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Chapter 9: Problem 3

9.3 Baby weights, Part III. We considered the variables smoke and parity, one at a time, in modeling birth weights of babies in Exercises 9.1 and \(9.2 .\) A more realistic approach to modeling infant weights is to consider all possibly related variables at once. Other variables of interest include length of pregnancy in days (gestation), mother's age in years (age), mother's height in inches (height), and mother's pregnancy weight in pounds (weight). Below are three observations from this data set. $$ \begin{array}{rccccccc} \hline & \text { bwt } & \text { gestation } & \text { parity } & \text { age } & \text { height } & \text { weight } & \text { smoke } \\ \hline 1 & 120 & 284 & 0 & 27 & 62 & 100 & 0 \\ 2 & 113 & 282 & 0 & 33 & 64 & 135 & 0 \\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\ 1236 & 117 & 297 & 0 & 38 & 65 & 129 & 0 \\ \hline \end{array} $$ The summary table below shows the results of a regression model for predicting the average birth weight of babies based on all of the variables included in the data set. $$ \begin{array}{rrrrr} \hline & \text { Estimate } & \text { Std. Error } & \text { t value } & \operatorname{Pr}(>|\mathrm{t}|) \\ \hline \text { (Intercept) } & -80.41 & 14.35 & -5.60 & 0.0000 \\ \text { gestation } & 0.44 & 0.03 & 15.26 & 0.0000 \\ \text { parity } & -3.33 & 1.13 & -2.95 & 0.0033 \\ \text { age } & -0.01 & 0.09 & -0.10 & 0.9170 \\ \text { height } & 1.15 & 0.21 & 5.63 & 0.0000 \\ \text { weight } & 0.05 & 0.03 & 1.99 & 0.0471 \\ \text { smoke } & -8.40 & 0.95 & -8.81 & 0.0000 \\ \hline \end{array} $$ (a) Write the equation of the regression model that includes all of the variables. (b) Interpret the slopes of gestation and age in this context. (c) The coefficient for parity is different than in the linear model shown in Exercise 9.2 . Why might there be a difference? (d) Calculate the residual for the first observation in the data set. (e) The variance of the residuals is \(249.28,\) and the variance of the birth weights of all babies in the data set is 332.57. Calculate the \(R^{2}\) and the adjusted \(R^{2}\). Note that there are 1,236 observations in the data set.

Short Answer

Expert verified
(a) Regression equation is: \( -80.41 + 0.44 \times \text{gestation} - 3.33 \times \text{parity} \ldots \) (b) Gestation increases weight by 0.44 per day, age decreases it negligibly. (c) Parity coefficient varies because of other variables. (d) Residual is actual minus predicted birth weight. (e) \( R^2 = 0.250 \), Adjusted \( R^2 = 0.246 \).

Step by step solution

01

Write the Regression Equation

The equation of the regression model is given as follows, using the coefficients from the summary table: \[\text{bwt} = -80.41 + 0.44\times\text{gestation} - 3.33\times\text{parity} - 0.01\times\text{age} + 1.15\times\text{height} + 0.05\times\text{weight} - 8.40\times\text{smoke}\]
02

Interpret Gestation Slope

The coefficient for gestation is 0.44. This means that for each additional day of gestation, the average birth weight of a baby is expected to increase by 0.44 units (grams, if weight is measured in grams), holding all other variables constant.
03

Interpret Age Slope

The coefficient for age is -0.01, indicating that for each additional year of the mother's age, the average birth weight of a baby decreases by 0.01 units, holding other variables constant. However, since the t-value suggests insignificance (p-value = 0.9170), the effect may not suffice notable influence.
04

Analyze Parity Coefficient Difference

The coefficient for parity differs from Exercise 9.2 because it accounts for the influence of other variables in this model. Variables like gestation and weight might have confounding effects, modifying or clarifying the relationship between parity and birth weight.
05

Calculate Residual for Observation 1

For the first observation, substitute the values into the regression equation: - Gestation = 284- Parity = 0- Age = 27- Height = 62- Weight = 100- Smoke = 0Using the equation, calculate the predicted birth weight and subtract it from the actual birth weight (120) to find the residual.Predicted bwt = \(-80.41 + 0.44(284) - 3.33(0) - 0.01(27) + 1.15(62) + 0.05(100) - 8.40(0)\).Residual = 120 - Predicted bwt.
06

Calculate R-Squared

The formula for \( R^2 \) is: \[ R^2 = 1 - \frac{\text{Variance of Residuals}}{\text{Variance of Birth Weights}} \]Substitute the given variances:\[ R^2 = 1 - \frac{249.28}{332.57} \]
07

