rochelle 发表于 2022-3-19 21:56:06

Greene and Touchstone conducted a study to relate birth weight and Estriol le...

Question:Greene and Touchstone conducted a study to relate birth weight and Estriol level in pregnant women as given in the table below:

iEstriol (mg/24hr) XiBirthweight (g/100) YiiEstriol (mg/24hr) XiBirthweight (g/100) Yi
1725171732
2925182532
3925192734
41227201534
51427211534
61627221535
71624231635
81430241934
91630251835
101631261736
111730271837
121931282038
132130292240
142428302539
151532312443
161632

1. Compute the slope and intercept of the estimated regression line based on above table.2. If low birthweight is defined as <span id="MathJax-Element-1-Frame" class="mjx-chtml MathJax_CHTML" tabindex="0" data-mathml="&#x2264;" role="presentation" style="box-sizing: border-box; display: inline-block; line-height: 0; font-size: 16.8px; overflow-wrap: normal; word-spacing: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding-top: 1px; padding-bottom: 1px; position: relative;">≤≤2500 g, for what estriol level would the predicted birth weight be 2500 g?3. What is the predicted birth weight if a pregnant woman has an estriol level of 15 mg/24hr?4. Fit a linear regression model based on the data using R.5. Test for the significance of the regression line derived for the birthweight - estriol data.
Linear RegressionLinear Regression is a method of modeling a dependent variable as a linear function of independent variables. When the number of independent variables is single, it is called Simple Linear Regression. Using the regression models, the unknown values of the dependent variable can be predicted using the known values of the independent variable.
Answer and Explanation:Since the answer to the question requires a lot of computation, we can either use a calculator or the R statistical software application.1. Slope and Intercept of the regression lineLet the regression line be<span id="MathJax-Element-2-Frame" class="mjx-chtml MathJax_CHTML" tabindex="0" data-mathml="Y=mX+b" role="presentation" style="box-sizing: border-box; display: inline-block; line-height: 0; font-size: 16.8px; overflow-wrap: normal; word-spacing: normal; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding-top: 1px; padding-bottom: 1px; position: relative;">Y=mX+bY=mX+b



rochelle 发表于 2022-3-19 21:56:22

where
m
m is the slope of the regression line and
b
b is the intercept.

To find the slope and intercept, we will compute a few summary statistics of
X
i
and
Y
i
Xi and Yi using R software first.

Number of observations
N
=
31
N=31
Mean Estriol
¯¯¯¯¯
X
=
∑
31
i
=
1
X
i
N
≈
17.23
X¯=∑i=131XiN≈17.23
Mean Birthweight
¯¯¯¯
Y
=
∑
31
i
=
1
Y
i
N
=
32
Y¯=∑i=131YiN=32
Standard Deviation of Estriol
S
X
=
∑
31
i
=
1
(
X
i
−
¯¯¯¯¯
X
)
2
N
−
1
≈
4.75
SX=∑i=131(Xi−X¯)2N−1≈4.75
Standard Deviation of Birthweight
S
Y
=
∑
31
i
=
1
(
Y
i
−
¯¯¯¯
Y
)
2
N
−
1
≈
4.74
SY=∑i=131(Yi−Y¯)2N−1≈4.74
Covariance between Estriol and Brithweght
S
X
Y
=
∑
31
i
=
1
(
X
i
−
¯¯¯¯¯
X
)
(
Y
i
−
¯¯¯¯
Y
)
N
−
1
≈
13.733
SXY=∑i=131(Xi−X¯)(Yi−Y¯)N−1≈13.733
Correlation between Estriol and Brithweght
r
=
S
X
Y
S
X
S
Y
≈
0.61
r=SXYSXSY≈0.61
Plugging in the above values into the formula for the slope of the regression line yields:
m
=
r
S
Y
S
X
≈
0.61
m=rSYSX≈0.61

Intercept of the regression line is
b
=
¯¯¯¯
Y
−
m
¯¯¯¯¯
X
=
32
−
0.61
∗
17.23
≈
21.52
b=Y¯−mX¯=32−0.61∗17.23≈21.52

Therefore the slope of the estimated regression line is approximate 0.61 and the intercept is 21.52.

2 For what estriol level would the predicted weight be 2500g?
From part 1, the estimated regression line is
Y
i
=
0.61
X
i
+
21.52
Yi=0.61Xi+21.52.

We need to find the Estriol level
X
i
Xi for which the predicted weight
Y
i
=
25
Yi=25

Plugging in
Y
i
=
25
Yi=25 into the estimated regression line and solving for
X
i
Xi yields:
25
=
0.61
X
i
+
21.52
⇒
X
i
≈
5.72
25=0.61Xi+21.52⇒Xi≈5.72
.

Therefore the estriol level for the predicted birth weight of 2500 g will be 5.72 mg/24hr.

3. Predicted Birth weight for the Estriol level of 15 mg/24hr.
From part 1, the estimated regression line is
Y
i
=
0.61
X
i
+
21.52
Yi=0.61Xi+21.52.

Plugging in
X
i
=
15
Xi=15 yields the predicted birth weight,
Y
i
=
0.61
⋅
15
+
21.52
≈
30.65
Yi=0.61⋅15+21.52≈30.65

Therefore the predicted birth weight for the Estriol level of 15 mg/24hr is 3065g.

4. Fit a linear regression model using R
Here's the program for the linear regression model using R statistical software:

# Assume data frame df contains the data in columns named Xi and Yi
# Linear Regression Model Fitting
mdl = lm(Yi ~ Xi, data = df)
summary(mdl)
The lm function is used to fit the regression line. The slope and intercept of the estimated regression line and significance values are obtained using the summary function.

Call:
lm(formula = Yi ~ Xi, data = df)
Residuals:
    Min      1QMedian      3Q   Max
-8.1200 -2.0381 -0.03813.35376.8800
Coefficients:
            Estimate Std. Error t value Pr(>|t|)
(Intercept)21.5234   2.6204   8.214 4.68e-09 ***
Xi            0.6082   0.1468   4.143 0.000271 ***
---
Signif. codes:0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
Residual standard error: 3.821 on 29 degrees of freedom
Multiple R-squared:0.3718,        Adjusted R-squared:0.3501
F-statistic: 17.16 on 1 and 29 DF,p-value: 0.0002712
5. Test for significance of regression line
We will test at the significance level of 0.05 that there does not exist any relationship between birth weight and estriol. So the null hypothesis is
m
=
0
m=0 i.e. the slope of the regression line is 0.

From the output of the model in part 4 above, the p-value is 0.000271 which is much less than 0.05. Hence the null hypothesis can be rejected.

Therefore the regression line is significant and there exists an approximately linear relationship between birth weight and estriol.
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