TEK_615_Predictive_Analytics_Mock_Exam

TEK_615_Predictive_Analytics_Mock_Exam

University

8 Qs

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TEK_615_Predictive_Analytics_Mock_Exam

TEK_615_Predictive_Analytics_Mock_Exam

Assessment

Quiz

Business

University

Medium

Created by

Ivan Sanchez

Used 2+ times

FREE Resource

8 questions

Show all answers

1.

MULTIPLE CHOICE QUESTION

1 min • 1 pt

Which of the following scenarios is most appropriate for applying a logistic regression model instead of a linear regression model?

Predicting the amount of daily demand for a product based on temperature and price.

Forecasting monthly transportation costs based on distance and fuel price.

Estimating the probability that a delivery will be late (yes/no) based on vehicle type and traffic level.

Modeling the relationship between advertising spend and number of orders.

2.

MULTIPLE CHOICE QUESTION

1 min • 1 pt

In the Carter Racing case, why is logistic regression preferred over linear regression to predict whether the engine will break or not?

Because logistic regression better captures time-dependent patterns.

Because the dependent variable is binary, and linear regression can yield probabilities outside the [0,1] range.

Because logistic regression always provides higher accuracy than linear regression.

Because the temperature variable is non-linear.

3.

MULTIPLE CHOICE QUESTION

2 mins • 1 pt

Which of the following statements best describes a key difference between statistical modeling and machine learning?

Statistical modeling is only used for images, while machine learning is used for numbers.

Statistical modeling focuses on prediction, while machine learning focuses on explaining relationships.

Statistical modeling assumes a model structure and is easier to interpret; machine learning focuses on prediction and can handle more complex patterns.

Machine learning requires strong assumptions about the data, while statistical modeling does not.

4.

MULTIPLE CHOICE QUESTION

1 min • 1 pt

Time series decomposition consists of splitting the time series into several components, each representing an underlying pattern category. The pattern categories observed in time series are:

Trend, forecast, noise and variance.

Trend, cycle, seasonality and noise.

Forecast, residuals, autocorrelation and seasonality.

Trend, cycle, seasonality and forecast.

5.

MULTIPLE CHOICE QUESTION

2 mins • 1 pt

Media Image

The following graph shows the time series of airlines revenues in the US. The green line shows the rolling mean with a time window of 4 periods.

Given that a stationary time-series is one whose properties do not depend on the time at which the series is observed, you can conclude that the shown time series is:

Stationary because there are changes in the covariance.

Non-stationary because the variance does not change over time.

Non-stationary because a trend is evident given the changes in the mean.

Stationary because there is trend showing changes in the mean.

6.

MULTIPLE CHOICE QUESTION

1 min • 1 pt

A suitable smoothing method able to capture the trend and seasonality of a time series is:

Exponential smoothing.

Moving average.

Holt’s linear trend.

Holt-Winters.

7.

MULTIPLE CHOICE QUESTION

2 mins • 1 pt

Media Image

Q7-8: Your gut feeling made you think that the best time-series model for predicting revenues is a MA(1). After running model fitting, you got the following plots about the residuals. Q7: Based on these outputs you realized that choosing a MA(1) is not good idea because:

The R2 of the model is too high.

 The residuals do not behave as white noise.

The AIC is too small.

The residuals follow a normal distribution with mean 0 and standard deviation of 1.

8.

MULTIPLE CHOICE QUESTION

2 mins • 1 pt

Media Image

You decided to run a grid search that selects the parameters for an optimized SARIMA model. Which of the following options would you choose based on the AIC.

SARIMA(1, 0, 0)x(0, 1, 1, 4) - AIC:68.49847059289476

SARIMA(0, 1, 0)x(0, 0, 1, 4) - AIC:110.8628967654575

SARIMA(0, 0, 1)x(0, 1, 1, 4) - AIC:69.2162637320265

SARIMA(0, 0, 1)x(1, 1, 1, 4) - AIC:76.36086579098165