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

137.8

Explanation:

Answer 2

`137.8-1.533(15.595)/(√(5))=127.108`

`137.8+1.533(15.595)/(√(5))=148.492`

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

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)(s)/(√(n))` (1)

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

`\bar X= \sum_(i=1)^n (x_i)/(n)` (2)

`s=\sqrt{(\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(15.595)/(√(5))=127.108`

`137.8+1.533(15.595)/(√(5))=148.492`

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