| Exam Board | OCR |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2006 |
| Session | January |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | T-tests (unknown variance) |
| Type | Pooled variance estimate calculation |
| Difficulty | Standard +0.3 This is a standard two-sample t-test question with pooled variance. Part (i) is routine calculation of pooled variance, part (ii) is textbook confidence interval construction, and part (iii) is a straightforward hypothesis test. All steps follow standard procedures taught in S3 with no novel insight required, making it slightly easier than average. |
| Spec | 5.05b Unbiased estimates: of population mean and variance5.05c Hypothesis test: normal distribution for population mean5.05d Confidence intervals: using normal distribution |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(s^2=(8 \times 0.7400+9 \times 0.8160)/17\) | M1 | Formula for pooled estimate |
| \(=0.7802\) | A1 | At least 4DP shown |
| \(0.780\) AG | ||
| (ii) Assumes braking distances have normal distributions | B1 | |
| Use \(\bar{x}_A - \bar{y}_B \pm t\sigma\) | M1 | Must be \(t\) value |
| \(t=2.567\) | A1 | |
| \(=\sqrt{(0.7802(1/10+1/9)}\) \((0.40584)\) | B1 | Allow 0.780 |
| \((1.518,3.602)\) | A1 | 5 |
| (iii) \(H_0:\mu_A-\mu_B=2\), \(H_1:\mu_A-\mu_B > 2\) | B1 | For both hypotheses |
| Use of CV, 1.740 | B1 | |
| EITHER: Test statistic \(=(2.56-2)/\sigma\) | M1 | Standardising, \(\sigma\) as above |
| \(=1.38\) | A1 | Rounding to 1.38 |
| OR: Critical region | ||
| \(\bar{x}_A - \bar{y}_B > 2+1.74 \times 0.4054\) | M1 | |
| \(=2.7054\) | A1 | 2.70 or 2.71 |
| Indication that test statistic is not in critical region and Insufficient evidence to accept claim and \(H_1\) | M1 | Not from different signs test statistic |
| A1 | 6 | critical value. A1 dep on correct \(H_0\) |
(i) $s^2=(8 \times 0.7400+9 \times 0.8160)/17$ | M1 | Formula for pooled estimate
$=0.7802$ | A1 | At least 4DP shown
$0.780$ AG | | | 2
(ii) Assumes braking distances have normal distributions | B1 |
Use $\bar{x}_A - \bar{y}_B \pm t\sigma$ | M1 | Must be $t$ value
$t=2.567$ | A1 |
$=\sqrt{(0.7802(1/10+1/9)}$ $(0.40584)$ | B1 | Allow 0.780
$(1.518,3.602)$ | A1 | 5 | art (1.52, 3.60)
(iii) $H_0:\mu_A-\mu_B=2$, $H_1:\mu_A-\mu_B > 2$ | B1 | For both hypotheses
Use of CV, 1.740 | B1 |
EITHER: Test statistic $=(2.56-2)/\sigma$ | M1 | Standardising, $\sigma$ as above
$=1.38$ | A1 | Rounding to 1.38
OR: Critical region | |
$\bar{x}_A - \bar{y}_B > 2+1.74 \times 0.4054$ | M1 |
$=2.7054$ | A1 | 2.70 or 2.71
Indication that test statistic is not in critical region and Insufficient evidence to accept claim and $H_1$ | M1 | Not from different signs test statistic
| A1 | 6 | critical value. A1 dep on correct $H_0$
---6 A company with a large fleet of cars compared two types of tyres, $A$ and $B$. They measured the stopping distances of cars when travelling at a fixed speed on a dry road. They selected 20 cars at random from the fleet and divided them randomly into two groups of 10 , one group being fitted with tyres of type $A$ and the other group with tyres of type $B$. One of the cars fitted with tyres of type $A$ broke down so these tyres were tested on only 9 cars. The stopping distances, $x$ metres, for the two samples are summarised by
$$n _ { A } = 9 , \quad \bar { x } _ { A } = 17.30 , \quad s _ { A } ^ { 2 } = 0.7400 , \quad n _ { B } = 10 , \quad \bar { x } _ { B } = 14.74 , \quad s _ { B } ^ { 2 } = 0.8160 ,$$
where $s _ { A } ^ { 2 }$ and $s _ { B } ^ { 2 }$ are unbiased estimates of the two population variances.\\
It is given that the two populations have the same variance.\\
(i) Show that an unbiased estimate of this variance is 0.780 , correct to 3 decimal places.
The population mean stopping distances for cars with tyres of types $A$ and $B$ are denoted by $\mu _ { A }$ metres and $\mu _ { B }$ metres respectively.\\
(ii) Stating any further assumption you need to make, calculate a $98 \%$ confidence interval for $\mu _ { A } - \mu _ { B }$.
The manufacturers of Type $B$ tyres assert that $\mu _ { B } < \mu _ { A } - 2$.\\
(iii) Carry out a significance test of this assertion at the $5 \%$ significance level.
\section*{[Question 7 is printed overleaf.]}
\hfill \mbox{\textit{OCR S3 2006 Q6 [13]}}