Hypothesis tests: correlation and Normal mean – Year 13 A level maths test answers to download as PDF

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1. Answer to Question 1 (4 marks)

   1. \(H_0:\rho=0\), \(H_1:\rho\ne0\) where \(\rho\) is the population correlation coefficient between temperature and ice lolly sales (2).
   2. \(0.58\gt 0.5760\), so reject \(H_0\) (1): there is sufficient evidence at the 5% level of correlation between temperature and ice lolly sales (1).

   The test is two-tailed because the owner only asks whether there is correlation, not its direction. The value 0.58 only just exceeds 0.5760, so the evidence is weak: with 12 days of data, a slightly lower \(r\) would not be significant. Hypotheses must use \(\rho\), not \(r\), to score in (a).

2. Answer to Question 2 (4 marks)

   Method: if \(X\sim N(\mu,\sigma^2)\), the mean of a random sample of size \(n\) satisfies \(\bar X\sim N\left(\mu,\dfrac{\sigma^2}{n}\right)\). The standard error is \(\dfrac{\sigma}{\sqrt n}\); standardise with \(Z=\dfrac{\bar X-\mu}{\sigma/\sqrt n}\). Mark scheme: 1 mark for part a), 1 mark for part b), 2 marks for part c).

   1. The standard error is \(\dfrac{15}{\sqrt{25}}=3\). \(\bar X\sim N(75,3^2)\) (1).
   2. Standardise: \(z=\dfrac{72-75}{3}=-1\). \(P(Z\lt -1)=0.1587\) (1). So \(P(\bar X\lt 72)\approx0.159\).
   3. Standardise: \(z=\dfrac{80-75}{3}=1.667\), then take the upper tail. \(P(\bar X\gt 80)=P(Z\gt 1.667)=0.0478\) (2). So \(P(\bar X\gt 80)\approx0.0478\).

3. Answer to Question 3 (5 marks)

   Method: state the hypotheses about \(\mu\), find the distribution of \(\bar X\) under \(H_0\), calculate the test statistic or \(p\)-value, compare with the significance level, then conclude in context. Mark scheme as shown in brackets.

   1. \(H_0:\mu=330\), \(H_1:\mu\lt 330\) (1). \(\bar X\sim N\left(330,\dfrac9{40}\right)\), standard error 0.4743; \(z=\dfrac{-1.1}{0.4743}=-2.319\) (1). \(p=0.0102\lt 0.05\) (1), so reject \(H_0\): sufficient evidence that the mean volume is less than 330 ml (1).
   2. \(0.0102\gt 0.01\), so at the 1% level we do not reject \(H_0\): the conclusion changes (1).

   So there is evidence at the 5% level, but not at the 1% level, that the bottles contain less than 330 ml on average.

4. Answer to Question 4 (4 marks)

   Method: the critical region is the set of values of \(\bar x\) that lead to rejecting \(H_0\). For a two-tailed test at 5%, put 2.5% in each tail and use \(z=\pm1.96\) with the standard error \(\dfrac{6}{\sqrt{16}}=1.5\).

   1. \(\bar X\sim N(80,1.5^2)\) (1). Critical values \(80\pm1.96\times1.5=77.06\) and \(82.94\) (1). Critical region: \(\bar x\le77.06\) or \(\bar x\ge82.94\) (1).
   2. \(83.5\gt 82.94\), in the critical region: reject \(H_0\) (1).

   Conclusion in context: there is evidence at the 5% level that the population mean is not 80.

5. Answer to Question 5 (3 marks)

   1. One-tailed: \(0.41\gt 0.3783\), reject \(H_0\) (1). Two-tailed: \(0.41\lt 0.4438\), do not reject \(H_0\) (1).
   2. The one-tailed test puts the whole 5% in one tail, so its critical value is smaller; the direction must be decided before looking at the data (1).

   So the same sample gives evidence of positive correlation with a one-tailed test but no evidence of correlation with a two-tailed test. Choosing the one-tailed version after seeing \(r=0.41\) would be cheating: the true significance level would no longer be 5%. The hypotheses for (a) are \(H_0:\rho=0\) and \(H_1:\rho\gt0\) or \(H_1:\rho\ne0\).

Total : 20 marks