An article includes the accompanying data on compression strength (lb) for a sam

Question

An article includes the accompanying data on compression strength (lb) for a sample of 12-oz aluminum cans filled with strawberry drink and another sample filled with cola. Does the data suggest that the extra carbonation of cola results in a higher average compression strength? Base your answer on a P-value. (Use alpha = 0.05.) State the relevant hypothesis. (Use mu_1 for the strawberry drink and mu_2 for the cola.) H_0 middot mu_1 - mu_2 = 0 H_a middot mu_1 - mu_2 plusminus 0 H_0 middot mu_1 - mu_2 = 0 H_a middot mu_1 - mu_2 Greaterthanorequalto 0 H_0 middot mu_1 - mu_2 = 0 H_a middot mu_1 - mu_2 > 0 H_0 middot mu_1 - mu_2 = 0 H_a middot mu_1 - mu_2

Explanation / Answer

a)

Formulating the null and alternative hypotheses,              
              
Ho:   u1 - u2   =   0  
Ha:   u1 - u2   <   0   [ANSWER, D]

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b)

At level of significance =    0.05          
As we can see, this is a    left   tailed test.      
Calculating the means of each group,              
              
X1 =    537          
X2 =    555          
              
Calculating the standard deviations of each group,              
              
s1 =    20          
s2 =    18          
              
Thus, the standard error of their difference is, by using sD = sqrt(s1^2/n1 + s2^2/n2):              
              
n1 = sample size of group 1 =    10          
n2 = sample size of group 2 =    10          
Thus, df = n1 + n2 - 2 =    18          
Also, sD =    8.508818954          
              
Thus, the t statistic will be              
              
t = [X1 - X2 - uD]/sD =    -2.115452226   [ANSWER, TEST STATISTIC]

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where uD = hypothesized difference =    0          
              
Also, using p values, as this is left tailed,              
              
p =    0.024297388   [ANSWER, P VALUE]

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As P < 0.05, we reject Ho.

Hence,

OPTION A: Reject Ho. This data suggests that cola has a higher average compression strength than the stawberry drink. [ANSWER, A]

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OPTION B: The distribution of compression strengths are approximately normal. [ANSWER]

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Hi! If you use another method/formula in calculating the degrees of freedom in this t-test, please resubmit this question together with the formula/method you use in determining the degrees of freedom. That way we can continue helping you! Thanks!

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