Question

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.

Answer 1

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 $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1-0.95}{2} = 0.025$

Now, we have to find z in the Ztable as such z has a pvalue of $1-\alpha$.

So it is z with a pvalue of $1-0.025 = 0.975$, so $z = 1.96$

Now, find M as such

$M = z*\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

$M = 1.96*\frac{0.48}{\sqrt{33}} = 0.164$

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 $\pi$, and a confidence level of $1-\alpha$, we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$.

The margin of error is given by:

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

For this problem, we have that:

$\pi = 0.05$

95% confidence level

So $\alpha = 0.05$, z is the value of Z that has a pvalue of $1 - \frac{0.05}{2} = 0.975$, so $Z = 1.96$.

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

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

$0.03 = 1.96\sqrt{\frac{0.05*0.95}{n}}$

$0.03\sqrt{n} = 1.96\sqrt{0.05*0.95}$

$\sqrt{n} = \frac{1.96\sqrt{0.05*0.95}}{0.03}$

$(\sqrt{n})^{2} = (\frac{1.96\sqrt{0.05*0.95}}{0.03})^{2}$

$n = 203$

The sample size that needs to be taken is 273.