Question

Analysis of the tensile strength of 100 samples of a new polymeric material, obtained from each of the two production lines of a company gave the following mean and standard deviation values (in MPa)

	Line 1	Line 2
sample size	100	100
mean	33.9	35.8
Standard Deviation	5.83	5.92

a. Use 5 % significant level can it be concluded that there is no difference in the tensile strength of the products obtained from the two production lines?

b. Using 95% confidence interval, estimate the difference in the mean strength of the polymers produced in the two production lines

c. The quality control engineer will only be concerned if there is a difference of more than 5 MPa in the products produced in the two lines. Based on your confidence interval in (b), is the statistical significance obtain in question (a) of practical significance to the Quality Control Engineer? Explain.

Answer 1

a) The null and alternative hypothesis

Test statistic

where

= 5.8752

Thus

= - 2.29

df =n1+n2-2 =100+100-2 =198

For 0.05 with df = 198 , two tailed critical value of t is

tc = 1.97 ( from t table)

Since calculated I t I > 1.97

Reject H0

There is sufficient evidence to conclude that there is difference in tensile strength of product obtained from two production lines .

b) The 95% confidence interval for the difference in mean tensile strength is

For 95% confidence with df =198 , tc =1.97

Thus , 95% confidence interval is

= (-3.54 , - 0.26 )

c) No , the statistical significance obtained in (a) is not of practical significance to the quality control engineer .

Because , the confidence interval of difference between mean tensile strength (-3.54 , - 0.26 ) does not contain the value -5 in it , which is minimum required diffe rence between the two tensile strength .

Note : If we construct the confidence interval for line 2 - line 1 , then would get (0.26 , 3.54 )

which again do not contain the value 5 in it .