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Mathematics 9709/72 — May/June 2017

Cambridge A-Level · Probability & Statistics 2 · worked solutions for every part, with the mark scheme

7
questions
50
marks
75
minutes

Topics Sampling and Estimation · The Poisson Distribution · Hypothesis Tests · Linear Combinations of Random Variables · Continuous Random Variables

Q14MSampling and EstimationFree sample

In a survey of 2000 randomly chosen adults, 1602 said that they owned a smartphone. Calculate an approximate 95% confidence interval for the proportion of adults in the whole population who own a smartphone.

DifficultyMedium-Easy
Worked solution

Approach

We estimate the population proportion pp using the sample proportion p^\hat{p}. For a large sample, an approximate 95% confidence interval is

p^±z×p^(1−p^)n\hat{p} \pm z \times \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}

with z=1.96z = 1.96 for 95% confidence.

Working

The sample proportion is

p^=16022000=0.801\hat{p} = \frac{1602}{2000} = 0.801

The standard error is

p^(1−p^)n=0.801×0.1992000=0.0000797≈0.00893\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} = \sqrt{\frac{0.801 \times 0.199}{2000}} = \sqrt{0.0000797} \approx 0.00893

The margin of error is

1.96×0.00893≈0.01751.96 \times 0.00893 \approx 0.0175

So the 95% confidence interval is

0.801±0.01750.801 \pm 0.0175

which gives

0.801−0.0175=0.7835and0.801+0.0175=0.81850.801 - 0.0175 = 0.7835 \quad \text{and} \quad 0.801 + 0.0175 = 0.8185

Answer

0.784 to 0.818 (3 sf)0.784 \text{ to } 0.818 \text{ (3 sf)}
Final answer

0.784 to 0.818 (3 sf)

Detailed explanation

Walkthrough

We want a 95% confidence interval for the population proportion pp of adults who own a smartphone. From the sample, the proportion is p^=1602/2000=0.801\hat{p} = 1602/2000 = 0.801. For large samples, the sampling distribution of p^\hat{p} is approximately normal, so the 95% confidence interval is p^±1.96×SE\hat{p} \pm 1.96 \times SE where SE=p^(1−p^)/nSE = \sqrt{\hat{p}(1-\hat{p})/n}. First compute the variance of p^\hat{p}: p^(1−p^)/n=0.801×0.199/2000=0.0000797\hat{p}(1-\hat{p})/n = 0.801 \times 0.199 / 2000 = 0.0000797. Then the standard error is the square root, about 0.00893. Multiply by 1.96 to get the margin of error 0.0175. Add and subtract from 0.801 to get the interval (0.7835, 0.8185), which rounds to 0.784 to 0.818 to 3 significant figures.

Key Takeaways

  • The confidence interval for a proportion uses the sample proportion p^\hat{p} and the standard error p^(1−p^)/n\sqrt{\hat{p}(1-\hat{p})/n}.
  • For 95% confidence, the z-value is 1.96.
  • The interval must be presented as a range (lower, upper), not just a single value.

Common Mistakes

  • Forgetting to take the square root of p^(1−p^)/n\hat{p}(1-\hat{p})/n before multiplying by z.
  • Using the wrong z-value (e.g. 1.645 for 90% or 2.576 for 99%).
  • Giving only one endpoint instead of the full interval.
  • Not rounding correctly to 3 significant figures.

Things to Be Careful About

  • The mark scheme requires the final answer as an interval; an unsupported single value would not earn the A1 mark.
  • The standard error must be computed correctly: p^(1−p^)/n\hat{p}(1-\hat{p})/n, not p^(1−p^)\hat{p}(1-\hat{p}) alone.
  • Allow the answer 0.783 to 0.819 as an alternative rounding.
Techniques used
calculate the sample proportioncompute the standard error of the proportionapply the 95% z-valueform the confidence interval

The rest of this paper

6 more questions
  • Q2The Poisson Distribution5M
  • Q3Sampling and Estimation · Hypothesis Tests8M
  • Q4Hypothesis Tests7M
  • Q5Continuous Random Variables8M
  • Q6The Poisson Distribution · Linear Combinations of Random Variables9M
  • Q7Linear Combinations of Random Variables9M
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