OCR Further Statistics 2018 March — Question 9 8 marks

Exam BoardOCR
ModuleFurther Statistics (Further Statistics)
Year2018
SessionMarch
Marks8
TopicLinear regression
TypePrediction with confidence or prediction intervals
DifficultyChallenging +1.2 This is a straightforward application of confidence intervals for predicted mean response in regression. Students must find the regression line from summary statistics, calculate the predicted mean at x=20, then apply the standard formula for confidence interval of mean prediction using given variance. While it requires multiple steps (regression coefficients, standard error calculation, critical value), each step follows a standard procedure taught in Further Statistics with no novel insight required. The given variance simplifies what would otherwise be a more complex calculation.
Spec5.05d Confidence intervals: using normal distribution5.09a Dependent/independent variables5.09b Least squares regression: concepts5.09c Calculate regression line

9 The values of a set of bivariate data \(\left( x _ { i } , y _ { i } \right)\) can be summarised by $$n = 50 , \sum x = 1270 , \sum y = 5173 , \sum x ^ { 2 } = 42767 , \sum y ^ { 2 } = 701301 , \sum x y = 173161 .$$ Ten independent observations of \(Y\) are obtained, all corresponding to \(x = 20\). It may be assumed that the variance of \(Y\) is 1.9 , independently of the value of \(x\). Find a \(95 \%\) confidence interval for the mean \(\bar { Y }\) of the 10 observations of \(Y\). \section*{END OF QUESTION PAPER}

AnswerMarks Guidance
\(S_{xx} = \frac{42767}{50} - \left(\frac{1270}{50}\right)^2 = 210.18\)M1 Correct method for \(b\)
\(S_{xy} = \frac{173161}{50} - \frac{1270}{50} \times \frac{5173}{50} = 835.336\)A1 Correct value of \(b\), in [3.97, 3.98]
\(b = S_{xy}/S_{xx} = 3.97438...\)A1 Correct value of \(a\), awrt 2.51
\(a = 103.46 - 25.4 \times b = 2.510...\)M1 Estimate \(y\)
\(y = 2.510 + 20 \times 3.974\)A1 Correct value
\(= 82.0\)M1 \(y \pm z\sigma^2/n\)
\(\hat{y} \pm 1.96\sqrt{\frac{1.9}{10}}\)A1 \(z = 1.96\)
\(= (81.14, 82.85)\)A1 Both correct, awrt 4 sf. Allow [81.1, 81.2] and [82.8, 82.9] only if all other numbers fully correct
$S_{xx} = \frac{42767}{50} - \left(\frac{1270}{50}\right)^2 = 210.18$ | M1 | Correct method for $b$

$S_{xy} = \frac{173161}{50} - \frac{1270}{50} \times \frac{5173}{50} = 835.336$ | A1 | Correct value of $b$, in [3.97, 3.98]

$b = S_{xy}/S_{xx} = 3.97438...$ | A1 | Correct value of $a$, awrt 2.51

$a = 103.46 - 25.4 \times b = 2.510...$ | M1 | Estimate $y$

$y = 2.510 + 20 \times 3.974$ | A1 | Correct value
$= 82.0$ | M1 | $y \pm z\sigma^2/n$

$\hat{y} \pm 1.96\sqrt{\frac{1.9}{10}}$ | A1 | $z = 1.96$

$= (81.14, 82.85)$ | A1 | Both correct, awrt 4 sf. Allow [81.1, 81.2] and [82.8, 82.9] only if all other numbers fully correct
9 The values of a set of bivariate data $\left( x _ { i } , y _ { i } \right)$ can be summarised by

$$n = 50 , \sum x = 1270 , \sum y = 5173 , \sum x ^ { 2 } = 42767 , \sum y ^ { 2 } = 701301 , \sum x y = 173161 .$$

Ten independent observations of $Y$ are obtained, all corresponding to $x = 20$. It may be assumed that the variance of $Y$ is 1.9 , independently of the value of $x$. Find a $95 \%$ confidence interval for the mean $\bar { Y }$ of the 10 observations of $Y$.

\section*{END OF QUESTION PAPER}

\hfill \mbox{\textit{OCR Further Statistics 2018 Q9 [8]}}