Question

The manager of a coffee shop called Perks is interested in determining if a new ordering system that a competitor's store, Cool Beans, has been using might make service time quicker on average at his store. Minor differences in mean service time are to be expected, but a decrease of more than 30 seconds using the new system would be practically significant to the manager of Perks and allow him to conclude the new technology would be useful for his business.

Let μ₁ be the mean service time at Perks and μ₂ be the mean service time at Cool Beans. The manager goes through the point-of-sale data and randomly selects a sample of 46 orders from the weekday 6-10 a.m. shift (Group 1) and records the service time. He offers double overtime to one of his employees to hang out inconspicuously in the Cool Beans parking lot to watch the drive-thru lane at Cool Beans, recording another 46 service times using its ordering system (Group 2). The results are summarized in the table below.

Group	Mean	Standard Deviation	n
Group 1 (Perks)	215	29	42
Group 2 (Cool Beans)	189	34	42

Calculate a 90% confidence interval for the difference in mean service time before and after implementing the new technology. Assume the population standard deviations are not equal (Case 2). Take all calculations toward the answer to three (3) decimal places, and report your answer to three (3) decimal places. Use the following website to help you calculate v: http://web.utk.edu/~cwiek/TwoSampleDoF

Now, using your answer, fill in the interpretation below. Refer to Group 1 as "before the change" and Group 2 as "after the change."

We can be ______% confident that the true difference in mean _____ between ______and _____ is between _____ and _____. Therefore, we _____ (can/cannot) conclude the mean service time for the two groups differs.

Answer 1

TRADITIONAL METHOD

given that,
mean(x)=215
standard deviation , s.d1=29
number(n1)=42
y(mean)=189
standard deviation, s.d2 =34
number(n2)=42
I.
stanadard error = sqrt(s.d1^2/n1)+(s.d2^2/n2)
where,
sd1, sd2 = standard deviation of both
n1, n2 = sample size
stanadard error = sqrt((841/42)+(1156/42))
= 6.895
II.
margin of error = t a/2 * (stanadard error)
where,
t a/2 = t -table value
level of significance, α = 0.1
from standard normal table, two tailed and
value of |t α| with min (n1-1, n2-1) i.e 41 d.f is 1.683
margin of error = 1.683 * 6.895
= 11.605
III.
CI = (x1-x2) ± margin of error
confidence interval = [ (215-189) ± 11.605 ]
= [14.395 , 37.605]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
mean(x)=215
standard deviation , s.d1=29
sample size, n1=42
y(mean)=189
standard deviation, s.d2 =34
sample size,n2 =42
CI = x1 - x2 ± t a/2 * Sqrt ( sd1 ^2 / n1 + sd2 ^2 /n2 )
where,
x1,x2 = mean of populations
sd1,sd2 = standard deviations
n1,n2 = size of both
a = 1 - (confidence Level/100)
ta/2 = t-table value
CI = confidence interval
CI = [( 215-189) ± t a/2 * sqrt((841/42)+(1156/42)]
= [ (26) ± t a/2 * 6.895]
= [14.395 , 37.605]

We can be 90% confident that the true difference in mean is between [14.395 , 37.605]
Therefore, we can conclude the mean service time for the two groups differs.

MINITAB PROCEDURE

Two-Sample T-Test and CI

Sample N Mean StDev SE Mean
1 42 215.0 49.0 7.6
2 42 189.0 34.0 5.2

Difference = μ (1) - μ (2)

Estimate for difference: 26.00
90% CI for difference: (10.67, 41.33)
T-Test of difference = 0 (vs ≠): T-Value = 2.83 P-Value = 0.006 DF = 73

NOTE: calculated confidence value from minitab is diffrent since degrees of freedom considered as other value rather when consider to written method procedure, hence the changes.