Data606-Homework7

Problem 7.24

a)The more calories in a Starbucks menu item, the more carbohydrates it is likely to have.
b)Calories is the explanatory variable and carbohydrates is the response variable.
c)We would fit a regression line to the data to decide if there is a linear relationship between the two variables.
d)The plot of the residuals shows heteroskadasticity. The variance is not stable for all ranges. This means a regression can’t be trusted across the interval and we would need other techniques to control for it.

Problem 7.26

a) height = 105.97 cm + .608 * shoulder girth

To calculate the regression line, b₁ = R * s_y/s_x = .679.41/10.37 = 0.6079749
y = b₀ + b₁ X —– 171.14 = b₀ + .708 107.2 —– b₀ = 171.14-.60797*107.2 = 105.965616
b) This regression means we expect someone’s height to be .608 cm more for every cm taller they are.
c) R² is .67² = 0.4489. This means the regression accounts for a moderate amount of the variation in height.
d) If someone has a shoulder girth of 100 cm, we expect that person to be 166.77 cm tall.
e)If that same person is actually 160 cm tall, the residual is -6.77 cm. The residual is the prediction error for a particular x_i, observation.
f) Suppose a one year old has a shoulder girth of 56. We would not be able to use this regression to predict height in that situation because it’s outside of the interval we fit the regression in.

Problem 7.30

a) The linear model is heart weight = -.357 g + 4.034 * body weight in kg.
b) The intercept moves the line to fit the regression up and down. It is the less interesting of the two variables.
c) The slope shows how much heavier we eaxpect a cat heart to weigh for each 1kg increase in total weight.
d) R² is 64.66% . It means we expect the regression to account for a majority of the variation.
e) The correlation coefficient is 0.8041144

Problem 7.40

a)3.9983 = 4.010 + b₁* (-.0883) — b₁ = 0.1325028

b) These data show that the relationship between teaching ratings and beauty rating is positive. In the last column f the presented output, it says the right tail of this t- score of 4.13 is about .0000.

c) Our conditions are met. The residuals appear to be fairly random. With 463 observations, the fact that the data is slightly left skewed does not prevent a t-test from being accepted. There does not appear to be any autocorrelation. We could use more tests to see if the residuals are random enough to accept our regression results.