Question 1 - A-Level Maths

CAIE Further Paper 4 2023 November — Question 1 6 marks

Exam Board	CAIE
Module	Further Paper 4 (Further Paper 4)
Year	2023
Session	November
Marks	6
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	T-tests (unknown variance)
Type	Single sample t-test
Difficulty	Standard +0.3 This is a straightforward one-sample t-test with clearly stated hypotheses (one-tailed test for improvement). Students must calculate sample mean and standard deviation from summary statistics, then apply the standard t-test procedure. While it requires multiple steps, each is routine for Further Statistics students, and the context makes the direction of the test obvious.
Spec	5.05c Hypothesis test: normal distribution for population mean

1 Maya is an athlete who competes in 1500-metre races. Last summer her practice run times had mean 4.22 minutes. Over the winter she has done some intense training to try to improve her times. A random sample of 10 of her practice run times, $x$ minutes, this summer are summarised as follows. $$\sum x = 42.05 \quad \sum x ^ { 2 } = 176.83$$ Maya's new practice run times are normally distributed. She believes that on average her times have improved as a result of her training. Test, at the $5 \%$ significance level, whether Maya's belief is supported by the data.

Show mark scheme Show mark scheme source

Question 1:

Answer	Marks	Guidance
Answer	Mark	Guidance
$H_0: \mu = 4.22 \quad H_1: \mu < 4.22$	B1
$\bar{x} = \frac{42.05}{10} [= 4.205]$, $\quad s^2 = \frac{1}{9}\left(176.83 - \frac{42.05^2}{10}\right)\left[= \frac{13}{12000} = 0.0010833\right]$	M1
$t = \dfrac{4.205 - 4.22}{\sqrt{\dfrac{s^2}{10}}}$	M1	Their $4.205$, their $s^2$
$t = -1.44$	A1	Condone sign
Tabular value $= 1.833$: \(\	{-1.44}\	< 1.833\), accept $H_0$
There is insufficient evidence to support Maya's belief.	A1	Correct conclusion in context, following correct work, level of uncertainty in language. A0 if hypotheses wrong way round or missing.
6

## Question 1:

| Answer | Mark | Guidance |
|--------|------|----------|
| $H_0: \mu = 4.22 \quad H_1: \mu < 4.22$ | B1 | |
| $\bar{x} = \frac{42.05}{10} [= 4.205]$, $\quad s^2 = \frac{1}{9}\left(176.83 - \frac{42.05^2}{10}\right)\left[= \frac{13}{12000} = 0.0010833\right]$ | M1 | |
| $t = \dfrac{4.205 - 4.22}{\sqrt{\dfrac{s^2}{10}}}$ | M1 | Their $4.205$, their $s^2$ |
| $t = -1.44$ | A1 | Condone sign |
| Tabular value $= 1.833$: $\|{-1.44}\| < 1.833$, accept $H_0$ | M1 | Compare *their* value with **correct** tabular value, **signs consistent**, allow 'not significant' |
| There is insufficient evidence to support Maya's belief. | A1 | Correct conclusion in context, following correct work, level of uncertainty in language. A0 if hypotheses wrong way round or missing. |
| | **6** | |

Show LaTeX source

1 Maya is an athlete who competes in 1500-metre races. Last summer her practice run times had mean 4.22 minutes. Over the winter she has done some intense training to try to improve her times. A random sample of 10 of her practice run times, $x$ minutes, this summer are summarised as follows.

$$\sum x = 42.05 \quad \sum x ^ { 2 } = 176.83$$

Maya's new practice run times are normally distributed. She believes that on average her times have improved as a result of her training.

Test, at the $5 \%$ significance level, whether Maya's belief is supported by the data.\\

\hfill \mbox{\textit{CAIE Further Paper 4 2023 Q1 [6]}}

This paper (6 questions)

View full paper

Q1 6 Q2 7 Q3 8 Q4 9 Q5 10 Q6 10