Question

An airline reservation system suffers a 10% rate of no-shows. A new reservation system is then instituted in hopes of REDUCING this no-show rate. After implementation of the new procedure, a SRS of 2500 reservations showed a 10.24% (256 out of 2500) no-show rate.
A) what is the population?
B) variable of interest?
C)what are the null and alternative hypotheses?
D)determine the value of the appropriate observed test statistic to perform this test (note: 10.24%=0.01024 & 10%=0.1000)
E)whats the p-value?

Answer 1

A) The population is the total number of reservations that took place at the airline reservation system. B) The variable of interest is the no-show rate C) Since the new reservation system instituted, aims to reduce the no-show rate. Hence the null and alternative hypotheses can be stated as follows: Null hypothesis: Ho: P = 0.10 i.e. the no-show rate of the airline reservation system is 10% Alternate hypothesis: Hj: P < 0.10 i.e. the no-show rate of the airline reservation system is less than 10%. D) Since the sample size is large, we will use "One sample z-test for population proportions" P-P. Hence formula of test statistic will be Z= P.Q Vn where Z=computed value of test statistic P=population proportion i.e. proportion of units belonging from a particular class in a population Por specified value of proportion n=sample size Qo=1-Po P is calculated according to the formula P=“, here a is the number of units belonging to a particular class. In the given problem, a = 256 and n = 2500 256 Hence, P= = = 0.1024 25000.10 Po = 0.10 (given) Qo=1 - 0.10 = 0.90 So, substituting the values of P, Po, Qo, and n in the formula of test statistic, we get 2_0.1024 -0.10 <=0.4 0.10x 0.90 V 2500 Therefore, the value of test statistic is 0.4 z E) Using the Standard Normal Z-table, at Z= 0.4, the observed p-value is area to the left side of 0.4 which is 0.6554 or 65.54%