It’s looking like a .5 correlation, moderate, positive, and linear relationship with high variability.

Explanatory would be our X axis, so calories. Response is Y axis, so carbs.

####Mostly to confirm (a), easy to make predictions and see residuals with a regression line.

You could use a least squares line. Wouldn’t be too good though. Constant variability is an issue, and its kinda skewed to the left.

Sx <- 10.37
Sy <- 9.41

R <- 0.67

b1 <- (Sy / Sx) * R

xhat <- 107.2
yhat <- 171.14

b0 <- yhat - b1 * xhat

The slope shows the increase in height as shoulder girth increases. The incerpet is the height with a shoulder girth of 0.

R2 <- .67^2
R2
## [1] 0.4489
height <- function(x) {
y  <- 105.97 + 0.61*x
return(y)
}

height(100)
## [1] 166.97
160 - height(100) 
## [1] -6.97

Looking back at 7.15 we see that the lowest shoulder girth was 85cm. We wouldn’t use our regression model to estimate this child’s height since he or she is out of our range.

a)y=4.034x???.357

b) The intercept is nonsensical because a cat can’t have 0 weight, and negative weight is also not possible. The intercept only serves to adjust the position of the line.

c) For every additional 1 kg in body weight, the heart will weigh an additional 4.034 grams on average

d) 65 percent of the varinace in heart weight is explained by body weight

e) r = 0.8041144

# calculate slope
b1 <- (4.010-3.9983) / (0-(-0.0883))
b1
## [1] 0.1325028

The slope b1 is positive (0.1325)indicate a positive correlation between beauty and teaching.

Linearity: The top left plot above shows that the data appear to be linear. There is no apparent curve or other pattern.

Nearly Normal Residuals: The two top plots show that the residuals appear to be nearly normal with just a slight left skew.

Constant Variability: The top left plot above shows no obvious pattern. The points are seem evenly spread out in the plot indicating constant variability.

Independent Observations: In this case we have no information about the independence of the observations so we have to assume that they are independent to use a least squares regression line.