Question

ASK YOUR TEACHER An article reported that for a sample of 42 kitchens with gas cooking appliances monitored during a one-week period, the sample mean CO2 level (ppm) was 654.16, and the sample standard deviation was 165.23. (a) Calculate and interpret a 95% (two-sided) confidence interval for true average CO2 level in the population of all homes from which the sample was selected. (Round your answers to two decimal places.)

Answer 1

Answer:

$654.16-2.02\frac{165.23}{\sqrt{42}}=602.66$    

$654.16+2.02\frac{165.23}{\sqrt{42}}=705.66$    

So on this case the 95% confidence interval would be given by (602.66;705.66)

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".

$\bar X=654.16$ represent the sample mean

$\mu$ population mean (variable of interest)

s=165.23 represent the sample standard deviation

n=42 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)

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:

$df=n-1=42-1=41$

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,41)".And we see that $t_{\alpha/2}=2.02$

Now we have everything in order to replace into formula (1):

$654.16-2.02\frac{165.23}{\sqrt{42}}=602.66$    

$654.16+2.02\frac{165.23}{\sqrt{42}}=705.66$    

So on this case the 95% confidence interval would be given by (602.66;705.66)