Question

With individual lines at its various windows, a post office finds that the standard deviation for normally distributed waiting times for customers on Friday afternoon is 7.2 minutes. The post office experiments with a single, main waiting line and finds that for a random sample of 25 customers, the waiting times for customers have a standard deviation of 3.5 minutes.

With a significance level of 5 percent, test the claim that a single line causes lower variation among waiting times (shorter waiting times) for customers.

what is the p-value?

Answer 1

Here we have given that

n= number of customers=25

s= sample standard deviation = 3.5 minutes

claim : To check weather a single line causes lower variation among waiting times (shorter waiting times) for customers i.e. minutes

The hypothesis are

minutes

v/s

minutes
Here we assume that the population is normally distributed.

Now,

Test statistic is as follows,

=

= 5.67

we get the Test statistics is 5.67

Now we can find the P-value

= level of significance=0.05

Degrees of freedom = n-1 = 25-1 =2

P-value= 0.99996 using excel = CHIDIST( chi-statistics=5.67,D.F=24)

Now,

Decision:

Here P-value > 0.05

that is here we fail to reject the Null hypothesis Ho

Conclusion:

we can conclude that there is not sufficient evidence to support the claim single line causes lower variation among waiting times (shorter waiting times) for customers i.e. minutes