CAIE FP2 2012 June — Question 6 7 marks

Exam BoardCAIE
ModuleFP2 (Further Pure Mathematics 2)
Year2012
SessionJune
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicT-tests (unknown variance)
TypeSingle sample t-test
DifficultyStandard +0.3 This is a straightforward one-sample t-test with all necessary values provided (sample mean, sum of squared deviations, hypothesized mean, and significance level). Students need to calculate the sample standard deviation, compute the t-statistic, compare with critical value from tables, and state a conclusion. While it requires careful calculation and knowledge of the t-test procedure, it's a standard textbook application with no conceptual challenges or novel problem-solving required, making it slightly easier than average.
Spec5.05c Hypothesis test: normal distribution for population mean

6 A random sample of 10 observations of a normal random variable \(X\) has mean \(\bar { x }\), where $$\bar { x } = 8.254 , \quad \Sigma ( x - \bar { x } ) ^ { 2 } = 0.912 .$$ Using a \(5 \%\) significance level, test whether the mean of \(X\) is greater than 8.05.

Question 6:
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(s^2 = 0.912/9 = 0.1013\) or \(38/375\) or \(0.3183^2\)M1 Estimate population variance (allow biased: \(0.0912\) or \(0.3020^2\))
\(H_0: \mu = 8.05\), \(H_1: \mu > 8.05\)B1 State hypotheses (B0 for \(\bar{x}\ldots\))
\(t = (\bar{x} - 8.05)/(s/\sqrt{10}) = 2.03\)M1 A1 Calculate value of \(t\) (to 3 sf)
\(t_{9,\,0.95} = 1.83[3]\)B1 State or use correct tabular \(t\) value (or can compare \(\bar{x}\) with \(8.05 + 0.184[5] = 8.23\))
Reject \(H_0\) if \(t >\) tabular valueM1 Valid method for reaching conclusion
\(2.03 > 1.83\) so mean is greaterA1 Conclusion (AEF, needs correct values)
Total: 7 marks
## Question 6:

| Answer/Working | Marks | Guidance |
|---|---|---|
| $s^2 = 0.912/9 = 0.1013$ or $38/375$ or $0.3183^2$ | M1 | Estimate population variance (allow biased: $0.0912$ or $0.3020^2$) |
| $H_0: \mu = 8.05$, $H_1: \mu > 8.05$ | B1 | State hypotheses (B0 for $\bar{x}\ldots$) |
| $t = (\bar{x} - 8.05)/(s/\sqrt{10}) = 2.03$ | M1 A1 | Calculate value of $t$ (to 3 sf) |
| $t_{9,\,0.95} = 1.83[3]$ | B1 | State or use correct tabular $t$ value (or can compare $\bar{x}$ with $8.05 + 0.184[5] = 8.23$) |
| Reject $H_0$ if $t >$ tabular value | M1 | Valid method for reaching conclusion |
| $2.03 > 1.83$ so mean is greater | A1 | Conclusion (AEF, needs correct values) |

**Total: 7 marks**

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6 A random sample of 10 observations of a normal random variable $X$ has mean $\bar { x }$, where

$$\bar { x } = 8.254 , \quad \Sigma ( x - \bar { x } ) ^ { 2 } = 0.912 .$$

Using a $5 \%$ significance level, test whether the mean of $X$ is greater than 8.05.

\hfill \mbox{\textit{CAIE FP2 2012 Q6 [7]}}