Question

You are told that a 95% CI for expected lead content when traffic flow is 15, based on a sample of n = 12 observations, is (461.2, 598.6). Calculate a CI with confidence level 99% for expected lead content when traffic flow is 15

Answer 1

Answer:

$\bar X = \frac{461.2+598.6}{2}=529.9$

$ME= \frac{Width}{2}= \frac{137.4}{2}= 68.7$

The margin of error is given by:

$ME = t_{\alpha/2} SE$

For 95% of confidence the value of the significance is $\alpha =0.05$ and $\alpha/2 = 0.025$, the degrees of freedom are given by:

$df = 12-1 = 11$

And then we can calculate the critical value for 95% with df = 11 and we got $t_{\alpha}= 2.20$

And then we can find the standard error:

$SE = \frac{ME}{t_{\alpha/2}}= \frac{68.7}{2.20}= 31.227$

$t_{\alpha/2}= 3.11$

And using the confidence interval formula we got:

$\bar X \pm t_{\alpha/2} SE$

$529.9 - 3.11*31.227 = 432.784$

$529.9 + 3.11*31.227 = 627.016$

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 represent the sample size  

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)

For the 95% confidence interval we know that the result is given by (461.2, 598.6), we can estimate the sample mean like this:

$\bar X = \frac{461.2+598.6}{2}=529.9$

Because the distribution is symmetrical, now we can estimate the width of the interval like this:

$Width = 598.6-461.2= 137.4$

And the margin of error is given by:

$ME= \frac{Width}{2}= \frac{137.4}{2}= 68.7$

The margin of error is given by:

$ME = t_{\alpha/2} SE$

For 95% of confidence the value of the significance is $\alpha =0.05$ and $\alpha/2 = 0.025$, the degrees of freedom are given by:

$df = 12-1 = 11$

And then we can calculate the critical value for 95% with df = 11 and we got $t_{\alpha}= 2.20$

And then we can find the standard error:

$SE = \frac{ME}{t_{\alpha/2}}= \frac{68.7}{2.20}= 31.227$

The standard error is given by $SE= \frac{s}{\sqrt{n}}$

Now we are interested for the 99% confidence interval, so then we need to find a new critical value for this confidence level and we got:

$t_{\alpha/2}= 3.11$

And using the confidence interval formula we got:

$\bar X \pm t_{\alpha/2} SE$

$529.9 - 3.11*31.227 = 432.784$

$529.9 + 3.11*31.227 = 627.016$