Answer:
The answers is " Option B".
Step-by-step explanation:
[tex]CI=\hat{Y}\pm t_{Critical}\times S_{e}[/tex]
Where,
[tex]\hat{Y}=[/tex] predicted value of lead content when traffic flow is 15.
[tex]\to df=n-1=8-1=7[/tex]
[tex]95\% \ CI\ is\ (463.5, 596.3) \\\\\hat{Y}=\frac{(463.5+596.3)}{2}\\\\[/tex]
[tex]=\frac{1059.8}{2}\\\\=529.9[/tex]
Calculating thet-critical value[tex]t_{ \{\frac{\alpha}{2},\ df \}}=-2.365[/tex]
The lower predicted value [tex]=529.9-2.365(Se)[/tex]
[tex]463.5=529.9-2.365(Se)\\\\529.9-463.5=2.365(Se)\\\\66.4=2.365(Se)\\\\Se=\frac{66.4}{2.365} \\\\Se=28.076[/tex]
When [tex]99\%[/tex] of CI use as the expected lead content: [tex]\to 529.9\pm t_{0.005,7}\times 28.076 \\\\=(529.9 \pm 3.499 \times 28.076)\\\\=(529.9 \pm 98.238)\\\\=(529.9-98.238, 529.9+98.238)\\\\=(431.662, 628.138)\\\\=(431.6, 628.1)[/tex]
