Question

the time (in minutes) taken for a local anaesthetic to take effect is associated with to the amount administered (in units). to investigate this association a researcher collected the following data.

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$$\begin{tabular}{|l|c|c|c|c|c|c|c|c|c|c|c|} \\hline amount & 0.5 & 0.6 & 0.7 & 0.8 & 0.9 & 1.0 & 1.1 & 1.2 & 1.3 & 1.4 & 1.5 \\\\ \\hline time & 3.7 & 3.6 & 3.4 & 3.3 & 3.2 & 3.0 & 2.9 & 2.7 & 2.5 & 2.3 & 2.1 \\\\ \\hline \\end{tabular}$$

the association between the variables amount and time is non-linear as can be seen from the scatterplot below. a squared transformation applied to the variable time will linearise the scatterplot.

a apply the squared transformation to the variable time and fit a least squares regression line to the transformed data. amount is the ev. write the equation of this line with the slope and intercept rounded to two significant figures.

b use the equation to predict the time for the anaesthetic to take effect when the dose is 0.4 units. give the answer rounded to one decimal place.

Explanation:

Transform the response variable

We apply a squared transformation to the variable \(y = \text{time}\). Let \(y^* = \text{time}^2\).
The explanatory variable is \(x = \text{amount}\).
We calculate \(y^*\) for each data point:

• For \(x = 0.5\), \(y^* = 3.7^2 = 13.69\)
• For \(x = 0.6\), \(y^* = 3.6^2 = 12.96\)
• For \(x = 0.7\), \(y^* = 3.4^2 = 11.56\)
• For \(x = 0.8\), \(y^* = 3.3^2 = 10.89\)
• For \(x = 0.9\), \(y^* = 3.2^2 = 10.24\)
• For \(x = 1.0\), \(y^* = 3.0^2 = 9.00\)
• For \(x = 1.1\), \(y^* = 2.9^2 = 8.41\)
• For \(x = 1.2\), \(y^* = 2.7^2 = 7.29\)
• For \(x = 1.3\), \(y^* = 2.5^2 = 6.25\)
• For \(x = 1.4\), \(y^* = 2.3^2 = 5.29\)
• For \(x = 1.5\), \(y^* = 2.1^2 = 4.41\)

Fit least squares regression line

We calculate the linear regression of \(y^*\) on \(x\):

• Number of data points \(n = 11\)
• Sum of \(x\): \(\sum x = 11.0\)
• Mean of \(x\): \(\bar{x} = 1.0\)
• Sum of \(y^*\): \(\sum y^* = 100.0\)
• Mean of \(y^*\): \(\bar{y}^* \approx 9.0909\)
• Sum of \(x^2\): \(\sum x^2 = 12.1\)
• Sum of \(x y^*\): \(\sum x y^* = 90.7\)

We compute the slope \(b\):

$$ b = \frac{\sum x y^* - n \bar{x} \bar{y}^*}{\sum x^2 - n \bar{x}^2} = \frac{90.7 - 11 \times 1.0 \times 9.0909}{12.1 - 11 \times 1.0^2} = \frac{90.7 - 100.0}{12.1 - 11.0} = \frac{-9.3}{1.1} \approx -8.4545 $$

We compute the intercept \(a\):

$$ a = \bar{y}^* - b \bar{x} \approx 9.0909 - (-8.4545) \times 1.0 \approx 17.545 $$

Rounding the slope and intercept to two significant figures:

• Slope: \(-8.5\)
• Intercept: \(18\)

The equation is:

$$ \text{time}^2 = 18 - 8.5 \times \text{amount} $$

Predict time for amount of 0.4

We substitute \(\text{amount} = 0.4\) into our regression equation:

$$ \text{time}^2 = 18 - 8.5 \times 0.4 = 18 - 3.4 = 14.6 $$

We solve for \(\text{time}\):

$$ \text{time} = \sqrt{14.6} \approx 3.821 $$

Rounding to one decimal place gives \(3.8\) minutes.

Answer:

Question a

$$ \text{time}^2 = 18 - 8.5 \times \text{amount} $$

Question b

\(3.8\) minutes