Martys Burger Bam has historically utilized a onewindow driv

Marty\'s Burger Bam has historically utilized a one-window drive thru service. However, they have recently been experimenting with two-window setups. Let denote the average time a customer spends in the drive thru lane under a one-window setup, and let fi: denote the average time a customer spends in the drive thru lane under a two-window setup. Marty has asked for your help in comparing the two configurations., so you have correctly formulated the following hypothesis test. To perform the test, you asked six customers to go to two Marty\'s Burger Bam locations (one using each configuration), place an identical order at each location and record the time they spent in the drive thru lane. The results are summarized in the table below. You intend to use a level of significance of 0.1. Compute the test statistic and determine the conclusion of the test using the critical region. Verify the result from part (b) using a confidence interval. What can you say about the efficiency of the two configurations?

Solution

a)
Set Up Hypothesis
Null , There Is No-Significance between them Ho: u1 = u2
Alternate, There Is Significance between them - H1: u1 != u2
Test Statistic
X(Mean)=132.683
Standard Deviation(s.d1)=64.632 ; Number(n1)=6
Y(Mean)=120.483
Standard Deviation(s.d2)=54.276; Number(n2)=6
we use Test Statistic (t) = (X-Y)/Sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =132.683-120.483/Sqrt((4177.29542/6)+(2945.88418/6))
to =0.35
| to | =0.35
Critical Value
The Value of |t | with Min (n1-1, n2-1) i.e 5 d.f is 2.015
We got |to| = 0.35408 & | t | = 2.015
Make Decision
Hence Value of |to | < | t | and Here we Do not Reject Ho
P-Value: Two Tailed ( double the one tail ) - Ha : ( P != 0.3541 ) = 0.738
Hence Value of P0.1 < 0.738,Here We Do not Reject Ho

b)
CI = x1 - x2 ± t a/2 * Sqrt ( sd1 ^2 / n1 + sd2 ^2 /n2 )
Where,
x1 = Mean of Sample 1, x2 = Mean of sample2
sd1 = SD of Sample 1, sd2 = SD of sample2
a = 1 - (Confidence Level/100)
ta/2 = t-table value
CI = Confidence Interval
Mean(x1)=132.683
Standard deviation( sd1 )=64.632
Sample Size(n1)=6
Mean(x2)=120.483
Standard deviation( sd2 )=54.276
Sample Size(n1)=6
CI = [ ( 132.683-120.483) ±t a/2 * Sqrt( 4177.295424/6+2945.884176/6)]
= [ (12.2) ± t a/2 * Sqrt( 1187.2) ]
= [ (12.2) ± 4.032 * Sqrt( 1187.2) ]
= [-126.73 , 151.13]