a) There appears to be a weak positive correlation between the amount of calories and the amounts of carbs in Starbucks menu items. The relation appears to be linear, but is somewhat weak, so the linearity is subjective.
b) The explanatory variable is calories and the response variable is carbs
c) We might have a diet app and want to estimate the number of carbs as new items appear on the menu, so points can be assessed for the item
d) Linearity is fine, as are indepedance and normality (at least conditionally normal). There appears to be heteroskedacity. I would be wary of that, but it doesn’t look like a dealbreaker because varinace appears to be constant once we get to 300 calories. A gamma distribution, for example, would have varinace contantly increasing.
a) \[ b\quad =\quad \frac { { s }_{ y } }{ { s }_{ x } } *\quad r \] \[{ b }_{ 0 }\quad =\quad \overline { y } \quad -\quad { b }_{ 1 }\quad *\quad \overline { x } \]
b1 <- 9.41 / 10.37 * .67
b0 <- 171.14 - b1* 107.2y = 0.6079749 x X + 105.9650878
b) According to the regression line, a person with a 0 cm sholder width will be 106 cm tall. Every additional cm in shoulder width will add .6 cm in height. The intercept is nonsensical because a 0 cm shoulder width is not possible.
c) \[{ R }^{ 2 }\] = 0.4489
44.98% of the variance in height can be explained by shoulder girth
d) \[height\quad =\quad 100*{ b }_{ 1 }\quad +\quad 105.96\] =166.77
e) residual = 205.17
f) For an individual with shouler girth of about 5 standard deviations away from the mean of the training data, the model might not be appropriate. This is extrapolation, which is generally not recommended. The training data is likely taken from adults, so the model won’t generalize to infants
a) \[y\quad =\quad 4.034x\quad -.357\]
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
a) slope = t x SE = 0.132986
b) The slope estimate is over 4 SE greater than 0. This appears to be convincing evidence to reject the null hypothesis that the slope is 0.
c)
-Linearity There is no pattern in the residuals or in the data. This is confirmed
-Normality of residuals The residuals are appropriatly scattered and the histogram and qq plots are approximately normal. This is confirmed
-Constant Variance of residuals There appears to be no difference in variance at the lower levels of professor rating compared with the upper. This is confirmed
-Indepednance of observations There is no pattern in the residuals based on order of data collection. This is confirmed