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)
Answer:
-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
[Please give me a Thumbs Up if you are satisfied with myanswer. If you are not, please comment on it, so I can edit theanswer. Thanks.]
"Our Prices Start at $11.99. As Our First Client, Use Coupon Code GET15 to claim 15% Discount This Month!!"
![](https://writinghelpe.com/wp-content/uploads/2022/08/save.jpg)