Question

QUESTION 6

1. An office manager is testing a new system that is intended to decrease the variance of the time customers wait before they relate to a service representative. Under the old system, a random sample of 18 customers had a variance of 250. Under the new system, a random sample of 9 customers had a standard deviation of 11. At α = 0.10, is there enough evidence to convince the manager that the new system has a lower variance? Assume both populations are normally distributed and randomly selected.

   1. What is the null hypotheses and alternative hypotheses: (2)
   2. What are the degrees of freedom for this distribution, what distribution are we going to use? (3)
   3. What I the critical value of this distribution: (1)
   4. What is the test statistic for out sample (2)
   5. Do we reject or accept the null hypotheses? Why? (2)

Answer 1

: variance of the time customers wait before they relate to a service representative under the old system

: variance of the time customers wait before they relate to a service representative under the new system

Null hypothesis :

Alternative hypothesis:

Distribution we are going to use : F-Distribution

Degrees of freedom for numerator : Sample size of old system sample = n1-1= 18-1=17

Degrees of freedom for denominator = Sample size of new system sample :n2-1 = 9-1 =8

Sample variance under the old system : s12 = 250

Sample standard deviation under the new system : s2 = 11; s22 = 11 x 11 =121

Critical value of the distribution :

Significance level : =0.10

For right tailed test :

F0.10 for Degrees of freedom for numerator :17 and Degrees of freedom for denominator:8 = 2.4458

Test Statistic

Test Statistic = 2.0661

As Value of the test statistic: is less than Critical Value i.e. ( 2.0661<2.4458 ); Fail to Reject Null Hypothesis.

Fail to reject the null hypothesis.

There is not sufficient evidence to conclude that the new system has a lower variance