An airline knows that on average about 5 percent of the people who make reservations on a certain flight will not show up. Conse
1 answer:
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Answer:
n=50
Step-by-step explanation:
1) 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 for the sample
population mean (variable of interest)
s=1.8 represent the sample standard deviation
n=25 represent the sample size
2) Solution to the problem
The confidence interval for the mean is given by the following formula:
(1)
We need to find the degrees of freedom given by:
Since the Confidence is 0.95 or 95%, the value of and
, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.025,0,1)".And we see that
Since we assume that we are taking a bigger sample then we can replace the t distribution with the normal standard distribution, and we can assume that th population deviation is 1.8. The margin of error is given by this formula:
(a)
And on this case we have that ME =0.5 and we are interested in order to find the value of n, if we solve n from equation (a) we got:
(b)
Replacing into formula (b) we got:
So the answer for this case would be n=50 rounded up to the nearest integer