Question

The deterioriation 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. A study reports the following data on tensile strength (psi) of liner specimen both when a certain fusion process was used and when the process was not used.

Strength with Fusion 3027 3356 3359 3297 3125 2910 2889 2902
Strength without Fusion 2748 2700 2655 2822 2511 3149 3257 3213 3220 2753
The authors of the study state that the fusion process increased the average tensile strength.
*Perform a hypothesis test (α=0.05) to see if the data supports this claim?

Answer 1

`t=\frac{(3108.25-2920)-0}{\sqrt{(219.528^2)/(8)+(295.395^2)/(10)}}}=1.550`

`p_v =P(t_(16)<1.55)=0.0703`

Comparing the p value with the significance level given
`\alpha=0.05` we see that
`p_v>\alpha` so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can't conclude that the mena for the Fusion process is significanlty higher than the mean for the without Fusion process.

Explanation:

Data given and notation

We can calculate the sample mean and deviation with the following formulas:

`\bar X=(\sum_(i=1)^n X_i)/(n)`

`s=\sqrt{(\sum_(i=1)^n (X_i-\bar X)^2)/(n-1)}`

Data given:

Strength with Fusion 3027 3356 3359 3297 3125 2910 2889 2902

Strength without Fusion 2748 2700 2655 2822 2511 3149 3257 3213 3220 2753

We got the following results:

`\bar X_(F)=3108.25` represent the mean for the with Fusion sample

`\bar X_(WF)=2920` represent the mean for the without Fusion sample

`s_(F)=219.528` represent the sample standard deviation for Fusion sample

`s_(WF)=295.395` represent the sample standard deviation for the without Fusion sample

`n_(F)=8` sample size for the group Fusion

`n_(WF)=10` sample size for the group without Fusion

t would represent the statistic (variable of interest)

Concepts and formulas to use

We need to conduct a hypothesis in order to check if the mean for the Fusion process is higher than the mean for the without Fusion process, the system of hypothesis would be:

Null hypothesis:
`\mu_(F)-\mu_(WF) \leq 0`

Alternative hypothesis:
`\mu_(F) - \mu_(WF)>0`

We don't have the population standard deviations, so for this case is better apply a t test to compare means, and the statistic is given by:

`t=\frac{(\bar X_(F)-\bar X_(WF))-\Delta}{\sqrt{(\sigma^2_(F))/(n_(F))+(\sigma^2_(WF))/(n_(WF))}}` (1)

And the degrees of freedom are given by
`df=n_F +n_(WF) -2=8+10-2=16`

t-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.

Calculate the statistic

With the info given we can replace in formula (1) like this:

`t=\frac{(3108.25-2920)-0}{\sqrt{(219.528^2)/(8)+(295.395^2)/(10)}}}=1.550`

P value

Since is a right tailed test the p value would be:

`p_v =P(t_(16)<1.55)=0.0703`

Comparing the p value with the significance level given
`\alpha=0.05` we see that
`p_v>\alpha` so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can't conclude that the mena for the Fusion process is significanlty higher than the mean for the without Fusion process.