# Use the output below to answer

Use the output below to answer the following questions regardingthe association between carbon dioxide levels and amount of coalused.

Estimate Std. Error. t value.   P r(> |t|

Intercept 6.166678. 2.171695.    2.84  0.00571

Coal 0.030171. 0.002976   10.14.  4.52e-16

-Create a 98% estimate for the independent variable.

– Is the explanatory variable significant at α = 0.01? –

Suppose the correlation is 0.7478824. Determine the value of thecoefficient of determination and explain what this valuerepresents.

– What is the response variable when the explanatory variable is700? (NOTE: This value is within the range of the explanatoryvariable)

-Create a 98% estimate for the independent variable

Critical t- score for 98% confidence is t = 2.4033

Lower limit of the confidence interval = b – t * SE = 0.030171 -2.4033 * 0.002976 = 0.0230

Upper limit of the confidence interval = b + t * SE = 0.030171 +2.4033 * 0.002976 = 0.0373

The 98% confidence interval for the slope is [0.0230,0.0373]

– Is the explanatory variable significant at α = 0.01?

p- value (4.52e-16) < α (0.01), so the explanatory variableis significant

– Suppose the correlation is 0.7478824. Determine the value ofthe coefficient of determination and explain what this valuerepresents

r^2 = 0.7478824^2 = 0.5593. This means about 55.93% of thevariation in y is explained by the variation in x using thismodel

– What is the response variable when the explanatory variable is700?

When x = 700, y = 0.030171 * 700 + 6.166678 = 27.286378

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