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Consider the following data that fit the quadratic modelE(y)=0+1x+2x2:

a. Construct a scatterplot for this data. Give the prediction equation and calculate R2based on the model above.

b. Interpret the value ofR2.

c. Justify whether the overall model is significant at the 1% significance level if the data result into a p-value of 0.000514.

Short Answer

Expert verified

a. Scatter plot, prediction equation is y=8.541667-2.25357x+0.386905x2and value of R2calculated here is 0.9516

b. The value ofR2here is 0.9516 which is a high value denoting that almost 95% of the variation in the variables is explained by the model. This means that the model is a good fit for the data.

c. At 1% significance level, it can be concluded that 120

Step by step solution

01

Scatterplot for the data

X

Y

0

8

1

7.3

2

5.6

3

5.9

4

5.2

5

6.5

6

8.8

7

12.1

SUMMARY OUTPUT

















Regression Statistics








Multiple R

0.975509








R Square

0.951618








Adjusted R Square

0.932265








Standard Error

0.586556








Observations

8

















ANOVA









df

SS

MS

F

Significance F




Regression

2

33.83476

16.91738

49.17163

0.000515




Residual

5

1.720238

0.344048






Total

7

35.555













Coefficients

Standard Error

t Stat

P-value

Lower 95%

Upper 95%

Lower 95.0%

Upper 95.0%

Intercept

8.541667

0.49366

17.30272

1.18E-05

7.272673

9.810661

7.272673

9.810661

X

-2.25357

0.329452

-6.84036

0.001019

-3.10046

-1.40669

-3.10046

-1.40669

X^2

0.386905

0.045254

8.549672

0.000361

0.270576

0.503233

0.270576

0.503233

On solving the second-order model equation in excel, the output generated is attached here. The prediction equation, therefore isy=8.541667-2.25357x+0.386905x2

The value ofR2 calculated here is 0.9516.

02

Interpretation of R2

The value ofR2 here is 0.9516 which is a high value denoting that almost 95% of the variation in the variables is explained by the model. This means that the model is a good fit for the data.

03

Goodness of fit

H0:1=2=0

Ha:At least one of the parameters1 or2 is non zero

Here, F test statistic=SSEn-(k+1)=1.7202385=0.3440

H0is rejected if p-value < . For =0.01, since 0.000514 < 0.01

Sufficient evidence to rejectH0 at 95% confidence interval.

Therefore,120.

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