b Let's start by making a .
As we can see, there is a negative linear association between the length of an organelle and the diameter of the cell. To find a linear regression between the length of the organelle and the diameter of the cell, we first have to enter the values into lists. Push STAT, choose Edit, and then enter the values in the first two columns.
To view the linear regression analysis of the dataset push STAT, scroll right to view the CALC options, and then choose the fourth option in the list, LinReg.
The linear regression is y=- 1.6+49.5 This tells us that as the length of the organelle increases by 1, the diameter of the cell decreases by 1.6.
From the regression, we see that the r-value is - 0.92. This means we have a strong negative correlation. Also, the R-squared is 0.86, which means 86 % of the variability in the diameter of the cell is explained by the length of the organelle. Let's add the linear regression to the scatterplot.
The upper and lower boundary have the same as the line of best fit, and they are equidistant from the line of best fit.
Upper boundary: y=- 1.6x+b_u
Lower boundary: y=- 1.6x+b_l
To determine the b_u and b_l, we have to find the observation that is farthest away from the line of best fit. In other words, we have to find the largest residual. Since the residual is the actual value minus the predicted value, we first want to find all of the predicted values.
c|l|c
x & - 1.6x+49.5 & Predicted
2 & - 1.6( 2)+49.5 & 46.3
4 &- 1.6( 4)+49.5 & 43.1
6 &- 1.6( 6)+49.5 & 39.9
7 & - 1.6( 7)+49.5 & 38.3
9 &- 1.6( 9)+49.5 & 35.1
With this information, we can calculate the residual and identify which one is the largest.
c|c|c|c
x & Actual & Predicted & Residual
2 & 46 & 46.3 & - 0.3
9 & 34 & 35.1& 1.1
7 & 36 & 38.3 & - 2.3
4 & 42 & 43.1 & - 1.1
4 & 46 & 43.1 & 2.9
9 & 36 & 35.1 & 0.9
6 & 42 & 39.9 & 2.1
2 & 45 & 46.3 & - 1.3
The greatest residual is 2.9. Since the residual is positive, we know that the upper boundary will be a straight line through the observation (4,46).
y=- 1.6x+b_u
Let's substitute this point in the equation and solve for b_u.
y=- 1.6x+b_u
46=- 1.6( 4)+b_l
b_l=52.4
Finally, to make sure that the linear association is appropriate, we will also include a residual plot.
As we can see, the residuals are scattered evenly about the x-axis, which means the linear association is appropriate.
