ISYE 6402 MIDTERM CERTIFICATION
SCRIPT 2026 QUESTIONS WITH SOLUTIONS
GRADED A+
◍ One objective in ANOVA is to determine which groups have statistically
significantly different means from each other. We can compare two groups
at a time, which is called pairwise comparison, and the Tukey method
(TukeyHSD()) will generate all of these for you. (T/F).
Answer: True.
◍ Goodness of fit describes how accurately a model fits the observed data by
minimizing its residuals. (T/F).
Answer: False. Goodness of fit is all about seeing if a regression fits a
model's assumptions, which is different than evaluating its performance
(which is what the question was describing). Remember: goodness-of-fit
relates to "does this satisfy the model's assumptions" (and therefore make
inference more reliable)?" while evaluating performance has to do with:
"how small are the residuals (e.g. how close are the fitted and/or predicted
values to the observation)?
◍ What is a bi-modal distribution?.
Answer: Here are key characteristics of a bi-modal distribution:Two Peaks:
The distribution has two prominent peaks or modes.Valley Between Peaks:
There is typically a dip or valley between the two modes.Separation: The
modes are separated by a range of values with lower frequency or
probability.
◍ SSE.
Answer: is the sum of square differences between the observations and the
individual sample means
,◍ In Box-Cox transformation, if the value of lambda is 1, we:.
Answer: Do not need to transform
◍ n-1 in ANOVA table in multiple regression.
Answer: The number of total degrees of freedom
◍ pooled variance estimator (Si^2).
Answer: are the sample variances of the data samplesof the response
variable. The formula is also called the mean square error inANOVA or
MSE.
◍ ifthe between-variability is larger than the within-variability.
Answer: We will find significant differences across the means
◍ conditional.
Answer: the interpretation of the t-test for statistical significance is ____ on
thepresence of other predicting variables to be in the model
◍ The F-test for equal means has a null hypothesis that the means are the
same: if you obtain a lower p-value than your threshold (normally .05), then
you reject the null hypothesis and reject that the means tested as part of the
F-test are equal. (T/F).
Answer: True. In general, the idea in statistics is that you should take the
less controversial assumption as your null hypothesis. When you are trying
to 'detect' or 'prove' that a difference exists between groups, you begin by
assuming that the groups are the same... and you place the burden on the
evidence to convince you otherwise.
◍ mu_k and sigma_k square for kth population.
Answer: true mean and variance for the responsevariable are ___
◍ Second Order Model.
Answer: we include the square of the predictors, so we include an X1
squared and X2 squared as additional predictors
◍ aov().
Answer: R command to fit an ANOVA model using the R statistical
, software
◍ Tukey Method.
Answer: compares all possible pairs of means
◍ Estimate confidence intervals.
Answer: for all the pairs of means, in order to identify which of the means
are not equal, or which of the means are statistically significantly different.
Specifically, we will consider a hypothesis test for each pair of means with
the null hypothesis that the means in the pair are equal versus the alternative
that are not equal. We will perform all the hypothesis tests across all pair
jointly
◍ j.
Answer: is the index within group
◍ ß1 hat.
Answer: is the estimated expected change in the response variable
associated withone unit of change in the predicting variable.
◍ The estimated versus predicted regression line for a given x*.
Answer: Have the same expectation
◍ pairwise comparison.
Answer: to determine which treatment means arebigger or smaller. One way
to do this is to compare all possible pairs; there are k(k-1)/2 unique pairs of
treatments
◍ obersvational studies.
Answer: Association statements are made in a ____ environment.
◍ How will the procedure change if we test whether the coefficient is equal to
a constant?.
Answer: We reject the null hypothesis if the t-value in absolute value is
larger than the t critical point for alpha over 2 with n-p-1 degrees of
freedom.If the p-value is small, for example smaller than .01, we reject the
null hypothesis that β0 is equal to the null value b.
SCRIPT 2026 QUESTIONS WITH SOLUTIONS
GRADED A+
◍ One objective in ANOVA is to determine which groups have statistically
significantly different means from each other. We can compare two groups
at a time, which is called pairwise comparison, and the Tukey method
(TukeyHSD()) will generate all of these for you. (T/F).
Answer: True.
◍ Goodness of fit describes how accurately a model fits the observed data by
minimizing its residuals. (T/F).
Answer: False. Goodness of fit is all about seeing if a regression fits a
model's assumptions, which is different than evaluating its performance
(which is what the question was describing). Remember: goodness-of-fit
relates to "does this satisfy the model's assumptions" (and therefore make
inference more reliable)?" while evaluating performance has to do with:
"how small are the residuals (e.g. how close are the fitted and/or predicted
values to the observation)?
◍ What is a bi-modal distribution?.
Answer: Here are key characteristics of a bi-modal distribution:Two Peaks:
The distribution has two prominent peaks or modes.Valley Between Peaks:
There is typically a dip or valley between the two modes.Separation: The
modes are separated by a range of values with lower frequency or
probability.
◍ SSE.
Answer: is the sum of square differences between the observations and the
individual sample means
,◍ In Box-Cox transformation, if the value of lambda is 1, we:.
Answer: Do not need to transform
◍ n-1 in ANOVA table in multiple regression.
Answer: The number of total degrees of freedom
◍ pooled variance estimator (Si^2).
Answer: are the sample variances of the data samplesof the response
variable. The formula is also called the mean square error inANOVA or
MSE.
◍ ifthe between-variability is larger than the within-variability.
Answer: We will find significant differences across the means
◍ conditional.
Answer: the interpretation of the t-test for statistical significance is ____ on
thepresence of other predicting variables to be in the model
◍ The F-test for equal means has a null hypothesis that the means are the
same: if you obtain a lower p-value than your threshold (normally .05), then
you reject the null hypothesis and reject that the means tested as part of the
F-test are equal. (T/F).
Answer: True. In general, the idea in statistics is that you should take the
less controversial assumption as your null hypothesis. When you are trying
to 'detect' or 'prove' that a difference exists between groups, you begin by
assuming that the groups are the same... and you place the burden on the
evidence to convince you otherwise.
◍ mu_k and sigma_k square for kth population.
Answer: true mean and variance for the responsevariable are ___
◍ Second Order Model.
Answer: we include the square of the predictors, so we include an X1
squared and X2 squared as additional predictors
◍ aov().
Answer: R command to fit an ANOVA model using the R statistical
, software
◍ Tukey Method.
Answer: compares all possible pairs of means
◍ Estimate confidence intervals.
Answer: for all the pairs of means, in order to identify which of the means
are not equal, or which of the means are statistically significantly different.
Specifically, we will consider a hypothesis test for each pair of means with
the null hypothesis that the means in the pair are equal versus the alternative
that are not equal. We will perform all the hypothesis tests across all pair
jointly
◍ j.
Answer: is the index within group
◍ ß1 hat.
Answer: is the estimated expected change in the response variable
associated withone unit of change in the predicting variable.
◍ The estimated versus predicted regression line for a given x*.
Answer: Have the same expectation
◍ pairwise comparison.
Answer: to determine which treatment means arebigger or smaller. One way
to do this is to compare all possible pairs; there are k(k-1)/2 unique pairs of
treatments
◍ obersvational studies.
Answer: Association statements are made in a ____ environment.
◍ How will the procedure change if we test whether the coefficient is equal to
a constant?.
Answer: We reject the null hypothesis if the t-value in absolute value is
larger than the t critical point for alpha over 2 with n-p-1 degrees of
freedom.If the p-value is small, for example smaller than .01, we reject the
null hypothesis that β0 is equal to the null value b.
