A study is conducted to compare the tensile strength of two

A study is conducted to compare the tensile strength of two types of roof coatings. It is thought that, on the average, butyl coatings are stronger than acrylic coatings. These data are gathered (?Urethane Foam-An Efficient Alternative,? J. Collett, Military Engineer, August 1983, pp. 348-350): Use alpha = 0.05 a. What are the hypotheses? b. Calculate the test statistic. c. What is the P value of the test? d. What is the decision about the null hypothesis?

Solution

Let sample 1 be Acrylic, and sample 2 be Butyl, and the difference ud = u1 - u2.
              
Formulating the null and alternative hypotheses,              
              
Ho:   u1 - u2   >=   0  
Ha:   u1 - u2   <   0 [PART A]
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At level of significance =    0.05          
As we can see, this is a    left   tailed test.      
Calculating the means of each group,              
              
X1 =    300.22          
X2 =    319.11          
              
Calculating the standard deviations of each group,              
              
s1 =    21.741895          
s2 =    17.86365          
              
Thus, the standard error of their difference is, by using sD = sqrt(s1^2/n1 + s2^2/n2):              
              
n1 = sample size of group 1 =    16          
n2 = sample size of group 2 =    16          
Thus, df = n1 + n2 - 2 =    30          
Also, sD =    7.034824045          
              
Thus, the t statistic will be              
              
t = [X1 - X2 - uD]/sD =    -2.685212861   [PART B]

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where uD = hypothesized difference =    0          
              
Now, the critical value for t is              
              
tcrit =    -1.697260887          
              
Thus, comparing t and tcrit, we decide to   WE REJECT THE NULL HYPOTHESIS.          
              
Also, using p values,              
              
p =    0.00584645 [PART C]

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Comparing this to the significance level,    WE REJECT THE NULL HYPOTHESIS.   [PART D]

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