| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2014 |
| Session | November |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Z-tests (known variance) |
| Type | Two-tail z-test |
| Difficulty | Moderate -0.8 This is a straightforward hypothesis testing question requiring only standard procedure: recognizing why a two-tail test is appropriate (testing for 'different from' rather than directional change) and comparing a given z-value (1.91) to critical values at 4% significance level. No calculation of the test statistic is needed, just interpretation and conclusion. This is routine application of textbook methodology with minimal problem-solving. |
| Spec | 2.05a Hypothesis testing language: null, alternative, p-value, significance2.05c Significance levels: one-tail and two-tail2.05e Hypothesis test for normal mean: known variance |
| Answer | Marks | Guidance |
|---|---|---|
| Part (i) | ||
| "Different" being investigated | B1 | Oe ("changed", "not equal to") |
| Must be "pop mean", not just "mean" Can be awarded in (i) | ||
| Part (ii) | ||
| \(H_0\): Pop mean (or \(\mu\)) in region same as elsewhere | B1 | oe |
| \(H_1\): Pop mean (or \(\mu\)) in region diff from elsewhere | ||
| \(1.91 < 2.054\) (or 2.055) or \(-1.91 > -2.054\) | M1 | or \(P(z > 1.91) = 0.0281 > 0.02\) or \(0.0562 > 0.04\) or \(0.972 < 0.98\) Accept 2.05 if nothing better seen. |
| No evidence that mean is different | A1 | inequality sign incorrect M1A0 no contradictions "accept \(H_0\)" provided \(H_0\) reasonably well defined |
| Total | [4] |
| **Part (i)** |
|---|
| "Different" being investigated | B1 | Oe ("changed", "not equal to") |
| | | Must be "pop mean", not just "mean" Can be awarded in (i) |
| **Part (ii)** |
|---|
| $H_0$: Pop mean (or $\mu$) in region same as elsewhere | B1 | oe |
| $H_1$: Pop mean (or $\mu$) in region diff from elsewhere | | |
| $1.91 < 2.054$ (or 2.055) or $-1.91 > -2.054$ | M1 | or $P(z > 1.91) = 0.0281 > 0.02$ or $0.0562 > 0.04$ or $0.972 < 0.98$ Accept 2.05 if nothing better seen. |
| No evidence that mean is different | A1 | inequality sign incorrect M1A0 no contradictions "accept $H_0$" provided $H_0$ reasonably well defined |
| **Total** | [4] |1 A researcher wishes to investigate whether the mean height of a certain type of plant in one region is different from the mean height of this type of plant everywhere else. He takes a large random sample of plants from the region and finds the sample mean. He calculates the value of the test statistic, $z$, and finds that $z = 1.91$.\\
(i) Explain briefly why the researcher should use a two-tail test.\\
(ii) Carry out the test at the $4 \%$ significance level.
\hfill \mbox{\textit{CAIE S2 2014 Q1 [4]}}