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.

Respuesta :

Answer:

[tex]137.8-1.533\frac{15.595}{\sqrt{5}}=127.108[/tex]    

[tex]137.8+1.533\frac{15.595}{\sqrt{5}}=148.492[/tex]    

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".

[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)

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

[tex]\bar X= \sum_{i=1}^n \frac{x_i}{n}[/tex] (2)  

[tex]s=\sqrt{\frac{\sum_{i=1}^n (x_i-\bar X)}{n-1}}[/tex] (3)  

The mean calculated for this case is [tex]\bar X=137.8[/tex]

The sample deviation calculated [tex]s=15.595[/tex]

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

[tex]df=n-1=5-1=4[/tex]

Since the Confidence is 0.80 or 80%, the value of [tex]\alpha=0.2[/tex] and [tex]\alpha/2 =0.1[/tex], 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 [tex]t_{\alpha/2}=1.533[/tex]

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

[tex]137.8-1.533\frac{15.595}{\sqrt{5}}=127.108[/tex]    

[tex]137.8+1.533\frac{15.595}{\sqrt{5}}=148.492[/tex]    

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

Answer:

137.8

Step-by-step explanation: