Moderate -0.3 This is a straightforward one-tail z-test with all information clearly provided (n=60, known σ=0.4, significance level stated). Students need only apply the standard hypothesis testing procedure: state hypotheses, calculate test statistic z = (2.4-2.3)/(0.4/√60) ≈ 1.94, compare to critical value 1.96, and conclude. No conceptual challenges or problem-solving required, just routine application of a standard technique, making it slightly easier than average.
2 In the past, the mean height of plants of a particular species has been 2.3 m . A random sample of 60 plants of this species was treated with fertiliser and the mean height of these 60 plants was found to be 2.4 m . Assume that the standard deviation of the heights of plants treated with fertiliser is 0.4 m .
Carry out a test at the \(2.5 \%\) significance level of whether the mean height of plants treated with fertiliser is greater than 2.3 m .
\(H_0\): Pop mean height \(= 2.3\); \(H_1\): Pop mean height \(> 2.3\)
B1
Not just 'mean'; allow \(\mu\)
\(\frac{2.4 - 2.3}{\frac{0.4}{\sqrt{60}}}\)
M1
For standardising, must have \(\sqrt{60}\)
\(1.936\) or \(1.937\) or \(1.94\)
A1
\('1.936' < 1.96\)
M1
Valid comparison with \(1.96\); or \(2.64\% > 2.5\%\) OE; accept \(1.936 < 2.24\) or \(2.64\% > 1.25\%\) OE if \(H_1\mu \neq 2.3\)
[Do not reject \(H_0\)] No evidence that (mean) height (with fertiliser) is more than without
A1 FT
FT *their* \(z\); in context, not definite; no FT for 2 tail test (max B0 M1 A1 M1 A0 3/5); accept critical values method \(2.401\) (M1 A1) \(2.4 < 2.401\); (M1) condone \(2.299\) (M1 A1) \(< 2.3\) (M1); A1 conclusion
## Question 2:
| Answer | Marks | Guidance |
|--------|-------|----------|
| $H_0$: Pop mean height $= 2.3$; $H_1$: Pop mean height $> 2.3$ | B1 | Not just 'mean'; allow $\mu$ |
| $\frac{2.4 - 2.3}{\frac{0.4}{\sqrt{60}}}$ | M1 | For standardising, must have $\sqrt{60}$ |
| $1.936$ or $1.937$ or $1.94$ | A1 | |
| $'1.936' < 1.96$ | M1 | Valid comparison with $1.96$; or $2.64\% > 2.5\%$ OE; accept $1.936 < 2.24$ or $2.64\% > 1.25\%$ OE if $H_1\mu \neq 2.3$ |
| [Do not reject $H_0$] No evidence that (mean) height (with fertiliser) is more than without | A1 FT | FT *their* $z$; in context, not definite; no FT for 2 tail test (max B0 M1 A1 M1 A0 3/5); accept critical values method $2.401$ (M1 A1) $2.4 < 2.401$; (M1) condone $2.299$ (M1 A1) $< 2.3$ (M1); A1 conclusion |
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2 In the past, the mean height of plants of a particular species has been 2.3 m . A random sample of 60 plants of this species was treated with fertiliser and the mean height of these 60 plants was found to be 2.4 m . Assume that the standard deviation of the heights of plants treated with fertiliser is 0.4 m .
Carry out a test at the $2.5 \%$ significance level of whether the mean height of plants treated with fertiliser is greater than 2.3 m .\\
\hfill \mbox{\textit{CAIE S2 2022 Q2 [5]}}