1. ## Two-Sample t Test

[SIZE=3]The deterioration of many municipal pipeline networks across the country is a growing concern. One technology proposed for pipeline
rehabilitation uses a flexible liner threaded through existing pipe.

The article "Effect of Welding on a High-Density Polyethylene Liner" (J. of Materials in Civil Engr., 1996: 94-100) reported the following data on tensile strength (psi) of liner specimens when a certain fusion process was used and when this process was not used.

No Fusion :: 2748, 2700, 2655, 2822, 2511, 3149, 3257, 3213, 3220, 2753

Yes Fusion :: 3027, 3356, 3359, 3297, 3125, 2910, 2889, 2902

Compare the mean tensile strengths at the 95% confidence level.

Perform a COMPLETE APPROPRIATE analysis. Answer the question of interest, check assumptions, etc.

My work so far:

For No Fusion: x̅ = 2902.8, n1 = 10, s^2 = 76875.956, s = 277.265

For Yes Fusion: ȳ= 3108.125, n2 = 8, s^2 = 42382.411, s = 205.870

Then,

Sp˛ = 61785.03

Sp = 248.566

S (
x̅ - ȳ ) = 113.953

The part that is confusing me is how to find the degrees of freedom and how to get the t-value.

2. ## Re: Two-Sample t Test

The variances appear to be different so unless you have a reason to believe the variance is equal assume that variances are unequal. Sample sizes are also unequal so you will need to use Welch's t-test.
The t statistic is calculated the same way as your normally would

The formula to calculate degrees of freedom is quite long so here's a link to it Student's t-test - Wikipedia, the free encyclopedia To be conservative you should round down the degrees of freedom. For the Welsh test the pooled standard deviation is not used, a different estimator of standard deviation is used.

3. ## Re: Two-Sample t Test

Okay, well, then this is what I did:

A = s1^2 / n1 = 76875.956/10 = 7687.6

B = s2^2 / n2 = 42382.411/8 = 5297.8

df = [ (A + B)^2 ] / [ (A^2)/(n1 - 1) + (B^2)/(n2 - 1) ] = 15.94

rounding df to be 15

then, t = (x̅ - ȳ) / S ( x̅ - ȳ ) = (2902.8 - 3108.125)/ 113.953 = -1.8

The part that I'm confused about is the 95% confidence level. Alpha = 0.05 at 95% confidence. So, why would I use that formula above to get t instead of t(0.025, 15) = 2.131??

4. ## Re: Two-Sample t Test

$t=\pm 2.131$ is the critical value of the t statistic, -1.8 is the t statistic calculated from the data. Since -1.8 isn't outside the range [-2.131,2.131] you cannot reject the null hypothesis, the means are not significantly different.

5. ## Re: Two-Sample t Test

Oh, okay, then that makes sense. The other thing that I have to do is a 95% confidence interval.

The formula for that is:

(x̅ - ȳ) ± t * S ( x̅ - ȳ ) = (2902.8 - 3108.125) ± (2.131)*(113.953) = -205.3 ± 242.834

So, the confidence interval would end up being: (-448.134, 37.534)

The only problem that I have with this is that I googled this problem to compare my answer and every website with this same exact problem had the confidence interval being: (-488, 38)

I cannot figure out what I am doing wrong. The rest of my calculations match theirs perfectly except for the confidence interval.

6. ## Re: Two-Sample t Test

Nevermind. I figured out the whole problem, so disregard my comment above. Thank you for your help earlier though!!