(See Solution) In the 1991 General Social Survey (an American survey) 886 respondents were asked, "Do you believe there is a life after death?" A logistic
Question: In the 1991 General Social Survey (an American survey) 886 respondents were asked, "Do you believe there is a life after death?" A logistic regression of their dichotomous responses (1=yes, 0=no) on age in years (X 1 ) and a nine-point church attendance scale (X 2 ) produced the following equation:
\[\ln \left( \frac{{{p}_{i}}}{1-{{p}_{i}}} \right)=0.812-0.005\times {{X}_{1i}}+0.224\times {{X}_{2i}}\]- Explain why we have used logistic regression. Give three reasons why ordinary linear regression (OLS) is inappropriate in this case.
- The standard errors of b 1 and b 2 are 0.005 and 0.035. Is either predictor significantly related to the belief in a life after death; if so, in which direction and at what significance level? The critical values for Z are ±1.65 (90% level), ±1.96 (95% level), ±2.58 (99% level).
- What is the probability that a person believes in a life after death who is 40 years old and attends church a couple of times a year (X 2 =2)?
-
What is the probability that a person believes in a life after death who is 25 years old and attends church nearly every day (X
2
=8)?
Use the following formula when you calculate the predicted probabilities in question 3c:
\[{{P}_{i}}=\frac{1}{1+{{e}^{-(a+{{b}_{1}}\times {{X}_{1i}}+{{b}_{2}}\times {{X}_{2i}})}}}\]
Adding dummy variables for Catholic (X 3 ) and Jew (X 4 ) to the logistic regression, treating Protestant as the omitted category, the following equation results:
\[\ln \left( \frac{{{p}_{i}}}{1-{{p}_{i}}} \right)=1.093-0.006\times {{X}_{1i}}+0.234\times {{X}_{2i}}-0.721\times {{X}_{3i}}-1.405\times {{X}_{4i}}\] - What is the probability that a person believes in a life after death who is Catholic, 65 years old, and attends church nearly every week (X 2 =6)?
- What is the probability that a person believes in a life after death who is Protestant, 25 years old, and never attends church (X 2 =0)?
Use the following formula when you calculate the predicted probabilities in questions 3d–3e:
\[{{P}_{i}}=\frac{1}{1+{{e}^{-(a+{{b}_{1}}\times {{X}_{1i}}+{{b}_{2}}\times {{X}_{2i}}+{{b}_{3}}\times {{X}_{3i}}+{{b}_{4}}\times {{X}_{4i}})}}}\]
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