink soft drinks than Micha’s sample contained.

C) Rana’s sample contained more students who drank exactly seven soft drink servings than Micha’s sample contained.

D) Rana’s sample contained more students who drank exactly eight soft drink servings than Micha’s sample contained.

▶️ Answer/Explanation
Solution

Ans: A

Margin of error = \( z \times \frac{\sigma}{\sqrt{n}} \). Same method implies difference in \( n \).

Larger \( n \) reduces margin of error.

Option A: Rana’s sample contained more students (correct).

Option B: More drinkers irrelevant to margin (incorrect).

Option C: Seven servings irrelevant to margin (incorrect).

Option D: Eight servings irrelevant to margin (incorrect).

Question Hard

Two different store owners in a shopping center estimated the percentage of all visitors who wear eyeglasses. They each selected a random sample of the shopping center visitors and recorded whether the visitors were wearing eyeglasses. The results from each sample are shown in the table below.

 Percentage of visitors wearing eyeglassesMargin of error
Sample A21%3%
Sample B21%2%

If the associated margin of error was calculated the same way for both samples, which of the following is the most likely reason that the result for Sample A has a larger margin of error?

A) Sample A included more visitors than Sample B.

B) Sample B included more visitors than Sample A.

C) Sample A included a greater percentage of visitors who were wearing eyeglasses than Sample B.

D) Sample B included a greater percentage of visitors who were wearing eyeglasses than Sample A.

▶️ Answer/Explanation
Solution

Ans: B

Margin of error decreases with larger sample size. Same percentage (21%) but different margins (3% vs. 2%) suggest Sample A had fewer visitors.

Question Hard

\( \frac{2}{x-2} + \frac{3}{x+5} = \frac{r x + t}{(x-2)(x+5)} \)

The equation above is true for all \( x > 2 \), where \( r \) and \( t \) are positive constants. What is the value of \( r t \) ?

A) -20

B) 15

C) 20

D) 60

▶️ Answer/Explanation
Solution

Ans: C

Combine left side over common denominator \( (x-2)(x+5) \):

\( \frac{2(x+5) + 3(x-2)}{(x-2)(x+5)} = \frac{2x + 10 + 3x – 6}{(x-2)(x+5)} = \frac{5x + 4}{(x-2)(x+5)} \)

So, \( r x + t = 5x + 4 \), thus \( r = 5 \), \( t = 4 \).

\( r t = 5 \times 4 = 20 \)

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