Standard +0.8 This is a two-sample confidence interval problem requiring calculation of sample means, unbiased variance estimates, standard error of difference, and correct application of the t-distribution. While the individual steps are standard, the multi-stage calculation with two samples, the need to pool or use appropriate degrees of freedom, and the 9-mark allocation indicate this is more demanding than a routine single-sample CI question, placing it moderately above average difficulty.
The number, \(x\), of beech trees was counted in each of \(50\) randomly chosen regions of equal size in beech forests in country \(A\). The number, \(y\), of beech trees was counted in each of \(40\) randomly chosen regions of the same equal size in beech forests in country \(B\). The results are summarised as follows.
$$\Sigma x = 1416 \quad \Sigma x^2 = 41100 \quad \Sigma y = 888 \quad \Sigma y^2 = 20140$$
Find a \(95\%\) confidence interval for the difference between the mean number of beech trees in regions of this size in country \(A\) and in country \(B\). [9]
wrongly here, giving zs = 1⋅60, only this A1 is lost)
Answer
Marks
Guidance
x – y ± z s
M1
Find confidence interval for difference
z = 1⋅96
Answer
Marks
Guidance
0.975
A1
Use appropriate tabular value
z s = 1⋅62
A1
Evaluate semi-interval length
6⋅12 ± 1⋅62 or [4⋅50, 7⋅74]
A1)
State confidence interval (in either form)
Question
Answer
Marks
OR:
Assume equal [population] variances
s2 = (49 s 2 + 39 s 2) / 88
A B
Answer
Marks
Guidance
or (41 100 – 14162/50 + 20 140 – 8882/40) / 88
(B1
OR:
State assumption
Find pooled estimate of common variance
(M1 A1 for s 2 and s 2 may be implied here)
A B
Answer
Marks
Guidance
= 16⋅2 or 4⋅022
B1
x – y ± z s √(1/50 + 1/40)
M1
Find confidence interval for difference
z = 1⋅96
Answer
Marks
Guidance
0.975
A1
Use appropriate tabular z-value
(or appropriate t-value from calculator or interpolation)
(t = 1⋅9873 or 1⋅99)
88, 0.975
Answer
Marks
Guidance
1⋅67 (1⋅70)
A1
Evaluate semi-interval length
6⋅12 ± 1⋅67 or [4⋅45, 7⋅79]
Answer
Marks
Guidance
(6⋅12 ± 1⋅70 or [4⋅42, 7⋅82])
A1)
Evaluate confidence interval (in either form)
Total:
9
Question 8:
8 | x = 28⋅32 and y = 22⋅2 | B1 | Find both sample means
s 2 = (41 100 – 14162/50) / 49 and
A
s 2 = (20 140 – 8882/40) / 39
B | M1 | Estimate both population variances
s 2 = 20⋅39 and s 2 = 164/15 or 10⋅93 (to 3 s.f.)
A B | A1 | (allow biased here: 19⋅98 and 10⋅66)
EITHER:
s2 = s 2/50 + s 2/40 = 0⋅681 or 0⋅8252
A B
(19⋅98/50 + 10⋅66/40 = 0⋅666 is M1 A0, max 8/9) | (M1 A1 | EITHER:
Estimate combined variance (if biased values used
wrongly here, giving zs = 1⋅60, only this A1 is lost)
x – y ± z s | M1 | Find confidence interval for difference
z = 1⋅96
0.975 | A1 | Use appropriate tabular value
z s = 1⋅62 | A1 | Evaluate semi-interval length
6⋅12 ± 1⋅62 or [4⋅50, 7⋅74] | A1) | State confidence interval (in either form)
Question | Answer | Marks | Guidance
OR:
Assume equal [population] variances
s2 = (49 s 2 + 39 s 2) / 88
A B
or (41 100 – 14162/50 + 20 140 – 8882/40) / 88 | (B1 | OR:
State assumption
Find pooled estimate of common variance
(M1 A1 for s 2 and s 2 may be implied here)
A B
= 16⋅2 or 4⋅022 | B1
x – y ± z s √(1/50 + 1/40) | M1 | Find confidence interval for difference
z = 1⋅96
0.975 | A1 | Use appropriate tabular z-value
(or appropriate t-value from calculator or interpolation)
(t = 1⋅9873 or 1⋅99)
88, 0.975
1⋅67 (1⋅70) | A1 | Evaluate semi-interval length
6⋅12 ± 1⋅67 or [4⋅45, 7⋅79]
(6⋅12 ± 1⋅70 or [4⋅42, 7⋅82]) | A1) | Evaluate confidence interval (in either form)
Total: | 9
The number, $x$, of beech trees was counted in each of $50$ randomly chosen regions of equal size in beech forests in country $A$. The number, $y$, of beech trees was counted in each of $40$ randomly chosen regions of the same equal size in beech forests in country $B$. The results are summarised as follows.
$$\Sigma x = 1416 \quad \Sigma x^2 = 41100 \quad \Sigma y = 888 \quad \Sigma y^2 = 20140$$
Find a $95\%$ confidence interval for the difference between the mean number of beech trees in regions of this size in country $A$ and in country $B$. [9]
\hfill \mbox{\textit{CAIE FP2 2017 Q8 [9]}}