N2020 P2 Q7

N2020 P2 Q7

Junior College 2
8 marks
Free

A study into germination of parsnip seeds in the 1980s produced the following data for the average number of days, \(d\), taken for a seed to germinate at different soil temperatures, \(t\), measured in degrees Fahrenheit.

\(t\)\(32\)\(41\)\(50\)\(59\)\(68\)
\(d\)\(172\)\(57\)\(27\)\(19\)\(14\)
  1. Sketch a scatter diagram of the data. State the product moment correlation coefficient between \(d\) and \(t\).[2]
  2. Lim thinks the data can be modelled by the regression equation \(d=-a+bu\), where \(u=\frac1t\). Find the values of \(a\) and \(b\) for Lim's model, giving the values correct to 3 significant figures. State the product moment correlation coefficient between \(d\) and \(u\).[3]
  3. The study also found that, at a soil temperature of 86 degrees Fahrenheit, parsnip seeds took an average of 32 days to germinate. Determine whether Lim’s model fits this additional data.[1]
  4. A temperature of \(F\) degrees Fahrenheit is equivalent to a temperature of \(C\) degrees Celsius, where \(C=\frac59(F-32)\). Write Lim's equation from part (ii) in terms of \(d\) and \(T\), where \(T\) is the temperature in degrees Celsius.[2]

Default solution

(i) Plot \((32,172),(41,57),(50,27),(59,19),(68,14)\), with \(t\) on the horizontal axis and \(d\) on the vertical axis. The correlation is \(r_{dt}=-0.848\).

(ii) With \(u=1/t\), least-squares regression gives \(d=-145.12547+9456.40257u\). Thus \(a=145\) and \(b=9.46\times10^3\) to 3 significant figures, and \(r_{du}=0.941\).

(iii) At \(t=86\), the model predicts about \(-35.2\) days, which is impossible and far from the observed \(32\). It does not fit the additional data.

(iv) Since \(F=\frac95T+32\), \(d=-145+\frac{9460}{\frac95T+32}\), using the rounded coefficients from part (ii).

Answer:(i) \(-0.848\); (ii) \(a=145,\ b=9.46\times10^3,\ r=0.941\); (iii) No; (iv) \(d=-145+\frac{9460}{\frac95T+32}\)

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