Answer:
d. -5.666 to -0.926
Step-by-step explanation:
Here a pivotal quantity is [tex]T = \frac{\hat{\beta_{1}}-\beta_{1}}{se(\hat{\beta_{1}})}[/tex] where [tex]\beta_{1}[/tex] is the true slope of the regression equation (unknown), [tex]\hat{\beta_{1}}[/tex] is its least square estimate and [tex]se(\hat{\beta_{1}})=1.097[/tex] is its estimated standard error. And T has t-distribution with n-2=15-2=13 degrees of freedom (n is the sample size). We have that [tex]\hat{\beta_{1}} = -3.30[/tex], and because we want the 95% confidence interval, we should use the 2.5th quantile of the t distribution with 13 df, this value is -2.16 and the 95% confidence intervale is given by [tex]\hat{\beta_{1}}\pm t_{0.025}se(\hat{\beta_{1}})[/tex], i.e., [tex]-3.30\pm (2.16)(1.097)[/tex]. Therefore, the 95% confidence interval explicitly is (-5.666, -0.926)