Standard +0.8 This is a standard two-sample t-test with small samples requiring calculation of summary statistics from raw data, construction of a confidence interval, and a one-tailed hypothesis test. While it involves multiple steps and careful handling of degrees of freedom, it follows a well-established procedure taught in Further Statistics modules without requiring novel insight or complex problem-solving beyond textbook methods.
9 The leaves from oak trees growing in two different areas \(A\) and \(B\) are being measured. The lengths, in cm , of a random sample of 7 oak leaves from area \(A\) are
$$6.2 , \quad 8.3 , \quad 7.8 , \quad 9.3 , \quad 10.2 , \quad 8.4 , \quad 7.2$$
Assuming that the distribution is normal, find a 95\% confidence interval for the mean length of oak leaves from area \(A\).
The lengths, in cm, of a random sample of 5 oak leaves from area \(B\) are
$$5.9 , \quad 7.4 , \quad 6.8 , \quad 8.2 , \quad 8.7$$
Making suitable assumptions, which should be stated, test, at the \(5 \%\) significance level, whether the mean length of oak leaves from area \(A\) is greater than the mean length of oak leaves from area \(B\). [9]
9 The leaves from oak trees growing in two different areas $A$ and $B$ are being measured. The lengths, in cm , of a random sample of 7 oak leaves from area $A$ are
$$6.2 , \quad 8.3 , \quad 7.8 , \quad 9.3 , \quad 10.2 , \quad 8.4 , \quad 7.2$$
Assuming that the distribution is normal, find a 95\% confidence interval for the mean length of oak leaves from area $A$.
The lengths, in cm, of a random sample of 5 oak leaves from area $B$ are
$$5.9 , \quad 7.4 , \quad 6.8 , \quad 8.2 , \quad 8.7$$
Making suitable assumptions, which should be stated, test, at the $5 \%$ significance level, whether the mean length of oak leaves from area $A$ is greater than the mean length of oak leaves from area $B$. [9]
\hfill \mbox{\textit{CAIE FP2 2012 Q9 [14]}}