Practise the method.
Four linked problems drawn from this topic’s subtopics. 12 parts · untimed · answers and worked solutions at the end.
Example 1
A service surveys two groups independently, randomly within each group. Use each group's observed proportion as a planning estimate of its group population proportion. Group A and B need not appear in the sample in their population proportions. A separate population-weighted planning estimate is 0.35; the population fraction w in group A is unknown.
Find the pooled sample proportion satisfied, ignoring the population weights. Give your answer to 4 decimal places.
| Group | Sample size | Satisfied |
|---|---|---|
| A | 80 | 16 |
| B | 120 | 84 |
Show worked answer
Answer: 0.5
- Pool counts, not the two proportions: 80+12016+84=200100.
- Therefore the answer is 0.5.
Example 2
A stable process independently produces a marked component with population probability p. For a sample of n,p^ is the marked proportion. Use a normal approximation to p^ without continuity correction. A planning rule requires both expected counts to be at least 10.
a Find SD(p^) under this population model. Give your answer to 4 decimal places.
Approximate P(p^<0.15). Give your answer to 4 decimal places.
| Quantity | Value |
|---|---|
| Population proportion p | 0.12 |
| Current sample size n | 150 |
| Comparison proportion | 0.15 |
Show worked answer
Answer: 0.8709
- E(p^)=p=0.12, and its SD is 0.0265329983.
- z=0.02653299830.15−0.12=1.1306675422.
- Use Φ(z). The approximation conditions hold because np=18 and n(1−p)=132.
- Therefore the answer is 0.8709.
Example 3
A random pilot sample records whether a parcel arrives on time. Use the pilot estimate only as a PLANNING value for the population proportion p in future independent samples; it is not known to equal the true p. Keep unrounded values.
a Estimate the population on-time proportion from the pilot sample. Give your answer to 4 decimal places.
b Using the unrounded pilot estimate as p, find the minimum integer size of a future sample whose proportion has SD at most 0.03.
For a separate future sample of 60, use p=16846. Check the rule Np ≥10 AND N(1−p)≥10 for a normal approximation.
| Outcome | Count |
|---|---|
| On time | 46 |
| Late | 122 |
Show worked answer
Answer: Both expected-count conditions pass
- Expected successes =60×16846=16.4285714286.
- Expected failures =60×(1−16846)=43.5714285714.
- Both expected-count conditions pass. Random sampling and independence are additional requirements; satisfying the count rule does not establish them.