| Exam Board | OCR |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2013 |
| Session | January |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear combinations of normal random variables |
| Type | Two-sample t-test (unknown variances) |
| Difficulty | Standard +0.3 This is a standard two-sample t-test with pooled variance calculation. Students need to recall the formula, compute the pooled standard deviation, find the appropriate t-critical value, and construct the confidence interval. The calculations are straightforward with given summary statistics, and the additional assumption (equal variances) is a standard requirement. Slightly easier than average as it's a direct application of a well-practiced procedure with no conceptual surprises. |
| Spec | 5.05c Hypothesis test: normal distribution for population mean5.05d Confidence intervals: using normal distribution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Est \(\sigma_A^2 = 62\), Est \(\sigma_B^2 = 127\) | B1 | Can be implied e.g. by 89.86. Allow 248/4 |
| \(\text{PEV} = (4\times62 + 3\times127)/7 = [89.86 \text{ or } 629/7]\) | M1 | (or \(\bar{x}_B - \bar{x}_A = 4.5\)) |
| \((\bar{x}_A - \bar{x}_B = -4.5)\) | ||
| \(SD = \sqrt{89.86(5^{-1}+4^{-1})}\) | B1 | |
| \(CI = (-4.5 - tSD,\ -4.5 + tSD)\) | M1 | Allow M from incorrect 4.5. Must be \(t\), not \(z\). |
| \(t = 1.415\) | B1 | |
| \((-13.50,\ 4.50)\) | A1 | Allow \((-4.50,\ 13.50)\) |
| [6] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Scores resulting from two schemes should have equal variances | B1 | Allow popn. variances equal. NOT 'Variances equal' |
| [1] |
# Question 3:
## Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Est $\sigma_A^2 = 62$, Est $\sigma_B^2 = 127$ | B1 | Can be implied e.g. by 89.86. Allow 248/4 |
| $\text{PEV} = (4\times62 + 3\times127)/7 = [89.86 \text{ or } 629/7]$ | M1 | (or $\bar{x}_B - \bar{x}_A = 4.5$) |
| $(\bar{x}_A - \bar{x}_B = -4.5)$ | | |
| $SD = \sqrt{89.86(5^{-1}+4^{-1})}$ | B1 | |
| $CI = (-4.5 - tSD,\ -4.5 + tSD)$ | M1 | Allow M from incorrect 4.5. Must be $t$, not $z$. |
| $t = 1.415$ | B1 | |
| $(-13.50,\ 4.50)$ | A1 | Allow $(-4.50,\ 13.50)$ |
| | **[6]** | |
## Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Scores resulting from two schemes should have equal variances | B1 | Allow popn. variances equal. NOT 'Variances equal' |
| | **[1]** | |
---3 Two reading schemes, $A$ and $B$, are compared by using them with a random sample of 9 five-year-old children. The children are divided into two groups, 5 allotted to scheme $A$ and 4 to scheme $B$, and the schemes are taught under similar conditions.\\
After one year the children are given the same test and their scores, $x _ { A }$ and $x _ { B }$, are summarised below. With the usual notation,
$$\begin{aligned}
& n _ { A } = 5 , \bar { x } _ { A } = 52.0 , \sum \left( x _ { A } - \bar { x } _ { A } \right) ^ { 2 } = 248 , \\
& n _ { B } = 4 , \bar { x } _ { B } = 56.5 , \sum \left( x _ { B } - \bar { x } _ { B } \right) ^ { 2 } = 381 .
\end{aligned}$$
It may be assumed that scores have normal distributions.\\
(i) Calculate an $80 \%$ confidence interval for the difference in population mean scores for the two methods.\\
(ii) State a further assumption required for the validity of the interval.
\hfill \mbox{\textit{OCR S3 2013 Q3 [7]}}