ISYE6414 Regression Midterm 2 Test Exam Questions
And Answers Verified 100% Correct
In Poisson regression, we assume a nonlinear relationship between the log rate and
the predicting variables. --CORRECT ANSWER--> FALSE: In Poisson
regression, we assume a linear relationship between the log rate and the predicting
variables.
Linearity Assumption: 𝑙𝑜𝑔(𝐸(𝑌|𝑥1,...,𝑥𝑝))=𝛽0+𝛽1𝑥1+...+𝛽𝑝𝑥𝑝
Interpret the coefficient for concCu. --CORRECT ANSWER--> A 1-unit increase
in the concentration of copper decreases the log odds of botrytis blight surviving by
0.27483 when sulfur stays fixed.
Suppose you wanted to test if the coefficient for concCu is equal to -0.3. What
zvalue would you use for this test? --CORRECT ANSWER--> z-value =
(estimated coefficient - null value)/standard error of estimated coefficient =
(0.27483+0.3)/0.01784 = 1.411
Construct an approximate 95% confidence interval for the coefficient of concS. -
CORRECT ANSWER--> 95% confidence interval = (estimated coefficient - z
critical point * standard error of estimated coefficient, estimated coefficient + z
critical point * standard error of estimated coefficient) = (-4.32735 - 1.96*0.26518,
-4.32735 + 1.96*0.26518) = (-4.847, -3.808)
What is the probability of survival for a botrytis blight sample exposed to a copper
concentration of 0.6 and a sulfur concentration of 0.6? --CORRECT ANSWER-->
p = e^(beta_0+beta_1 * x1+beta_2 * x2)/1+e^(beta_0+beta_1 * x1+beta_2 * x2) =
e^(3.58770-4.32735 * 0.6-0.27483 * 0.6)/1+e^(3.58770-4.32735 * 0.6-0.27483 *
0.6) = 0.696
The p-value for testing the overall regression can be obtained from which of the
following? --CORRECT ANSWER--> 1-pchisq(419.33,2). The chi-square test
statistic is the difference between the null deviance (718.76) and the residual
deviance (299.43), which is 419.33. The degrees of freedom is the difference
between the null deviance degrees of freedom (19) and the residual deviance
degrees of freedom (17), which is 2 (the number of predicting variables in the
model).
, The p-value for a goodness-of-fit test using the deviance residuals for the
regression can be obtained from which of the following? --CORRECT ANSWER-
> 1-pchisq(299.43,17). The goodness of fit test uses the residual deviance (299.43)
and corresponding degrees of freedom (17) as the test statistic for the chi-squared
test.
Log odds Percent Change vs. Factor Change by --CORRECT ANSWER--> e^beta
is the factor it changes by. (exp(1.084) = 2.956 in this case)
And e^beta - 1 when beta is positive OR 1 - e^beta when beta is negative is the
percentage of change. (exp(1.084)-1) = 1.956 in this case.
DF: Simple Linear Regression --CORRECT ANSWER--> ->hypothesis testing for
the βs: n-2 DF (n = sample size)
We lose 2 degrees of freedom from replacing β0 and β1 with β^0 and β^1 to obtain
the residuals which replace the error terms.
DF: ANOVA --CORRECT ANSWER--> ->hypothesis testing for the μs: N-k
DF_errors, k-1 DF_treatment, and N-1 DF_total (N = total sample size) (k =
number of groups)
DF_errors: We have N-k DF because N is the sample size and the k population
means are replaced with their sample means.
DF_treatment: We have k-1 DF because k is the sample size (we have k sample
means) and the overall population mean is replaced with the overall sample mean.
DF_total: We have N-1 DF because N is the sample size and the overall population
mean is replaced with the overall sample mean.
DF_total = DF_errors + DF_treatment
DF: Multiple Linear Regression --CORRECT ANSWER--> ->hypothesis testing
for the βs: n-p-1 DF (n = sample size) (p = number of predicting variables)
This is consistent with simple linear regression (p=1).
