| Exam Board | OCR |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2007 |
| Session | January |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | T-tests (unknown variance) |
| Type | Two-sample t-test equal variance |
| Difficulty | Standard +0.3 This is a standard two-sample t-test from S3 with small samples requiring routine application of the test procedure. While it involves stating assumptions and performing calculations, it's a textbook exercise with no novel insight required. The small sample sizes and straightforward setup make it slightly easier than average A-level questions, but the S3 content (hypothesis testing with normal distributions) places it just above average difficulty overall. |
| Spec | 5.05c Hypothesis test: normal distribution for population mean |
| Answer | Marks | Guidance |
|---|---|---|
| (i) Assumes breaking strengths have normal normal distributions; Equal variances | B1, B1 | 2 |
| (ii) \(H_0: \mu_T = \mu_U\); \(H_1: \mu_T > \mu_U\) where \(\mu_T, \mu_U\) are means for treated and untreated thread | B1 | For both hypotheses |
| \(\bar{x}_T = 18.05\), \(\bar{x}_U = 17.26\) | B1 | May be implied below by 0.79 |
| \(s_T^2 = 0.715\), \(s_U^2 = 0.738\) | B1 | Allow biased, 0.596, 0.590 if \(s^2\) |
| \(s^2 = (5 \times 0.715 + 4 \times 0.738)/9\); EITHER: \((18.05 - 17.26)/\sqrt{s^2(1/5+1/6)} = 1.532\) | M1 | \(=(6 \times 0.596 + 5 \times 0.590)/9\); With pooled variance est. |
| Compare correctly with 1.383 | A1 | |
| Reject \(H_0\) and accept there is sufficient evidence that mean has increased so that the treatment has been successful | M1 | |
| OR: \(\bar{x}_T - \bar{x}_U \ge ks\sqrt{1/5 + 1/6}; = 0.713\); \(0.79 > 0.713\), reject \(H_0\) etc | M1A1 | Allow \(>\) or \(=\) |
| M1A1√ | 8 Or equivalent; Ft 0.713 |
**(i)** Assumes breaking strengths have normal normal distributions; Equal variances | B1, B1 | 2
**(ii)** $H_0: \mu_T = \mu_U$; $H_1: \mu_T > \mu_U$ where $\mu_T, \mu_U$ are means for treated and untreated thread | B1 | For both hypotheses
$\bar{x}_T = 18.05$, $\bar{x}_U = 17.26$ | B1 | May be implied below by 0.79
$s_T^2 = 0.715$, $s_U^2 = 0.738$ | B1 | Allow biased, 0.596, 0.590 if $s^2$
$s^2 = (5 \times 0.715 + 4 \times 0.738)/9$; EITHER: $(18.05 - 17.26)/\sqrt{s^2(1/5+1/6)} = 1.532$ | M1 | $=(6 \times 0.596 + 5 \times 0.590)/9$; With pooled variance est.
Compare correctly with 1.383 | A1 |
Reject $H_0$ and accept there is sufficient evidence that mean has increased so that the treatment has been successful | M1 |
OR: $\bar{x}_T - \bar{x}_U \ge ks\sqrt{1/5 + 1/6}; = 0.713$; $0.79 > 0.713$, reject $H_0$ etc | M1A1 | Allow $>$ or $=$
| M1A1√ | 8 Or equivalent; Ft 0.7133 A new treatment of cotton thread, designed to increase the breaking strength, was tested on a random sample of 6 pieces of a standard length. The breaking strengths, in grams, were as follows.
$$\begin{array} { l l l l l l }
17.3 & 18.4 & 18.6 & 17.2 & 17.5 & 19.3
\end{array}$$
The breaking strengths of a random sample of 5 similar pieces of the thread which had not been treated were as follows.
\section*{$\begin{array} { l l l l l } 18.6 & 17.2 & 16.3 & 17.4 & 16.8 \end{array}$}
A test of whether the treatment has been successful is to be carried out.\\
(i) State what distributional assumptions are needed.\\
(ii) Carry out the test at the $10 \%$ significance level.
\hfill \mbox{\textit{OCR S3 2007 Q3 [10]}}