| Exam Board | Edexcel |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2013 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Central limit theorem |
| Type | Hypothesis test for mean |
| Difficulty | Standard +0.3 This is a straightforward two-sample hypothesis test with known variances using the normal distribution. The setup is clearly guided (hypotheses, significance level given), calculations are routine (difference of means, standard error formula), and part (b) is standard bookwork about CLT justifying normality. Slightly above average due to the 'difference greater than 1kg' framing rather than simple equality, but still a standard S3 exercise. |
| Spec | 5.05a Sample mean distribution: central limit theorem5.05c Hypothesis test: normal distribution for population mean |
6. Fruit-n-Veg4U Market Gardens grow tomatoes. They want to improve their yield of tomatoes by at least 1 kg per plant by buying a new variety. The variance of the yield of the old variety of plant is $0.5 \mathrm {~kg} ^ { 2 }$ and the variance of the yield for the new variety of plant is $0.75 \mathrm {~kg} ^ { 2 }$. A random sample of 60 plants of the old variety has a mean yield of 5.5 kg . A random sample of 70 of the new variety has a mean yield of 7 kg .
\begin{enumerate}[label=(\alph*)]
\item Stating your hypotheses clearly test, at the $5 \%$ level of significance, whether or not there is evidence that the mean yield of the new variety is more than 1 kg greater than the mean yield of the old variety.
\item Explain the relevance of the Central Limit Theorem to the test in part (a).
\end{enumerate}
\hfill \mbox{\textit{Edexcel S3 2013 Q6 [11]}}