Answer to Question #264057 in Statistics and Probability for Zahid

Question

A certain polymer is used for evacuation systems

for aircraft. It is important that the polymer be resistant

to the aging process. Twenty specimens of the

polymer were used in an experiment. Ten were assigned

randomly to be exposed to an accelerated batch

aging process that involved exposure to high temperatures

for 10 days. Measurements of tensile strength of

the specimens were made, and the following data were

recorded on tensile strength in psi:

No aging: 227 222 218 217 225

218 216 229 228 221

Aging: 219 214 215 211 209

218 203 204 201 205

(a) Do a dot plot of the data.

(b) From your plot, does it appear as if the aging process

has had an effect on the tensile strength of this

polymer? Explain.

(c) Calculate the sample mean tensile strength of the

two samples.

(d) Calculate the median for both. Discuss the similarity

or lack of similarity between the mean and

median of each group.

Answer 1

(a) The data is plotted in the dot plot shown in the figure below The yellow dots represent the tensile strength

data for No aging group and green dots represent the tensile strength data for Aging group.

Sample size = n = 10 (for both groups) 

(b) As we can see in the dot plot that green dots (which represent sample undergone Aging process) are more to

the left, which means that tensile strength of specimens from that group is lower than the other group.

Hence it appears as the aging process has had an effect on the tensile strength of this polymer and the tensile

strength reduces due to aging process. 

(c) Let us denote sample mean tensile strength of the two samples by $\bar{x}_{N o . \text { Aging }} and \bar{x}_{\text {Aging, }},$ then

 

$\begin{aligned} \bar{x}_{\text {No.Aging }} &=\frac{227+222+218+217+225+218+216+229+228+221}{10}\\=\frac{2221}{10}=222.1 \\ \bar{x}_{\text {Aging }} &=\frac{219+214+215+211+209+218+203+204+201+205}{10}\\=\frac{2099}{10}=209.9 \end{aligned}$

 

(d) Let us denote median for both samples by $\tilde{x}_{\text {No.Aging }} \ and \ \tilde{x}_{\text {Aging }}.$

Since sample size (n=10) for both samples is an even number, hence expression for the median is given by:

 

$\operatorname{median}=\frac{x_{n / 2}+x_{n / 2+1}}{2}=\frac{x_{5}+x_{6}}{2}$

 

Let us arrange the data for no aging group in ascending order :

 $\begin{aligned} &216,217,218,218,221,222,225,227,228,229 \\ &\tilde{x}_{\text {No.Aging }}=\frac{x_{5}+x_{6}}{2}=\frac{221+222}{2}=221.5 \end{aligned}$

Let us arrange the data for aging group in ascending order :

 

$\begin{gathered} 201,203,204,205,209,211,214,215,218,219 \\ \tilde{x}_{\text {Aging }}=\frac{x_{5}+x_{6}}{2}=\frac{209+211}{2}=210 \end{gathered}$

Considering each group separately, each group's mean and median are very similar to each other. As we can see from the dot plot that for both samples data is almost symmetrically distributed. But since the data for Aging group is more to the left hence it's measure of centers (both mean and median) are less than those for no aging group.