I need help answering number 6 with work shown so I\'ll know please The compress

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I need help answering number 6 with work shown so I'll know please The compressive strength, in kilopascals, was measured for concrete blocks from five different batches of concrete, both three and six days after pouring. The data are as follows: Are these paired or independent samples? Construct and interpret a 95 percentage confidence interval for the difference between the mean strengths of blocks cured three days and blocks cured six days. (Use the same data as #6) A civil engineer believes there is no difference between the mean strengths of blocks cured three days and blocks cured six days. Use the alpha = 0.05 level of significance to test the hypothesis. State the null and alternative hypotheses. Perform the test State the conclusion

Explanation / Answer

6.

A)

This is paired, as the same batches are used after 3 and 6 days.

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b)

Note that              
              
Lower Bound = X - t(alpha/2) * s / sqrt(n)              
Upper Bound = X + t(alpha/2) * s / sqrt(n)              
              
where              
alpha/2 = (1 - confidence level)/2 =    0.025          
X = sample mean =    -33.2          
t(alpha/2) = critical t for the confidence interval =    2.776445105          
s = sample standard deviation =    15.49838701          
n = sample size =    5          
df = n - 1 =    4          
Thus,              
              
Lower bound =    -52.44378918          
Upper bound =    -13.95621082          
              
Thus, the confidence interval is              
              
(   -52.44378918   ,   -13.95621082   ) [ANSWER]

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7.

A)

Let ud = u2 - u1.  

where u2 = after 3 dyas
u1 = after 6 days
          
Formulating the null and alternative hypotheses,              
              
Ho:   ud   =   0  
Ha:   ud   =/   0   [ANSWER]

b)

At level of significance =    0.05          

As we can see, this is a    two   tailed test.      
              
Calculating the standard deviation of the differences (third column):              
              
s =    19.48348174          
              
Thus, the standard error of the difference is sD = s/sqrt(n):              
              
sD =    8.713277921          
              
Calculating the mean of the differences (third column):              
              
XD =    -33.2          
              
As t = [XD - uD]/sD, where uD = the hypothesized difference =    0   , then      
              
t =    -3.810276718          
              
As df = n - 1 =    4          
              
Then the critical value of t is              
              
tcrit =    +/-   2.776445105      
              
As |t| > 2.776, WE REJECT THE NULL HYPOTHESIS.          
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c)

Thus, there is significant difference between the mean strengths of blocks after 3 days and 6 days. [CONCLUSION]

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