Question

Suppose that the manager of the MileagePlus frequent flier program is promoted and consequently another individual is hired to replace him. Also suppose that United publishes in internal documentation that the average number of Premier Qualifying Miles (PQM) earned by individuals who travel for work at least once a month is 45,000 with a standard deviation of 5,000 miles. Further suppose that the new manager desires to test the claims that United has made to see if the statistics have changed.

a) First, are these statistics given by United describing the parent population or a sample ?

b) Define appropriate null and alternative hypothesis

c) Suppose that the new manager takes a sample of 50 such United customers. What is the probability that a sample of size 50 provides a sample mean within + - 1,000 miles of the 45,000 mile figure provided by United?

d) What is the probability that a sample of 50 such United customers provides a sample mean wiithin + - 500 miles?

e) Suppose that the new manager's sample has a sample average of 47,500 miles. Compute the 95% confidence interval for the population mean.

f) Based on the confidence interval you computed in part e), does this sample provide evidence for or against United's claim that the average number of Premier Qualifying Miles (PQM) earned is 45,000 ?

g) If the new manager wants to determine a specific p-value for the likelihood that the null hypothesis is true based on the sample collected in part c), should he use Z- or t- scores for the test statistic?

h) Determine and interpret the p-value

Answer 1

(a)

The statistics are given from a parent population.

Our parameter is the population mean . The new manager wants to test whether the statistics have changed,i.e. , if the mean is still 45000 or not.

(c)

To calculate the probability we need it's distribution. We assume that the population follows a normal distribution.We have a sample of 50

, therefore using Central Limit Theorem

(

P(45000-1000 < < 45000+1000) =

    )

    P(Z<-1.41) =1 - P(Z<1.41)

= 2*0.92073 - 1 (values obtained using noral probability table)

Final Ans: 0.84146

(d)

Using the steps same as above we have

P( 45000 - 500 < < 45000 + 500 ) = P (44500 < < 45500)

Converting this to Z we get

Final Ans : 0.5223 ( P(Z < 0.71) = 0.76115)

(e)

and

95% Confidence interval for population mean

Final Ans :

(f)

We have a claim that population mean is 45000 , but since 95% confidence interval does not include this value. Hence we cn conclude that at it provides evidence against the manager's claim of the average being 45000.

(g)

The manager should use a z- test since we have the population statistics, that is our population standard deviation is known and also the sample size is large.

(h)

Since our alternative hypotheses test the inequality on both sides we will use a 2-sided z-test

Our z -score will be = 3.5355

p-value = 0.000407 (can be calculated online or using excel function 'normdist')

Decision criteria: Reject if p-value < 5%

Decision : 0.000407 < 5% hence we reject and conclude that the population average is not 45000 at 5% level of significance.