A civil engineer is analyzing the compressive strength of concrete. Compressive

Question

A civil engineer is analyzing the compressive strength of concrete. Compressive strength is approximately normally distributed with variance sigma2=1000psi2. A random sample of 12 specimens has a mean compressive strength of 3255.43psi.

A. Test the hypothesis that mean compressive strength is 3500psi. Use a fixed test with alpha=0.01.

B. What is the smallest level of significance at which you would be willing to reject the hypothesis?

C. Construct a 95% two-sided CI on mean compressive strength.

D. Construct a 99% two-sided CI on mean compressive strength. Compare the width of this confidence interval with the width of the one from part C.

Please show all work! Thank you!

Explanation / Answer

a)

Formulating the null and alternative hypotheses,              
              
Ho:   u   =   3500  
Ha:    u   =/   3500  
              
As we can see, this is a    two   tailed test.      
              
Thus, getting the critical t,              
df = n - 1 =    11          
tcrit =    +/-   3.105806516      
              
Getting the test statistic, as              
              
X = sample mean =    3255.43          
uo = hypothesized mean =    3500          
n = sample size =    12          
s = standard deviation =    31.6227766          
              
Thus, t = (X - uo) * sqrt(n) / s =    -26.79130118          
              
Also, the p value is              
              
p =    2.27813E-11          
              
Comparing t and tcrit (or, p and significance level), we   REJECT THE NULL HYPOTHESIS.          

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b)

The P value here is too small that for almost any practical level of significance (even at 0.001), we reject the null hypothesis.
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c)

Note that              
Margin of Error E = t(alpha/2) * s / sqrt(n)              
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 =    3255          
t(alpha/2) = critical t for the confidence interval =    2.20098516          
s = sample standard deviation =    31.6227766          
n = sample size =    12          
df = n - 1 =    11          
Thus,              
Margin of Error E =    20.09215368          
Lower bound =    3234.907846          
Upper bound =    3275.092154          
              
Thus, the confidence interval is              
              
(   3234.907846   ,   3275.092154   ) [ANSWER]

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d)

Note that              
Margin of Error E = t(alpha/2) * s / sqrt(n)              
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 =    0.005          
X = sample mean =    3255          
t(alpha/2) = critical t for the confidence interval =    3.105806516          
s = sample standard deviation =    31.6227766          
n = sample size =    12          
df = n - 1 =    11          
Thus,              
Margin of Error E =    28.3520048          
Lower bound =    3226.647995          
Upper bound =    3283.352005          
              
Thus, the confidence interval is              
              
(   3226.647995   ,   3283.352005   ) [ANSWER]

This is wider as the confidence level is larger.

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