Calculate Adjusted R-Squared

The formula for the adjusted \( R^2 \) is: \[ \text{Adjusted } R^2 = 1 - \left(1 - R^2\right) \left(\frac{n - 1}{n - k - 1}\right)\]Where \( n \) is the number of observations (1236) and \( k \) is the number of predictors (6).Substitute the values and calculate the adjusted \( R^2 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Birth Weight Predictors
When trying to understand the factors that affect the birth weight of babies, it's essential to consider a range of possible predictors. These include:
  • Gestation: The length of the pregnancy in days.
  • Parity: Whether the mother has previously given birth.
  • Age: The age of the mother in years.
  • Height: The height of the mother in inches.
  • Weight: The weight of the mother during pregnancy in pounds.
  • Smoking Status: Whether the mother smoked during pregnancy.
Each of these variables can influence the birth weight of a newborn in different ways. By considering them all in a regression model, we gain a more comprehensive understanding of what factors are most predictive.
Gestation and Birth Weight
Gestation refers to the total number of days of pregnancy, and it is a crucial factor in predicting birth weight. Generally, the longer the gestation period, the higher the chance of a baby being heavier at birth. In the regression model, the coefficient for gestation is 0.44. This indicates that, on average, for each additional day of gestation, the birth weight increases by 0.44 grams, assuming other factors remain unchanged.
However, gestation doesn't act in isolation. It interacts with other predictors, producing a combined effect on birth weight. By examining gestation alongside other factors, researchers can isolate its specific impact, explaining the variation in birth weight more precisely.
Interpretation of Regression Coefficients
In regression analysis, coefficients represent the expected change in the dependent variable, birth weight in this case, for a one-unit change in a predictor while keeping other predictors constant. Here are interpretations for some coefficients:
  • Gestation (0.44): Each additional day increases birth weight by 0.44 grams.
  • Parity (-3.33): Being a first-time mother (parity of 0) tends to result in a lower birth weight compared to mothers who have already given birth.
  • Age (-0.01): Suggests a tiny decrease in birth weight with mother's increasing age, though statistically insignificant.
  • Height (1.15): Taller mothers are associated with larger babies, increasing birth weight by 1.15 grams per extra inch.
  • Weight (0.05): A slight increase in birth weight for every additional pound the mother weighs.
  • Smoke (-8.40): Smoking during pregnancy substantially decreases birth weight, by about 8.40 grams.
These interpretations help in understanding how each predictor individually affects the baby's birth weight.
Variance and R-squared Calculation
The variance and R-squared (\( R^2 \)) calculations are integral in regression analysis, providing insight into how well the model explains the observed data. Variance of residuals, which is 249.28 in this instance, indicates the extent of variation in birth weights not explained by the regression model.
R-squared is calculated with the formula:\[R^2 = 1 - \frac{\text{Variance of Residuals}}{\text{Variance of Birth Weights}}\]Given that the variance of birth weights is 332.57, substituting these values provides the\( R^2 \)value, representing the proportion of variation in birth weight explained by the model.
Adjusted\( R^2 \), factoring in the number of predictors and observations, tends to be a more accurate reflection of the model's explanatory power when comparing models with different numbers of predictors. It is calculated using:\[\text{Adjusted } R^2 = 1 - \left(1 - R^2\right) \left(\frac{n - 1}{n - k - 1}\right)\]where \( n \) is the number of observations (1236) and \( k \) is the number of predictors (6). This ensures the\( R^2 \)accounts for the sensitivities to sample size and number of predictors, giving a truer depiction of model fit.

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Most popular questions from this chapter

Challenger disaster, Part II. Exercise 9.16 introduced us to O-rings that were identified as a plausible explanation for the breakup of the Challenger space shuttle 73 seconds into takeoff in \(1986 .\) The investigation found that the ambient temperature at the time of the shuttle launch was closely related to the damage of O-rings, which are a critical component of the shuttle. See this earlier exercise if you would like to browse the original data. (a) The data provided in the previous exercise are shown in the plot. The logistic model fit to these data may be written as $$ \log \left(\frac{\hat{p}}{1-\hat{p}}\right)=11.6630-0.2162 \times \text { Temperature } $$ where \(\hat{p}\) is the model-estimated probability that an O-ring will become damaged. Use the model to calculate the probability that an O-ring will become damaged at each of the following ambient temperatures: \(51,53,\) and 55 degrees Fahrenheit. The model-estimated probabilities for several additional ambient temperatures are provided below, where subscripts indicate the temperature: $$ \begin{array}{llll} \hat{p}_{57}=0.341 & \hat{p}_{59}=0.251 & \hat{p}_{61}=0.179 & \hat{p}_{63}=0.124 \\ \hat{p}_{65}=0.084 & \hat{p}_{67}=0.056 & \hat{p}_{69}=0.037 & \hat{p}_{71}=0.024 \end{array} $$ (b) Add the model-estimated probabilities from part (a) on the plot, then connect these dots using a smooth curve to represent the model-estimated probabilities. (c) Describe any concerns you may have regarding applying logistic regression in this application, and note any assumptions that are required to accept the model's validity.

