Regression and Inferential Statistic

Regressionand Inferential Statistic

Regressionand Inferential Statistic

Thetype of hypothesis used by the Airline advertisement team, in thiscase, is the null hypothesis. It is because the null hypothesis hasthe claim that a population parameter is equal to some number. It isalways treated to be correct until proven otherwise. For our case,the population parameter under consideration is the average number oftime in minutes that is taken when the airplane gets to the gate inrelation to when the passengers will have to receive their bags.

Basingon the data provided, our null hypothesis becomes:

Ho:u ≤20.

Thatis, the average number of minutes taken is equal to or less than 20minutes. It is because the phrase “within” 20 minutes implies anytime span to the upper limit of 20 minutes. It, therefore, means thatunder no circumstance whatsoever will the time taken in the event ofpassengers being availed their bags exceed 20 minutes.

Toprove the claim otherwise, we focus then on the alternativehypothesis which is the hypothesis against which the null hypothesisis rejected. So, we seek to disapprove the claim that passengersalways receive their bags within a time span of 20 minutes. Theantonym of it will be that passengers get their bags in a period morethan 20 minutes after the plane arrives at the gate.

Forour case, the alternative hypothesis becomes:

HA:u ˃ 20, Thatis, the time that will be taken for the passengers to get their bagsavailed to them exceeds 20 minutes.

Illustration:Suppose that we conducted some survey to prove the validity of theclaim that passengers get their bags within 20 minutes. A study wascarried out for a week with a sample size of 30 passengers daily andfound out that the average time taken for the passengers to get theirbags was 25 minutes. Will there be sufficient evidence to supportthat claim say at 5% level of significance? Lets us use standarddeviation (σ = 8 minutes)

Solution:

Uo=20 minutes, X = 25 minutes, n = 30 passengers, σ = 8 minutes, α =0.05

Step1: Formulation of hypothesis

Ho:u ≤20.

HA:u ˃ 20

Step2: The critical test statistic

Zα= Z0.05+ 1.645 from the Z- table

Step3: The test statistic

butwe have σ /√n

=25-201.46

=3.423

Step4: The decision

Sincethe computed Z- value is greater than the critical test statistic,that is: 3.423 ˃ 1.645, we reject the null hypothesis above.

Step5: The conclusion

Fromthe above test, there is adequate evidence to reject the claim thatthe average time for the passengers to get their bags once the planegets to the gate is within 20 minutes. That is, the average timetaken is more than 20 minutes.

Fromthe above illustration, we can see how the null hypothesis can deducethat it is not likely the case that the passengers will be spendingwithin a time frame of 20 minutes to get their bags, but thereinstances that the time frame may be exceeded.

Therefore,the statement above is as a result of applying the null hypothesis inthe advertisement to win the passenger`s trust as far as timemanagement is concerned.

References

Weiss,N. A. (2016). Introductory Statistics (10th Ed.). New York, NY:Pearson Education.

Gareth,J. (2013). An Introduction to Statistical is learning withapplications.

Medhi,T. (1982). Stochastic process.