Answer
Confidence interval: $0.4889\lt β_1\lt2.8995$
We are 95% confident that as the total length increases by 1 cm, the weight increases between 0.4889 and 2.8995 kg.
Work Step by Step
$n=12$, so:
$d.f.=n-2=10$
$level~of~confidence=(1-α).100$%
$95$% $=(1-α).100$%
$0.95=1-α$
$α=0.05$
$t_{\frac{α}{2}}=t_{0.025}=2.228$
(According to Table VI, for d.f. = 10 and area in right tail = 0.025)
$Lower~bound=b_1-t_{\frac{α}{2}}\frac{s_e}{\sqrt {Σ(x_i-x ̅)^2}}$
$Upper~bound=b_1+t_{\frac{α}{2}}\frac{s_e}{\sqrt {Σ(x_i-x ̅)^2}}$
Now, see the results obtained in the MINITAB in item (b).
We can find the lower and upper bounds using the results from MINITAB. Use $\frac{s_e}{\sqrt {Σ(x_i-x ̅)^2}}=SE~Coef$
$Lower~bound=b_1-t_{\frac{α}{2}}(SE~Coef)=1.6942-2.228\times0.541=0.4889$
$Upper~bound=b_1+t_{\frac{α}{2}}(SE~Coef)=1.6942+2.228\times0.541=2.8995$
