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

The answers is " Option B".

Explanation:

`CI=\hat{Y}\pm t_(Critical)* 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}=((463.5+596.3))/(2)\\\\`

`=(1059.8)/(2)\\\\=529.9`

Calculating thet-critical value
`t_{ \{(\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=(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)* 28.076 \\\\=(529.9 \pm 3.499 * 28.076)\\\\=(529.9 \pm 98.238)\\\\=(529.9-98.238, 529.9+98.238)\\\\=(431.662, 628.138)\\\\=(431.6, 628.1)`