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 = 8 observations, is (465.3, 594.5). What parameter is this interval estimating?

Answer 1

Answer:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$  

The interval on this case is provided (465.3, 594.5) we can calculate th sample mean on this case:

$\bar X = \frac{465.3 +594.5}{2}= 529.9$

The margin of error for this case would be:

$ME = \frac{594.5-465.3}{2}= 64.6$

And for this case the parameter of interest is the true mean for the xpected lead content.

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X$ represent the sample mean for the sample  

$\mu$ population mean (variable of interest)

s represent the sample standard deviation

n=8 represent the sample size  

Confidence = 0.95 or 95%

Solution to the problem

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$   (1)

The interval on this case is provided (465.3, 594.5) we can calculate th sample mean on this case:

$\bar X = \frac{465.3 +594.5}{2}= 529.9$

The margin of error for this case would be:

$ME = \frac{594.5-465.3}{2}= 64.6$

And for this case the parameter of interest is the true mean for the xpected lead content.