Respuesta :
Answer:
The estimation for the slope on this case is [tex] \hat m = 0.740[/tex]
And the associated standard error is [tex] SE_{m}= 0.047[/tex]
For the critical value we need to find the degrees of freedom given by:
Since we don't have the sample size we can assume it large enough to consider the normal approximation to the t distribution and on this case the critical value for 95% of confidence would be [tex] t_{\alpha/2} =1.96[/tex]
[tex] 0.740 -1.96*0.047= 0.648[/tex]
[tex] 0.740 +1.96*0.047= 0.832[/tex]
The confidence interval would be between (0.648, 0.832)
Step-by-step explanation:
We are assuming that the estimation was using the least squares method
For this case we need to calculate the slope with the following formula:
[tex]m=\frac{S_{xy}}{S_{xx}}[/tex]
Where:
[tex]S_{xy}=\sum_{i=1}^n x_i y_i -\frac{(\sum_{i=1}^n x_i)(\sum_{i=1}^n y_i)}{n}[/tex]
[tex]S_{xx}=\sum_{i=1}^n x^2_i -\frac{(\sum_{i=1}^n x_i)^2}{n}[/tex]
Nowe we can find the means for x and y like this:
[tex]\bar x= \frac{\sum x_i}{n}=\frac{720}{12}=60[/tex]
[tex]\bar y= \frac{\sum y_i}{n}=\frac{324}{12}=27[/tex]
And we can find the intercept using this:
[tex]b=\bar y -m \bar x [/tex]
The confidence interval for this case is given by this formula:
[tex] \hat m \pm t SE_{m}[/tex]
The estimation for the slope on this case is [tex] \hat m = 0.740[/tex]
And the associated standard error is [tex] SE_{m}= 0.047[/tex]
For the critical value we need to find the degrees of freedom given by:
Since we don't have the sample size we can assume it large enough to consider the normal approximation to the t distribution and on this case the critical value for 95% of confidence would be [tex] t_{\alpha/2} =1.96[/tex]
And replacing we got:
[tex] 0.740 -1.96*0.047= 0.648[/tex]
[tex] 0.740 +1.96*0.047= 0.832[/tex]
The confidence interval would be between (0.648, 0.832)
