Answer:
1) The 95% confidence interval for the mean check-in time is between 2.003 hours and 2.331 hours.
2) The sample size that needs to be taken is 273.
Step-by-step explanation:
1)
We have that to find our level, that is the subtraction of 1 by the confidence interval divided by 2. So:
Now, we have to find z in the Ztable as such z has a pvalue of .
So it is z with a pvalue of , so
Now, find M as such
In which is the standard deviation of the population and n is the size of the sample.
The lower end of the interval is the sample mean subtracted by M. So it is 2.167 – 0.164 = 2.003 hours
The upper end of the interval is the sample mean added to M. So it is 2.167 + 0.164 = 2.331 hours.
The 95% confidence interval for the mean check-in time is between 2.003 hours and 2.331 hours.
2)
In a sample with a number n of people surveyed with a probability of a success of , and a confidence level of , we have the following confidence interval of proportions.
In which
z is the zscore that has a pvalue of .
The margin of error is given by:
For this problem, we have that:
95% confidence level
So , z is the value of Z that has a pvalue of , so .
What is the sample size that needs to be taken for 0.03 desired margin of error?
This is n when M = 0.03. So
The sample size that needs to be taken is 273.