Question

The mean arrival rate of flights at Philadelphia International Airport is 195 flights or less per hour with a historical standard deviation of 13 flights. To increase arrivals, a new air traffic control procedure is implemented. In the next 30 days, the arrival rate per day is given in the data vector below called flights. Air traffic control manager wants to test if there is sufficient evidence that arrival rate has increased.
flights <- c(210, 215, 200, 189, 200, 213, 202, 181, 197, 199,
193, 209, 215, 192, 179, 196, 225, 199, 196, 210,
199, 188, 174, 176, 202, 195, 195, 208, 222, 221)
a) Find sample mean and sample standard deviation of arrival rate using R functions mean() and sd().
b) Is this a left-tailed, right-tailed or two-tailed test? Formulate the null and alternative hypothesis.
c) What is the statistical decision at the significance level α = .01?

Answer 1

Answer:

a) The sample mean is M=200.

The sample standard deviation is s=13.19.

b) Right-tailed. The null and alternative hypothesis are:

$H_0: \mu=195\\\\H_a:\mu> 195$

c) At a significance level of 0.01, there is  notenough evidence to support the claim that the arrival rate is significantly higher than 195.

Step-by-step explanation:

We start by calculating the sample and standard deviation.

The sample size is n=30.

The sample mean is M=200.

The sample standard deviation is s=13.19.

$M=\dfrac{1}{n}\sum_{i=1}^n\,x_i\\\\\\M=\dfrac{1}{30}(210+215+200+. . .+221)\\\\\\M=\dfrac{6000}{30}\\\\\\M=200\\\\\\s=\sqrt{\dfrac{1}{n-1}\sum_{i=1}^n\,(x_i-M)^2}\\\\\\s=\sqrt{\dfrac{1}{29}((210-200)^2+(215-200)^2+(200-200)^2+. . . +(221-200)^2)}\\\\\\s=\sqrt{\dfrac{5048}{29}}\\\\\\s=\sqrt{174.07}=13.19$

This is a hypothesis test for the population mean.

The claim is that the arrival rate is significantly higher than 195.  As we are interested in only the higher tail for a significant effect, this is a right-tailed test.

Then, the null and alternative hypothesis are:

$H_0: \mu=195\\\\H_a:\mu> 195$

The significance level is 0.01.

The standard deviation of the population is known and has a value of σ=13.

We can calculate the standard error as:

$\sigma_M=\dfrac{\sigma}{\sqrt{n}}=\dfrac{13}{\sqrt{30}}=2.373$

Then, we can calculate the z-statistic as:

$z=\dfrac{M-\mu}{\sigma_M}=\dfrac{200-195}{2.373}=\dfrac{5}{2.373}=2.107$

This test is a right-tailed test, so the P-value for this test is calculated as:

$\text{P-value}=P(z>2.107)=0.018$

As the P-value (0.018) is bigger than the significance level (0.01), the effect is not significant.

The null hypothesis failed to be rejected.

At a significance level of 0.01, there is  notenough evidence to support the claim that the arrival rate is significantly higher than 195.