cs-5821/hw2/answers

1. Describe the null hypotheses to which the p-values given in Table
   3.4 correspond. Explain what conclusions you can draw based on
   these p-values. Your explanation should be phrased in terms of
   sales , TV , radio , and newspaper , rather than in terms of the
   coefficients of the linear model.


	P-values that are very small indicate that the model for that
	predictor is likely to account for a significant amount of the
	association between the predictor and the response. If that is
	true, then, we reject the null hypothesis, and conclude that a
	relationship exists between the predictor and the response. The
	p-values computed from the response of sales to marketing budget
	for each marketing paradigm indicate will give us insight into
	which of these predictors have a strong relationship with sales
	of this product.

    TV marketing and radio marketing both have a strong relationship
    to sales, according to their linear regression p-values, but
    newspaper advertising does not appear to be effective, given
    that the linear model does not account for much of the variation
    in sales across that domain. We can conclude that cutting back
    on newspaper advertising will likely have little effect on the
    sales of the product, and that increasing TV and radio
    advertising budgets likely will have an effect. Furthermore, we
    can see that radio advertising spending has a stronger
    relationship with sales, as the best-fit slope is significantly
    more positive than the best fit for TV advertising spending, so
    increasing the radio advertising budget will likely be more
    effective.



3. Suppose we have a data set with five predictors, X₁ = GPA, X₂ =
   IQ, X₃ = Gender (1 for Female and 0 for Male), X₄ = Interaction
   between GPA and IQ, and X₅ = Interaction between GPA and Gender.
   The response is starting salary after graduation (in thousands of
   dollars). Suppose we use least squares to fit the model, and get
   β₀ = 50, β₁ = 20, β₂ = 0.07, β₃ = 35, β₄ = 0.01, β₅ = −10.

    This is the model: ŷ = 50 + 20 X₁ + 0.07 X₂ + 35 X₃ + 0.01 X₄ +
        -10 X₅

        For fixed IQ and GPA, we can infer that the starting salary
        for a female sharing an IQ and GPA with her male counterpart
        will make (35*1 - 10*(GPA*1)) more starting salary units
        than her male counterpart. This means that at very low
        GPAs(maybe this includes people who didn't attend school?),
        males have a lower starting wage, and as GPA grows, males
        make a larger starting salary from that point, overtaking
        females at GPA=3.5. Therefore,

    (a) Which answer is correct, and why? →  iii. For a fixed value
        of IQ and GPA, males earn more on average than females
        provided that the GPA is high enough.
            
            This one is correct.

    (b) Predict the salary of a female with IQ of 110 and a GPA of
    4.0.

        ŷ = 50 + 20*4.0 + 0.07*110 + 35*1 + 0.01*(4.0*110) - 10*(4.0*1)

        → ŷ = 137.1 salary units

    (c) True or false: Since the coefficient for the GPA/IQ
    interaction term is very small, there is very little evidence of
    an interaction effect. Justify your answer.

        False. There is still a noticeable effect because the
        coefficient for IQ's effect alone is only 7 times greater
        than the coefficient of the interaction term. So, this term
        holds significant weight compared to the overall
        response of the model to IQ.


4. I collect a set of data (n = 100 observations) containing a
   single predictor and a quantitative response. I then fit a linear
   regression model to the data, as well as a separate cubic
   regression, i.e. Y = β₀ + β₁ X + β₂ X² + β₃ X³ + ε.

	(a) Suppose that the true relationship between X and Y is
	linear, i.e. Y = β₀ + β₁ X + ε. Consider the training residual
	sum of squares (RSS) for the linear regression, and also the
	training RSS for the cubic regression. Would we expect one to be
	lower than the other, would we expect them to be the same, or is
	there not enough information to tell? Justify your answer.

        For the training data, the cubic regression might return a
        better RSS than the linear regression, but this would only
        be because the cubic is fitting points that are varied
        according to the ε random error. It also may not, depending
        on how that random error expressed itself in this case.

	(b) Answer (a) using test rather than training RSS.

        For the test error, the RSS will almost certainly be greater
        for the cubic model than the linear model, because the
        random error ε will likely express itself in a way that is
        inconsistent with the noise that the cubic model adopted
        during its training. The linear model will be more likely to
        have a lower RSS the more test data is used against the
        models.

	(c) Suppose that the true relationship between X and Y is not
	linear, but we don’t know how far it is from linear. Consider
	the training RSS for the linear regression, and also the
	training RSS for the cubic regression. Would we expect one to be
	lower than the other, would we expect them to be the same, or is
	there not enough information to tell? Justify your answer. (d)
	Answer (c) using test rather than training RSS.

        The cubic model will pick up more information because of its
        additional degrees of freedom. If the true relationship is
        more complex than linear, then the cubic model will likely
        have a lower RSS over the linear model. If the model is less
        complex than linear (E.G. perhaps it is just a constant
        scalar relationship) then the linear model will still be
        more likely to have a smaller RSS, because the cubic will
        again pick up information from the ε noise that is not
        inherent in the real relationship.





8. This question involves the use of simple linear regression on the Auto
data set.

    (a) Use the lm() function to perform a simple linear regression
    with mpg as the response and horsepower as the predictor. Use
    the summary() function to print the results. Comment on the
    output. For example:

        There is definitely a correlation between horsepower and
        mpg. The RSE is ~4.9, which is not insignificant and does
        indicate that the response may not be truly linear, but it
        is small enough relative to the mpg magnitude that it's
        clear a relationship exists. The R² statistics corroborates
        this by indicating (it has a small value at ~0.6) that a
        large proportion of the mpg variability is explained by the
        model. mpg has a negative correlation with horsepower,
        indicated by the negative coefficient on the horsepower
        factor.


        For example, for a vehicle with 98 horsepower, one can
        expect with 95% confidence that the mpg will be within 23.97
        and 24.96, if the vehicles follow our model. However, after
        incorporating the irreducible error, the prediction turns
        out to be much less precise, with a 95% prediction interval
        spanning 14.8 to 34.1. Some of this variability may also be
        reduced by using a quadratic model, from visual inspection
        of the plot.

    (b) Plot the response and the predictor. Use the abline() function
    to display the least squares regression line.

        Attached.

    (c) Use the plot() function to produce diagnostic plots of the least
    squares regression fit. Comment on any problems you see with
    the fit.

        Attached. From these four plots it's clear there is a lot of
        variability that remains unexplained by the linear model.
        The standardized residuals plotted against the fitted values
        shows clearly that the variability is strong, with values
        consistenly lying outside 1 standardized residual unit, but
        still within a tight range that doesn't extend past 3, which
        is often considered an approximate threshold to indicate
        values that aren't explained well by the model. There are
        many points with high leverage, and these values have less
        residual by default, of course, and in both of these graphs
        we see a few points (323, 330) that are rearing their ugly
        heads. These seems to be the bit of "uptick" toward the
        higher end of the horsepower scale that would probably be
        picked up by a quadratic fit.