Question

Most air travelers now use e-tickets. Electronic ticketing allows passengers to not worry about a paper ticket, and it costs the airline companies less to handle than paper ticketing. However, in recent times the airlines have received complaints from passengers regarding their e-tickets, particularly when connecting flights and a change of airlines were involved.
To investigate the problem, an independent watchdog agency contacted a random sample of 20 airports and collected information on the number of complaints the airport had with e-tickets for the month of March.
The information is reported below.

14, 14, 16, 12, 12, 12, 15, 15, 14, 13, 14, 13, 16, 15, 14, 13, 12, 13, 10, 13

At the .05 significance level, can the watchdog agency conclude the mean number of complaints per airport is less than 15 per month?

Answer 1

Answer:

$t=\frac{13.5-15}{\frac{1.504}{\sqrt{20}}}=-4.46$    

$p_v =P(t_{(19)}$    

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is lower than 15 months at at 5% of signficance.    

Step-by-step explanation:

Data given and notation    

Data: 14, 14, 16, 12, 12, 12, 15, 15, 14, 13, 14, 13, 16, 15, 14, 13, 12, 13, 10, 13

We can calculate the mean and deviation with the following formulas:

$\bar X = \frac{\sum_{i=1}^n X_i}{n}$

$s = \sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}$

$\bar X=13.5$ represent the sample mean

$s=1.504$ represent the sample standard deviation    

$n=20$ sample size    

$\mu_o =15$ represent the value that we want to test  

$\alpha=0.05$ represent the significance level for the hypothesis test.    

z would represent the statistic (variable of interest)    

$p_v$ represent the p value for the test (variable of interest)    

State the null and alternative hypotheses.    

We need to conduct a hypothesis in order to check if the mean is less than 15 :    

Null hypothesis:$\mu \geq 15$    

Alternative hypothesis:$\mu < 15$    

Since we don't know the population deviation, is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:    

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)    

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".    

Calculate the statistic    

We can replace in formula (1) the info given like this:    

$t=\frac{13.5-15}{\frac{1.504}{\sqrt{20}}}=-4.46$    

P-value    

First we need to calculate the degrees of freedom given by:

$df=n-1=20-1=19$

Since is a left tailed test the p value would be:    

$p_v =P(t_{(19)}$    

Conclusion    

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is lower than 15 months at at 5% of signficance.