Question

Suppose you are investigating the relationship between two variables, traffic flow and expected lead content, where traffic flow is a predictor of lead content. You find the 95% CI for expected lead content when traffic flow is 15, based on a sample of n= 10 observations, is (461.7, 598.1).
Required:
What parameter is this interval estimating?

Answer 1

Answer:

The answers is " Option B".

Step-by-step explanation:

$CI=\hat{Y}\pm t_{Critical}\times S_{e}$

Where,

$\hat{Y}=$ predicted value of lead content when traffic flow is 15.

$\to df=n-1=8-1=7$

 $95\% \ CI\ is\ (463.5, 596.3) \\\\\hat{Y}=\frac{(463.5+596.3)}{2}$

     $=\frac{1059.8}{2}\\\\=529.9$

Calculating thet-critical value$t_{ \{\frac{\alpha}{2},\ df \}}=-2.365$

The lower predicted value $=529.9-2.365(Se)$

$463.5=529.9-2.365(Se)\\\\529.9-463.5=2.365(Se)\\\\66.4=2.365(Se)\\\\Se=\frac{66.4}{2.365} \\\\Se=28.076$

When $99\%$ of CI use as the expected lead content: $\to 529.9\pm t_{0.005,7}\times 28.076 \\\\=(529.9 \pm 3.499 \times 28.076)\\\\=(529.9 \pm 98.238)\\\\=(529.9-98.238, 529.9+98.238)\\\\=(431.662, 628.138)\\\\=(431.6, 628.1)$