a)The relationship between number of calories and amount of calbohydarates in grams that food menu items contain seems to have positive correlation.
b)Explanatory = Calories, Response = Carbs
c)Fitting a regression line will enable predicting the number of carbs in gram for a given number of calories.
d)No, these data do not satisfy for fitting a least square line.

Ans:a)

mean.shoulder <- 107.20
sd.shoulder <- 10.37
mean.height <- 171.14
sd.height <- 9.41

R <- 0.67

b1 <- R * (sd.height / sd.shoulder)
b1
## [1] 0.6079749
b0 <- mean.height - b1 * mean.shoulder
b0
## [1] 105.9651

\[ \hat{height}=105.965+0.608∗shoulder \]

b)The slope is 0.6079749 – So, for every 10 cm increase in soulder girth, there will be an aditional 6.08cm to the height. But not all values makes sense as we plug them into linear regression equation. for ex: with a shoulder width 0cm (hypothetically) still indicates a height of 105.965 which doesn’t makes sense.
c)calculate R^2 - About 45% of the variability in height is account for by shoulder girth

R^2
## [1] 0.4489

d)random prediction

random.student.s <- 100
student.height <- b0 + (b1 * random.student.s)
student.height
## [1] 166.7626

e)The residual means- negative

160-student.height
## [1] -6.762581

f)No value is outside the observered values.

a)below equation \[ \hat{y}=−0.357+4.034∗body.weight \]

b)Expected heart rate of a cat with 0 kg of body weight is -0.357, Since these are not meaningful values, the y-intercept serves to adjust the height of the regression line.

c)For each additional kg increase in body weight, we expect an additional 4.034 grams in the heart weight of a cat.

d)About 65% of the variability in weight is accounted for by body weight

e)0.8041144

R2 <- 0.6466
sqrt(R2)
## [1] 0.8041144

a)below is model summary:

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

b)Yes. The slope is positive and the p-value is near 0
c) Linearity: It’s not certain that we are seeing a linear trend.
Nearly normal residuals: The histogram of the residuals is left-skewed. Residual values don’t follow the qq line.
Constant variability: As the beauty score increases, the residual variability appears to decrease.