Unoccupied seats on flights cause airlines to lose revenue. Suppose a large airline wants to estimate its average number of unoccupied seats per flight over the past year. 225 flight records are randomly selected and the number of unoccupied seats is noted, with a sample mean of 11.6 seats and a standard deviation of 4.1 seats. Calculate a 90 percent confidence interval for μ, the mean number of unoccupied seats per flight during the past year.

So on this case the 90% confidence interval would be given by (11.148;12.052)    

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".

represent the sample mean

population mean (variable of interest)

s=4.1 represent the sample standard deviation

n=225 represent the sample size  

Solution to the problem

The confidence interval for the mean is given by the following formula:

  (1)

In order to calculate the critical value we need to find first the degrees of freedom, given by:

Since the Confidence is 0.90 or 90%, the value of and , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.05,224)".And we see that

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

So on this case the 90% confidence interval would be given by (11.148;12.052)    

answer: no

step-by-step explanation:

you are only given the cost not the amount of pounds.

if you are saying purchasing 3 social media ads and one newspaper ad counts as one purchase then there are:

6 purchases

if you are asking how many of each then:

18 social media ads

6 newspaper ads