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 :

Answer:

[tex] \bar X = \frac{461.2+598.6}{2}=529.9[/tex]

[tex] ME= \frac{Width}{2}= \frac{137.4}{2}= 68.7[/tex]

The margin of error is given by:

[tex] ME = t_{\alpha/2} SE[/tex]

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

[tex] df = 12-1 = 11[/tex]

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

And then we can find the standard error:

[tex] SE = \frac{ME}{t_{\alpha/2}}= \frac{68.7}{2.20}= 31.227[/tex]

[tex] t_{\alpha/2}= 3.11[/tex]

And using the confidence interval formula we got:

[tex] \bar X \pm t_{\alpha/2} SE[/tex]

[tex] 529.9 - 3.11*31.227 = 432.784[/tex]

[tex] 529.9 + 3.11*31.227 = 627.016[/tex]

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".

[tex]\bar X[/tex] represent the sample mean for the sample  

[tex]\mu[/tex] 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:

[tex]\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex]   (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:

[tex] \bar X = \frac{461.2+598.6}{2}=529.9[/tex]

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

[tex] Width = 598.6-461.2= 137.4[/tex]

And the margin of error is given by:

[tex] ME= \frac{Width}{2}= \frac{137.4}{2}= 68.7[/tex]

The margin of error is given by:

[tex] ME = t_{\alpha/2} SE[/tex]

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

[tex] df = 12-1 = 11[/tex]

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

And then we can find the standard error:

[tex] SE = \frac{ME}{t_{\alpha/2}}= \frac{68.7}{2.20}= 31.227[/tex]

The standard error is given by [tex] SE= \frac{s}{\sqrt{n}}[/tex]

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:

[tex] t_{\alpha/2}= 3.11[/tex]

And using the confidence interval formula we got:

[tex] \bar X \pm t_{\alpha/2} SE[/tex]

[tex] 529.9 - 3.11*31.227 = 432.784[/tex]

[tex] 529.9 + 3.11*31.227 = 627.016[/tex]

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