A
The relationship between calories and carbs is linear where the more calories there are, the more carbs there should be. The Residual scatterplot is closer to the zero line when calories are less than 300. It slowly expands away from the line the higher the calories there are. The residual histogram show a normal distribution that’s unimodal where the majority of the residuals are near or at zero.
B
Explanatory variable is calories.
Response variable is carbs
C
A regression line show us if the linear model is the best fit for our data. Since the residuals are the difference between the expected and observerd values, we want the residuals to be as close to the line as possible.
D
This data does not meet the requirements
Linearity - It does show a linear trend
Normal Residuals - Residuals seem normal without any major outliers
Constant Variability - The variability of points around the lest squares line does not remain constant. The higher the calories, the further away from the line the residuals are.
A
\(\hat {height} = B_0 + B_1(shoulder \ girth)\)
B
Slope of the least squares line:
\(b_1 = \frac{s_y}{s_x}R = \frac{10.37}{9.41}*(.67)\)
\(b_1 = .738\)
Point-slope equation:
\(y - y_0 = b_1(x-x_0)\)
\(y - 171.14 = 0.738(x - 107.2)\)
\(height = 91.99 + 0.738(shoulder \ girth)\)
C
\(R^2 = 0.67^2 = 0.4489\)
About 45% of the variance in height is explained by the shoulder girth.
D
\(height = 91.99 + 0.738(shoulder \ girth)\)
\(y = 165.79cm\)
E
165.79 - 160 = 5.79
The residual is 5.79cm away from the average mean. This means the linear model overestimated his height.
F
If the data meets the conditions for a least square line, then yes, we should be able to predict the height of the child from the shoulder girth. If the data is skewed, the distribution isn’t normal or the residual variablility isn’t constant, more predictor variables should be used or a more sophisticated linear model should be used or the height should not be predicted at all.
A
\(\hat{heart \ wt} = -0.357 + 4.034(body \ wt)\)
B
If the body weight was 0, the heart would weight -0.357g. Obviously, this isn’t a realistic value, but only is used to adjust the regression line.
C
For every 4.034kg of body weight, there is a predicted 1g of heart weight.
D
\(R^2\) value means that the body weight explains 64.66% of the variance in height weight for this data.
E
sqrt(.6466)## [1] 0.8041144A
\(T = \frac{B_1 - 0}{SE}\)
\(4.13 = \frac{B_1}{0.0322}\)
\(B_1 = 0.133\)
B
With a slope ration this low, even though there is a postive trend, it is so slight that it is hard to judge if there is a positive correlation or not. But the p-value is so low that we can conclude that there is a positive trend between beauty and the teacher evalutions.
D
Linearity
It is hard to tell linearity from these plots.
Nearly Normal Residuals
The residuals seem nearly normal. There are some outliers in the scatterplots but aren’t drastic enough to cause concern.
Constant Variability
The variability of the residuals look constant. The distribution is centered around the zero line with maybe a slight skew to the left.
Independency
We can assume these observations were independent of one another.