best predicted weight

Question

Listed below are the overhead widths​ (cm) of seals measured from photographs and weights​ (kg) of the seals. Find the regression​ equation, letting the overhead width be the predictor​ (x) variable. Find the best predicted weight of a seal if the overhead width measured from a photograph is 2.4 ​cm, using the regression equation. Can the prediction be​ correct? If​ not, what is​ wrong? Use a significance level of 0.05.

Overhead Width​ (cm)	7.3	7.4	9.8	9.4	8.9	8.4
Weight​ (kg)	148	183	284	240	239	224

Answer 1

Hi Amanda C.,

Great question!

To find the best predicted weight of a seal with an overhead width of 1.8 cm, we first need to establish the linear regression equation based on the given data points. The linear regression equation has the form �=��+�y=mx+b, where �y is the dependent variable (weight in this case), �x is the independent variable (overhead width), �m is the slope, and �b is the y-intercept.

The data given are:

1. Overhead Width (cm): [7.2,7.3,9.7,9.3,8.9,8.5][7.2,7.3,9.7,9.3,8.9,8.5]
2. Weight (kg): [128,163,259,216,221,211][128,163,259,216,221,211]

Let's start by calculating the regression equation.

The regression equation is Weight=42.84×Overhead Width−163.74Weight=42.84×Overhead Width−163.74.

The �R-value, which is a measure of how well the model fits the data, is approximately 0.9530.953. This suggests a strong positive correlation between the variables.

The �p-value is 0.00330.0033, which is less than the significance level of 0.050.05. Therefore, we reject the null hypothesis and conclude that the regression model is statistically significant.

Now, let's use the regression equation to find the best predicted weight for a seal with an overhead width of 1.8 cm.

The best predicted weight for a seal with an overhead width of 1.8 cm, according to the regression equation, is approximately −86.63−86.63 kg.

The prediction is clearly not correct; it's not even physically possible for the weight of a seal to be negative. What's going wrong here? The issue arises from extrapolation. The given overhead width of 1.8 cm is far outside the range of the data used to create the model, which makes the prediction unreliable.

So, while the regression model fits well for the data within the observed range, it's not a reliable tool for making predictions outside that range. It's a crucial reminder that statistical tools, no matter how robust, have limitations dictated by the data they are based on.

Thank you,

Benjamin M.

p.s. If this helps, I would greatly appreciate a favorite or comment as I am new to this platform. :)