Statistical inference – university Year 1 maths test answers to download as PDF

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1. Answer to Question 1 (4 marks)

   1. The sample is large (\(n=50\)), so the normal approximation can be used with \(s\) in place of \(\sigma\). The standard error is \(\dfrac{4.5}{\sqrt{50}}=0.636\) (1). The interval is \(31.2\pm1.96\times0.636=31.2\pm1.25\) (1), that is \((29.95,\ 32.45)\) minutes (1).
   2. If the sampling procedure were repeated many times, about 95% of the intervals constructed in this way would contain the true mean commuting time (1). It does not mean that \(\mu\) has a 95% probability of lying in this particular interval.

2. Answer to Question 2 (4 marks)

   Since \(\sigma\) is known, use a \(z\)-test. The standard error is \(\dfrac{10}{\sqrt{40}}=1.58\), so \(z=\dfrac{78.5-75}{1.58}=\dfrac{3.5}{10/\sqrt{40}}=2.21\) (2). The alternative is two-sided, so \(p=2(1-\Phi(2.21))\approx0.027\), which is less than \(0.05\) (1). Equivalently, \(2.21>1.96\). We reject \(H_0\): at the 5% level there is evidence that the school's mean mark differs from 75, and the sample suggests it is higher (1).

   A 95% confidence interval, \(78.5\pm1.96\times1.58=(75.4,\,81.6)\), excludes 75, which agrees with rejecting \(H_0\).

3. Answer to Question 3 (4 marks)

   \(\sigma\) is unknown and the sample is small, so use a \(t\)-test. The estimated standard error is \(\dfrac{1.2}{\sqrt9}=0.4\), so \(t=\dfrac{10.8-10}{0.4}=\dfrac{0.8}{1.2/3}=2.0\) (2). There are \(9-1=8\) degrees of freedom and the test is one-sided, so the critical value is \(t_{8,\,0.95}=1.860\) (1). Since \(2.0>1.860\), the statistic is in the rejection region: we reject \(H_0\) and conclude that there is evidence at the 5% level that \(\mu>10\) (1).

4. Answer to Question 4 (4 marks)

   1. Each expected frequency is (row total \(\times\) column total) / grand total. The row totals are 60 and 60 and the column totals 70 and 50, so the expected frequencies are \(\dfrac{60\times70}{120}=35\) and \(\dfrac{60\times50}{120}=25\) in each row: \(35,\ 25,\ 35,\ 25\) (1).
   2. Every \(|O-E|=5\), so \(\chi^2=\dfrac{25}{35}+\dfrac{25}{25}+\dfrac{25}{35}+\dfrac{25}{25}\approx3.43\) (2). A \(2\times2\) table has \((2-1)(2-1)=1\) degree of freedom, with critical value 3.841. Since \(3.43\lt 3.841\), we do not reject independence: the data do not show a significant link between attending and passing (1).

5. Answer to Question 5 (4 marks)

   1. The gradient is \(b=\dfrac{S_{xy}}{S_{xx}}=\dfrac{63}{42}=1.5\) (1). The means are \(\bar x=\dfrac{40}8=5\) and \(\bar y=\dfrac{96}8=12\), and the line passes through \((\bar x,\bar y)\), so \(a=12-1.5\times5=4.5\). The regression line is \(y=4.5+1.5x\) (1).
   2. At \(x=6\): \(y=4.5+1.5\times6=13.5\) (1). This is interpolation, so it is reliable. At \(x=20\), far outside the observed range 1 to 9, the prediction would be extrapolation: the linear pattern may not continue, so it is unreliable (1).

Total : 20 marks