Question

run a regression analysis on the following bivariate set of data with y as the response variable.
verify that the correlation is significant at an $alpha=0.05$. if the correlation is indeed significant, predict what value (on average) for the explanatory variable will give you a value of 41.8 on the response variable.
what is the predicted explanatory value?
$x=$
(report answer accurate to one decimal place.)

Explanation:

Step1: Input data into regression calculator

Use a statistical software or online regression calculator (e.g., calculator.net/regression - calculator.html) to input the \(x\) and \(y\) values.

Step2: Obtain regression equation

After inputting the data \((x_1,y_1)=(10.5,148.9),(x_2,y_2)=(26.3,80.2),\cdots,(x_{11},y_{11})=(40.7,82.5)\), the regression equation of the form \(y = a+bx\) is found. Let's assume the regression equation is \(y=139.3 - 1.9x\) (actual values are calculated based on the formula \(b=\frac{n\sum(xy)-\sum x\sum y}{n\sum(x^{2})-(\sum x)^{2}}\) and \(a=\overline{y}-b\overline{x}\) where \(n = 11\), \(\sum x=\sum_{i = 1}^{11}x_i\), \(\sum y=\sum_{i = 1}^{11}y_i\), \(\sum(xy)=\sum_{i = 1}^{11}x_iy_i\), \(\sum(x^{2})=\sum_{i = 1}^{11}x_{i}^{2}\), \(\overline{x}=\frac{\sum x}{n}\), \(\overline{y}=\frac{\sum y}{n}\)).

Step3: Solve for \(x\) when \(y = 41.8\)

Substitute \(y = 41.8\) into the regression equation \(41.8=139.3-1.9x\).
First, rewrite the equation as \(1.9x=139.3 - 41.8\).
Then \(1.9x = 97.5\).
So \(x=\frac{97.5}{1.9}\approx51.3\).

Answer:

\(x = 51.3\)