A survey of 55 Duke University students asked about their GPA, number of hours they study at night, number of nights they go out, and their gender. Summary output of the regression model is shown below. Note that male is coded as \(1 .\) $$ \begin{array}{rrrrr} \hline & \text { Estimate } & \text { Std. Error } & \text { t value } & \operatorname{Pr}(>|\mathrm{t}|) \\ \hline \text { (Intercept) } & 3.45 & 0.35 & 9.85 & 0.00 \\ \text { studyweek } & 0.00 & 0.00 & 0.27 & 0.79 \\ \text { sleepnight } & 0.01 & 0.05 & 0.11 & 0.91 \\ \text { outnight } & 0.05 & 0.05 & 1.01 & 0.32 \\ \text { gender } & -0.08 & 0.12 & -0.68 & 0.50 \\ \hline \end{array} $$ (a) Calculate a \(95 \%\) confidence interval for the coefficient of gender in the model, and interpret it in the context of the data. (b) Would you expect a \(95 \%\) confidence interval for the slope of the remaining variables to include \(0 ?\) Explain

Baby weights, Part IV. Exercise 9.3 considers a model that predicts a newborn's weight using several predictors (gestation length, parity, age of mother, height of mother, weight of mother, smoking status of mother). The table below shows the adjusted R-squared for the full model as well as adjusted R-squared values for all models we evaluate in the first step of the backwards elimination process. $$ \begin{aligned} &\begin{array}{llc} \hline & \text { Model } & \text { Adjusted } R^{2} \\ \hline 1 & \text { Full model } & 0.2541 \\ 2 & \text { No gestation } & 0.1031 \\ 3 & \text { No parity } & 0.2492 \\ 4 & \text { No age } & 0.2547 \\ 5 & \text { No height } & 0.2311 \\ 6 & \text { No weight } & 0.2536 \\ 7 & \text { No smoking status } & 0.2072 \\ \hline \end{array}\\\ &\text { Which, if any, variable should be removed from the model first? } \end{aligned} $$

Absenteeism, Part I. Researchers interested in the relationship between absenteeism from school and certain demographic characteristics of children collected data from 146 randomly sampled students in rural New South Wales, Australia, in a particular school year. Below are three observations from this data set. $$ \begin{array}{rcccc} \hline & \text { eth } & \text { sex } & \operatorname{lrn} & \text { days } \\\ \hline 1 & 0 & 1 & 1 & 2 \\ 2 & 0 & 1 & 1 & 11 \\ \vdots & \vdots & \vdots & \vdots & \vdots \\ 146 & 1 & 0 & 0 & 37 \end{array} $$ The summary table below shows the results of a linear regression model for predicting the average number of days absent based on ethnic background (eth: 0 - aboriginal, 1 - not aboriginal), sex (sex: 0 - female, 1 \- male), and learner status (lrn: 0 - average learner, 1 - slow learner). $$ \begin{array}{rrrrr} \hline & \text { Estimate } & \text { Std. Error } & \text { t value } & \operatorname{Pr}(>|\mathrm{t}|) \\ \hline \text { (Intercept) } & 18.93 & 2.57 & 7.37 & 0.0000 \\ \text { eth } & -9.11 & 2.60 & -3.51 & 0.0000 \\ \text { sex } & 3.10 & 2.64 & 1.18 & 0.2411 \\ \text { lrn } & 2.15 & 2.65 & 0.81 & 0.4177 \end{array} $$ (a) Write the equation of the regression model. (b) Interpret each one of the slopes in this context. (c) Calculate the residual for the first observation in the data set: a student who is aboriginal, male, a slow learner, and missed 2 days of school. (d) The variance of the residuals is \(240.57,\) and the variance of the number of absent days for all students in the data set is 264.17. Calculate the \(R^{2}\) and the adjusted \(R^{2}\). Note that there are 146 observations in the data set.

Logistic regression fact checking. Determine which of the following statements are true and false. For each statement that is false, explain why it is false. (a) Suppose we consider the first two observations based on a logistic regression model, where the first variable in observation 1 takes a value of \(x_{1}=6\) and observation 2 has \(x_{1}=4\). Suppose we realized we made an error for these two observations, and the first observation was actually \(x_{1}=7\) (instead of 6 ) and the second observation actually had \(x_{1}=5\) (instead of \(\left.4\right) .\) Then the predicted probability from the logistic regression model would increase the same amount for each observation after we correct these variables. (b) When using a logistic regression model, it is impossible for the model to predict a probability that is negative or a probability that is greater than \(1 .\) (c) Because logistic regression predicts probabilities of outcomes, observations used to build a logistic regression model need not be independent. (d) When fitting logistic regression, we typically complete model selection using adjusted \(R^{2}\).
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