CAIE Further Paper 4 2021 November — Question 6 10 marks

Exam BoardCAIE
ModuleFurther Paper 4 (Further Paper 4)
Year2021
SessionNovember
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicT-tests (unknown variance)
TypeTwo-sample t-test equal variance
DifficultyStandard +0.8 This is a two-sample t-test requiring calculation of summary statistics from raw data, correct identification of hypotheses, appropriate test selection, and statement of assumptions. While methodical, it requires multiple computational steps, proper interpretation of 'greater than' for a one-tailed test, and understanding of when two-sample tests apply (not paired despite the title). The 10-mark allocation and need to state validity assumptions elevate this above routine exercises.
Spec5.05c Hypothesis test: normal distribution for population mean5.05d Confidence intervals: using normal distribution

6 A scientist is investigating the masses of a particular type of fish found in lakes \(A\) and \(B\). He chooses a random sample of 10 fish of this type from lake \(A\) and records their masses, \(x \mathrm {~kg}\), as follows.
0.9
1.8
1.8
1.9
2.1
2.4
2.6
2.2
2.5
3.0 The scientist also chooses a random sample of 12 fish of this type from lake \(B\), but he only has a summary of their masses, \(y \mathrm {~kg}\), as follows. $$\sum y = 24.48 \quad \sum y ^ { 2 } = 53.75$$ Test at the \(10 \%\) significance level whether the mean mass of fish of this type in lake \(A\) is greater than the mean mass of fish of this type in lake \(B\). You should state any assumptions that you need to make for the test to be valid.
[0pt] [10]
If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.

Question 6:
AnswerMarks Guidance
AnswerMarks Guidance
\(H_0: \mu_A = \mu_B\) and \(H_1: \mu_A > \mu_B\)B1
\(\sum x = 21.2\), \(\sum x^2 = 47.92\) \(s_x^2 = \frac{1}{9}\left(47.92 - \frac{21.2^2}{10}\right) = 0.33067 = \frac{124}{375}\) \(s_y^2 = \frac{1}{11}\left(53.75 - \frac{24.48^2}{12}\right) = 0.34644 = \frac{9527}{27500}\)M1 A1
Pooled method: \(s^2 = \frac{9 \times 0.33067 + 11 \times 0.34644}{10 + 12 - 2} = 0.3393\)M1 A1
\(t = \frac{2.12 - 2.04}{s\sqrt{\frac{1}{10} + \frac{1}{12}}} = 0.321\)M1 A1 Accept \(0.321 - 0.322\)
Critical value: \(1.325\); \(0.321 < 1.325\) accept \(H_0\)M1 Compare their value with \(1.325\) and correct FT conclusion
Insufficient evidence that mean of \(A\) is greater than mean of \(B\)A1 Correct conclusion in context, following correct work. Level of uncertainty in language is used
EITHER: Distributions are normal and equal variances OR: Distributions are normalB1 Assumptions consistent with method used. Accept 'population is normal'. 'Distribution of difference of means is normal'
Alternative method:
AnswerMarks Guidance
AnswerMarks Guidance
\(H_0: \mu_A = \mu_B\) and \(H_1: \mu_A > \mu_B\)B1
\(s_x^2 = 0.33067\), \(s_y^2 = 0.34644\)M1 A1 M1 for one correct unsimplified; A1 for both correct
\(s^2 = \frac{0.33067}{10} + \frac{0.34644}{12} = 0.061936\)M1 A1
\(t = \frac{2.12 - 2.04}{s} = 0.321(5)\)M1 A1 Accept \(0.321 - 0.322\)
Critical value: \(1.325\); \(0.321 < 1.325\) accept \(H_0\)M1 Compare their value with \(1.325\) and correct FT conclusion
Insufficient evidence that mean of \(A\) is greater than mean of \(B\)A1 Correct conclusion in context, following correct work. Level of uncertainty in language is used
EITHER: Distributions are normal and equal variances OR: Distributions are normalB1 Assumptions consistent with method used. Accept 'population is normal'. 'Distribution of difference of means is normal'
Total10
## Question 6:

| Answer | Marks | Guidance |
|--------|-------|----------|
| $H_0: \mu_A = \mu_B$ and $H_1: \mu_A > \mu_B$ | **B1** | |
| $\sum x = 21.2$, $\sum x^2 = 47.92$ $s_x^2 = \frac{1}{9}\left(47.92 - \frac{21.2^2}{10}\right) = 0.33067 = \frac{124}{375}$ $s_y^2 = \frac{1}{11}\left(53.75 - \frac{24.48^2}{12}\right) = 0.34644 = \frac{9527}{27500}$ | **M1 A1** | |
| **Pooled method:** $s^2 = \frac{9 \times 0.33067 + 11 \times 0.34644}{10 + 12 - 2} = 0.3393$ | **M1 A1** | |
| $t = \frac{2.12 - 2.04}{s\sqrt{\frac{1}{10} + \frac{1}{12}}} = 0.321$ | **M1 A1** | Accept $0.321 - 0.322$ |
| Critical value: $1.325$; $0.321 < 1.325$ accept $H_0$ | **M1** | Compare their value with $1.325$ and correct FT conclusion |
| Insufficient evidence that mean of $A$ is greater than mean of $B$ | **A1** | Correct conclusion in context, following correct work. Level of uncertainty in language is used |
| EITHER: Distributions are normal and equal variances OR: Distributions are normal | **B1** | Assumptions consistent with method used. Accept 'population is normal'. 'Distribution of difference of means is normal' |

**Alternative method:**

| Answer | Marks | Guidance |
|--------|-------|----------|
| $H_0: \mu_A = \mu_B$ and $H_1: \mu_A > \mu_B$ | **B1** | |
| $s_x^2 = 0.33067$, $s_y^2 = 0.34644$ | **M1 A1** | M1 for one correct unsimplified; A1 for both correct |
| $s^2 = \frac{0.33067}{10} + \frac{0.34644}{12} = 0.061936$ | **M1 A1** | |
| $t = \frac{2.12 - 2.04}{s} = 0.321(5)$ | **M1 A1** | Accept $0.321 - 0.322$ |
| Critical value: $1.325$; $0.321 < 1.325$ accept $H_0$ | **M1** | Compare their value with $1.325$ and correct FT conclusion |
| Insufficient evidence that mean of $A$ is greater than mean of $B$ | **A1** | Correct conclusion in context, following correct work. Level of uncertainty in language is used |
| EITHER: Distributions are normal and equal variances OR: Distributions are normal | **B1** | Assumptions consistent with method used. Accept 'population is normal'. 'Distribution of difference of means is normal' |
| **Total** | **10** | |
6 A scientist is investigating the masses of a particular type of fish found in lakes $A$ and $B$. He chooses a random sample of 10 fish of this type from lake $A$ and records their masses, $x \mathrm {~kg}$, as follows.\\
0.9\\
1.8\\
1.8\\
1.9\\
2.1\\
2.4\\
2.6\\
2.2\\
2.5\\
3.0

The scientist also chooses a random sample of 12 fish of this type from lake $B$, but he only has a summary of their masses, $y \mathrm {~kg}$, as follows.

$$\sum y = 24.48 \quad \sum y ^ { 2 } = 53.75$$

Test at the $10 \%$ significance level whether the mean mass of fish of this type in lake $A$ is greater than the mean mass of fish of this type in lake $B$. You should state any assumptions that you need to make for the test to be valid.\\[0pt]
[10]\\

If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.\\

\hfill \mbox{\textit{CAIE Further Paper 4 2021 Q6 [10]}}