| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2009 |
| Session | November |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | T-tests (unknown variance) |
| Type | Single sample confidence interval t-distribution |
| Difficulty | Easy -1.2 Part (i) is a basic conceptual question about sampling bias (digits 1-4 appear twice as often as 5-9). Part (ii) is a routine confidence interval calculation with known variance using z-distribution, requiring only formula substitution with given values—simpler than typical A-level questions as variance is provided and no hypothesis testing is involved. |
| Spec | 2.01c Sampling techniques: simple random, opportunity, etc5.05d Confidence intervals: using normal distribution |
| Answer | Marks | Guidance |
|---|---|---|
| (i) Doubling only gives even numbers so odd numbers not included | B1 [1] | Needs to be aware that odd numbers aren't included |
| (ii) \(98\% \text{ CI} = 1.62 \pm 2.326 \times \frac{\sqrt{0.0052}}{\sqrt{5}} = 1.62 \pm 0.0750 = (1.54, 1.70)\) | M1, B1, A1 [3] | Correct shape with \(\sqrt{5}\) seen in denom and their evaluated mean and sd (condone unbiased estimate of sample variance). 2.326 seen correct answer, cwo. Accept 1.55 and 1.70 |
(i) Doubling only gives even numbers so odd numbers not included | B1 [1] | Needs to be aware that odd numbers aren't included
(ii) $98\% \text{ CI} = 1.62 \pm 2.326 \times \frac{\sqrt{0.0052}}{\sqrt{5}} = 1.62 \pm 0.0750 = (1.54, 1.70)$ | M1, B1, A1 [3] | Correct shape with $\sqrt{5}$ seen in denom and their evaluated mean and sd (condone unbiased estimate of sample variance). 2.326 seen correct answer, cwo. Accept 1.55 and 1.701 There are 18 people in Millie's class. To choose a person at random she numbers the people in the class from 1 to 18 and presses the random number button on her calculator to obtain a 3-digit decimal. Millie then multiplies the first digit in this decimal by two and chooses the person corresponding to this new number. Decimals in which the first digit is zero are ignored.\\
(i) Give a reason why this is not a satisfactory method of choosing a person.
Millie obtained a random sample of 5 people of her own age by a satisfactory sampling method and found that their heights in metres were $1.66,1.68,1.54,1.65$ and 1.57 . Heights are known to be normally distributed with variance $0.0052 \mathrm {~m} ^ { 2 }$.\\
(ii) Find a $98 \%$ confidence interval for the mean height of people of Millie's age.
\hfill \mbox{\textit{CAIE S2 2009 Q1 [4]}}