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9231 P44 - Nov 2025 - Q1 - 6 marks
6621
Local residents are concerned about the speed of cars travelling through their village. They record the speed, in \(\mathrm{kmh}^{-1}\), of a random sample of 13 cars travelling through their village. The recorded speeds are as follows.
40
53
59
42
43
48
62
67
46
82
66
45
70
(a) Construct a \(90 \%\) confidence interval for the population mean speed of cars passing through the village. (b) State an assumption that is necessary for the confidence interval found in \(\mathbf{1}(\mathbf{a})\) to be valid.
Solution
Answer:
(a) The 90% confidence interval for the population mean speed is \(49.2 \lt \mu \lt 62.1\) \(\mathrm{kmh}^{-1}\), to 3 significant figures.
(b) The population of car speeds passing through the village is normally distributed.
(a) Since the population variance is unknown and the sample size is small, use a \(t\)-interval. For the sample, \(n=13\), \(\sum x=723\) and \(\sum x^2=42261\). Hence the sample mean is \(\bar{x}=\frac{723}{13}=55.615\ldots\). The unbiased sample variance is \(s^2=\frac{1}{n-1}\left(\sum x^2-\frac{(\sum x)^2}{n}\right)=\frac{1}{12}\left(42261-\frac{723^2}{13}\right)=170.923\ldots\), so \(s=13.073\ldots\). For a 90% confidence interval with \(12\) degrees of freedom, \(t_{0.95,12}=1.782\). Therefore the confidence interval is \(\bar{x}\pm t\frac{s}{\sqrt{n}}=55.615\ldots\pm 1.782\sqrt{\frac{170.923\ldots}{13}}\). This gives \(55.615\ldots\pm 6.462\ldots\), so the interval is \((49.153\ldots,62.077\ldots)\). Thus the 90% confidence interval is \(49.2 \lt \mu \lt 62.1\) \(\mathrm{kmh}^{-1}\), to 3 significant figures.
(b) The required assumption is that the population distribution of car speeds is normal.
