Moderate -0.3 This is a straightforward confidence interval calculation requiring standard formulas for sample mean, standard deviation, and the appropriate z-value. It's slightly easier than average because it's a direct application of a single technique with no conceptual complications, though the 94% confidence level (requiring z = 1.881) adds minor difficulty compared to the standard 95% case.
100 randomly chosen adults each throw a ball once. The length, \(l\) metres, of each throw is recorded. The results are summarised below.
$$n = 100 \qquad \sum l = 3820 \qquad \sum l^2 = 182200$$
Calculate a 94% confidence interval for the population mean length of throws by adults. [6]
Unless otherwise indicated, marks once gained cannot subsequently be lost, e.g. wrong working following a correct form of answer is ignored (isw).
4
3820
[= 38.2]
Answer
Marks
100
B1
100182200 1 38202
−'38.2'2 or 182200−
Answer
Marks
Guidance
99 100 99 100
M1
Use of biased (362.76) scores M0
12092
= or 366.424 or 366 (3 sf)
Answer
Marks
Guidance
33
A1
Accept SD=19.1422 or 19.1(3sf)
'366.424'
‘38.2’ ± z×
Answer
Marks
Guidance
100
M1
Expression of the correct form must be a z-value.
z = 1.881 or 1.882
B1
Seen.
34.6 to 41.8 (3 sf)
A1
Allow use of biased giving (34.6,41.8)
Must be an interval.
6
Answer
Marks
Guidance
Question
Answer
Marks
Question 4:
4 | Unless otherwise indicated, marks once gained cannot subsequently be lost, e.g. wrong working following a correct form of answer is ignored (isw).
4 | 3820
[= 38.2]
100 | B1
100182200 1 38202
−'38.2'2 or 182200−
99 100 99 100 | M1 | Use of biased (362.76) scores M0
12092
= or 366.424 or 366 (3 sf)
33 | A1 | Accept SD=19.1422 or 19.1(3sf)
'366.424'
‘38.2’ ± z×
100 | M1 | Expression of the correct form must be a z-value.
z = 1.881 or 1.882 | B1 | Seen.
34.6 to 41.8 (3 sf) | A1 | Allow use of biased giving (34.6,41.8)
Must be an interval.
6
Question | Answer | Marks | Guidance
100 randomly chosen adults each throw a ball once. The length, $l$ metres, of each throw is recorded. The results are summarised below.
$$n = 100 \qquad \sum l = 3820 \qquad \sum l^2 = 182200$$
Calculate a 94% confidence interval for the population mean length of throws by adults. [6]
\hfill \mbox{\textit{CAIE S2 2021 Q4 [6]}}