Mathematics, 09.03.2020 21:11 Ayori144

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?

Answers: 2

Show answers

Answers

Answer from: dontcareanyonemo

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

$p_v =P(t_{(19)}$

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

Answer from: Quest

i think is 3 and 6. this took me 10 min to think.hope to and hope that my answer is brainliest.(follow me )

Answer from: Quest

The first one is correct

Another question on Mathematics

Mathematics, 21.06.2019 16:20

Which best explains why this triangle is or is not a right triangle?

Answers: 1

Answer

Mathematics, 21.06.2019 18:00

Look at arnold's attempt to solve the equation for b: 3b = 12 b = 3 · 12 b = 36 describe the mistake that arnold made.

Answers: 2

Answer

Mathematics, 21.06.2019 22:00

Identify the expression equivalent to 4(x + x + 7) − 2x + 8 − 4 by substituting x = 1 and x = 2.

Answers: 2

Answer

Mathematics, 21.06.2019 22:30

Solve: 25 points find the fifth term of an increasing geometric progression if the first term is equal to 7−3 √5 and each term (starting with the second) is equal to the difference of the term following it and the term preceding it.

Answers: 1

Answer

You know the right answer?

Most air travelers now use e-tickets. Electronic ticketing allows passengers to not worry about a pa...

Questions