And Answers Verified 100% Correct
In Poisson regression, we assume a nonlinear relationship between the log rate and
the predicting variables. --CORRECT ANSWER--> FALSE: In Poisson
regression, we assume a linear relationship between the log rate and the predicting
variables.
Linearity Assumption: 𝑙𝑜𝑔(𝐸(𝑌|𝑥1,...,𝑥𝑝))=𝛽0+𝛽1𝑥1+...+𝛽𝑝𝑥𝑝
Interpret the coefficient for concCu. --CORRECT ANSWER--> A 1-unit increase
in the concentration of copper decreases the log odds of botrytis blight surviving by
0.27483 when sulfur stays fixed.
Suppose you wanted to test if the coefficient for concCu is equal to -0.3. What
zvalue would you use for this test? --CORRECT ANSWER--> z-value =
(estimated coefficient - null value)/standard error of estimated coefficient =
(0.27483+0.3)/0.01784 = 1.411
Construct an approximate 95% confidence interval for the coefficient of concS. -
CORRECT ANSWER--> 95% confidence interval = (estimated coefficient - z
critical point * standard error of estimated coefficient, estimated coefficient + z
critical point * standard error of estimated coefficient) = (-4.32735 - 1.96*0.26518,
-4.32735 + 1.96*0.26518) = (-4.847, -3.808)
What is the probability of survival for a botrytis blight sample exposed to a copper
concentration of 0.6 and a sulfur concentration of 0.6? --CORRECT ANSWER-->
p = e^(beta_0+beta_1 * x1+beta_2 * x2)/1+e^(beta_0+beta_1 * x1+beta_2 * x2) =
e^(3.58770-4.32735 * 0.6-0.27483 * 0.6)/1+e^(3.58770-4.32735 * 0.6-0.27483 *
0.6) = 0.696
The p-value for testing the overall regression can be obtained from which of the
following? --CORRECT ANSWER--> 1-pchisq(419.33,2). The chi-square test
statistic is the difference between the null deviance (718.76) and the residual
deviance (299.43), which is 419.33. The degrees of freedom is the difference
between the null deviance degrees of freedom (19) and the residual deviance
degrees of freedom (17), which is 2 (the number of predicting variables in the
model).
, The p-value for a goodness-of-fit test using the deviance residuals for the
regression can be obtained from which of the following? --CORRECT ANSWER-
> 1-pchisq(299.43,17). The goodness of fit test uses the residual deviance (299.43)
and corresponding degrees of freedom (17) as the test statistic for the chi-squared
test.
Log odds Percent Change vs. Factor Change by --CORRECT ANSWER--> e^beta
is the factor it changes by. (exp(1.084) = 2.956 in this case)
And e^beta - 1 when beta is positive OR 1 - e^beta when beta is negative is the
percentage of change. (exp(1.084)-1) = 1.956 in this case.
DF: Simple Linear Regression --CORRECT ANSWER--> ->hypothesis testing for
the βs: n-2 DF (n = sample size)
We lose 2 degrees of freedom from replacing β0 and β1 with β^0 and β^1 to obtain
the residuals which replace the error terms.
DF: ANOVA --CORRECT ANSWER--> ->hypothesis testing for the μs: N-k
DF_errors, k-1 DF_treatment, and N-1 DF_total (N = total sample size) (k =
number of groups)
DF_errors: We have N-k DF because N is the sample size and the k population
means are replaced with their sample means.
DF_treatment: We have k-1 DF because k is the sample size (we have k sample
means) and the overall population mean is replaced with the overall sample mean.
DF_total: We have N-1 DF because N is the sample size and the overall population
mean is replaced with the overall sample mean.
DF_total = DF_errors + DF_treatment
DF: Multiple Linear Regression --CORRECT ANSWER--> ->hypothesis testing
for the βs: n-p-1 DF (n = sample size) (p = number of predicting variables)
This is consistent with simple linear regression (p=1).
