Question

Noise levels at 5 airports were measured in decibels yielding the following data: 152,154,139,124,120 Construct the 80% confidence interval for the mean noise level at such locations. Assume the population is approximately normal. Step 1 of 4 : Calculate the sample mean for the given sample data. Round your answer to one decimal place.

Answer 1

Answer:

$137.8-1.533\frac{15.595}{\sqrt{5}}=127.108$    

$137.8+1.533\frac{15.595}{\sqrt{5}}=148.492$    

So on this case the 80% confidence interval would be given by (127.108;148.492)    

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)

In order to calculate the mean and the sample deviation we can use the following formulas:  

$\bar X= \sum_{i=1}^n \frac{x_i}{n}$ (2)  

$s=\sqrt{\frac{\sum_{i=1}^n (x_i-\bar X)}{n-1}}$ (3)  

The mean calculated for this case is $\bar X=137.8$

The sample deviation calculated $s=15.595$

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

$df=n-1=5-1=4$

Since the Confidence is 0.80 or 80%, the value of $\alpha=0.2$ and $\alpha/2 =0.1$, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.1,4)".And we see that $t_{\alpha/2}=1.533$

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

$137.8-1.533\frac{15.595}{\sqrt{5}}=127.108$    

$137.8+1.533\frac{15.595}{\sqrt{5}}=148.492$    

So on this case the 80% confidence interval would be given by (127.108;148.492)