1. A sample of 33 airline passengers found that the average check-in time is 2.167. Based on long-term data, the population standard deviation is known to be 0.48. Find a 95% confidence interval for the mean check-in time (2 Points). 2. If 5% of parts produced in a factory are defective, what is the sample size that needs to be taken for 0.03 desired margin of error? Use 95% CL.

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.

answer: x=60

answer: #1 ads up by 8 and i don't know #2