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gtnhenbr [62]
3 years ago
15

Using the same data set, four models are estimated using the same response variable, however, the number of explanatory variable

s differs. Which of the following models provides the best fit? Model 1 Model 2 Model 3 Model 4 Multiple R 0.993 0.991 0.936 0.746 R Square 0.987 0.982 0.877 0.557 Adjusted R Square 0.982 0.978 0.849 0.513 Standard Error 4,043 4,463 11,615 20,878 Observations 12 12 12 12 Multiple Choice Model 1 Model 2 Model 3 Model 4
Mathematics
1 answer:
kirill [66]3 years ago
8 0

Answer:

Model A

Step-by-step explanation:

Given the table :

___________M 1 ____ M 2 ____ M 3 ____M 4

Multiple R _ 0.993 ___ 0.991 ___0.936__ 0.746

R Square __0.987___ 0.982 ___0.877 __0.557

Adj R² ____ 0.982___ 0.978 __ 0.849 ___0.513

S E_______ 4,043 __ 4,463 ___11,615 __20,878 Observations_ 12 _____ 12 _____ 12 ____12

Based on the detains of the model given, we could use the R value, R² and standard error values to evaluate the performance of the different models.

The best model will be one with Correlation Coefficient (R value) closet to 1. The model with the highest R value will also have the highest Coefficient of determination, R² value. The a best model is one which has a low a standard error value.

From the table, Model A has the highest R and R² values. It also has the lowest standard error value. Hence, we can conclude that model A provides the best fit.

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Answer:

(a) The mean or expected value of <em>X </em>is 2.2.

(b) The standard deviation of <em>X</em> is 1.3.

Step-by-step explanation:

Let <em>X</em> = number of times the traffic light is red when a commuter passes through the traffic lights.

The probability distribution of <em>X</em> id provided.

The formula to compute the mean or expected value of <em>X </em>is:

\mu=E(X)=\sum x.P(X=x)

The formula to compute the standard deviation of <em>X </em>is:

\sigma=\sqrt{E(X^{2})-(E(X))^{2}}

The formula of E (X²) is:

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(a)

Compute the expected value of <em>X</em> as follows:

E(X)=\sum x.P(X=x)\\=(0\times0.06)+(1\times0.25)+(2\times0.35)+(3\times0.15)+(4\times0.13)+(5\times0.06)\\=2.22\\\approx2.2

Thus, the mean or expected value of <em>X </em>is 2.2.

(b)

Compute the value of E (X²) as follows:

E(X^{2})=\sum x^{2}.P(X=x)\\=(0^{2}\times0.06)+(1^{2}\times0.25)+(2^{2}\times0.35)+(3^{2}\times0.15)+(4^{2}\times0.13)+(5^{2}\times0.06)\\=6.58

Compute the standard deviation of <em>X</em> as follows:

\sigma=\sqrt{E(X^{2})-(E(X))^{2}}\\=\sqrt{6.58-(2.22)^{2}}\\=\sqrt{1.6516}\\=1.285\\\approx1.3

Thus, the standard deviation of <em>X</em> is 1.3.

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