Moderate -0.3 This is a straightforward one-tail z-test with all information clearly provided: known population parameters, large sample size (n=150), and explicit significance level. Students simply need to state hypotheses, calculate the test statistic z = (25.0-24.0)/(4.8/√150) ≈ 2.55, compare to critical value z₀.₀₂ = 2.054, and conclude. It's slightly easier than average because it's a textbook application with no complications or interpretation challenges.
2 The heights of a certain type of plant have a normal distribution. When the plants are grown without fertilizer, the population mean and standard deviation are 24.0 cm and 4.8 cm respectively. A gardener wishes to test, at the \(2 \%\) significance level, whether Hiergro fertilizer will increase the mean height. He treats 150 randomly chosen plants with Hiergro and finds that their mean height is 25.0 cm . Assuming that the standard deviation of the heights of plants treated with Hiergro is still 4.8 cm , carry out the test.
\(H_0\): Pop mean = 24.0; \(H_1\): Pop mean > 24.0
B1
Allow '\(\mu\)' but not just 'mean'
\(\frac{25-24}{4.8} \div \sqrt{150}\)
M1
Standardise, with \(\sqrt{150}\). Ignore cc. Accept sd/var mixes. OR find \(x_{crit}\)
\(= 2.55(2)\)
A1
For correct z or area or \(x_{crit}\)
Comp z = 2.054 or 2.055; Evidence that Hiergro has incr hts
M1 A1ft [5]
Valid comparison (z values/areas/x values); Correct conclusion No contradictions (Note 2 tail test can score B0 M1 A1 M1 A1ft) (z = 2.326) A1ft)
$H_0$: Pop mean = 24.0; $H_1$: Pop mean > 24.0 | B1 | Allow '$\mu$' but not just 'mean'
$\frac{25-24}{4.8} \div \sqrt{150}$ | M1 | Standardise, with $\sqrt{150}$. Ignore cc. Accept sd/var mixes. OR find $x_{crit}$
$= 2.55(2)$ | A1 | For correct z or area or $x_{crit}$
Comp z = 2.054 or 2.055; Evidence that Hiergro has incr hts | M1 A1ft [5] | Valid comparison (z values/areas/x values); Correct conclusion No contradictions (Note 2 tail test can score B0 M1 A1 M1 A1ft) (z = 2.326) A1ft)
2 The heights of a certain type of plant have a normal distribution. When the plants are grown without fertilizer, the population mean and standard deviation are 24.0 cm and 4.8 cm respectively. A gardener wishes to test, at the $2 \%$ significance level, whether Hiergro fertilizer will increase the mean height. He treats 150 randomly chosen plants with Hiergro and finds that their mean height is 25.0 cm . Assuming that the standard deviation of the heights of plants treated with Hiergro is still 4.8 cm , carry out the test.
\hfill \mbox{\textit{CAIE S2 2012 Q2 [5]}}