Small sample test for the difference between two means

[Poke]

Homework Statement

The article “Variance Reduction Techniques: Experimental Comparison and Analysis for Single Systems” (I.Sabuncuoglu,M. Fadiloglu, and S. Celik, IIE Transactions, 2008:538–551) describes a study of the effectiveness of the method of Latin Hypercube Sampling in reducing the variance of estimators of the mean time-in-system for queueing models. For the M/M/1 queueing model, ten replications of the experiment yielded an average reduction of 6.1 with a standard deviation of 4.1. For the serial line model, ten replications yielded anaveragereductionof6.6withastandarddeviationof 4.3. Can you conclude that the mean reductions differ between the two models?

Homework Equations

nx = 10 ; X= 6.1 ; Sx = 4.1
ny = 10 ; Y= 6.6 ; Sy = 4.3

H0: X-Y = 0 ; H1: X-Y ≠0 ; Δ0 = 0

As we cannot assume standard deviations to be equal, we use the formula that find student's t and the degree of freedom v.

The Attempt at a Solution

It is found that v = 17.959, rounded down to 17
with t calculated as 0.2661

As H1: X-Y ≠0, P is the sum of cutoffs
We are to find Alpha, based on v = 17 and t' = 2t = 0.5322
It is founded that alpha between 0.25 and 0.40

All of which higher than 5%, so H0 is not rejected, and cannot conclude the mean reductions differ between the two models.My approaches correct, as well as the answer make sense? Thanks!

 

[FactChecker]

Not sure why you doubled t to t' = 2t = 0.5322 rather than using t directly. Do you have a reference for that? (I don't think that Welch's t-test does that.)

Your final conclusion of no significant difference certainly makes sense. A rough "guestimate" of the 95% confidence interval on the first sample puts the second sample mean well within the interval.