Important Notes: (i) Show all work in the spaces provided. Part marks may be giv
ID: 3130331 • Letter: I
Question
Important Notes: (i) Show all work in the spaces provided. Part marks may be given for partially correct solutions. (ii) Full credit may not be given for correct answers if they are not adequately justified (iii) Calculator is permitted The manager of a grocery store has taken a random sample of 100 customers. The average length of time it took the customers in the sample to check out was x =3.1 minutes with a standard deviation of s = 0.5 minutes. The manager wants to test to determine whether the mean waiting time (mu) of all customers is significantly less than 3 minutes. State the null and alternative hypotheses. Find the value of the appropriate test statistic. What is the p-value? At.the 5% level of significance, what is you, conclusion?Explanation / Answer
let X denote the length of the customers to check out
assuption is that X~N(mu,sigma2) where sigma is unknown
the manager wants to test whether mu is significantly less than 3 minutes.
a) hence the null hypothesis is H0: mu=3 and the alternative hypothesis is H1: mu<3
b) to test the hypothesis there is a random sample of size n=100 with sample mean=xbar=3.1 minutes and standard deviation=s=0.5
hence the test statistic for the test of the above hypothesis is
T=(xbar-3)*sqrt(n)/s which under H0 follows a t distribution with degrees of freedom n-1
hence the value of the test statistic is t=(3.1-3)*sqrt(100)/0.5=2 [answer]
c) since the alternative hypothesis is left sided hence the p value is
p=P[T<2] where T follows a t distribution with degrees of freedom=n-1=99
so p=0.975880 [answer] [using MINITAB]
d) at 5% level of significance we have alpha=level of significance=0.05
now p>alpha
hence H0 is accepted and the conclusion is that the mean waiting time of all customers is not significantly less than 3 minutes. [answer]