5.28 Assuming the compressive strength of lunar concrete is normally distributed

Question

5.28

Assuming the compressive strength of lunar concrete is normally distributed, Find a 95% confidence interval for the true mean compressive strength of this lunar concrete. Find a 99% lower confidence interval for the true mean compressive strength of this lunar concrete. The strength of a steel beam is measured in the size of deflection (in micrometers, nm) that results when subjecting the beam to a force of 10,000 pounds. The strength of steel beams is believed to be normally distributed. Suppose a materials engineer selects a random sample of 5 beams, tests them, and finds their deflections as follows (in /xm): Find a 95% confidence interval for the true mean strength of beams of this kind (as measured in nm of deflection). How many more beams would need to be tested in order to construct a 95% confidence interval which would estimate the true mean strength with a maximum error of plusminus 2 in deflection?

Explanation / Answer

a)

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 =    0.01          
X = sample mean =    73.2          
t(alpha/2) = critical t for the confidence interval =    3.746947388          
s = sample standard deviation =    3.701351105          
n = sample size =    5          
df = n - 1 =    4          
Thus,              
              
Lower bound =    66.99769846          
Upper bound =    79.40230154          
              
Thus, the confidence interval is              
              
(   66.99769846   ,   79.40230154   ) [ANSWER]

***********************

b)

Note that      
      
n = z(alpha/2)^2 s^2 / E^2      
      
where      
      
alpha/2 = (1 - confidence level)/2 =    0.025  
      
Using a table/technology,      
      
z(alpha/2) =    1.959963985  
      
Also,      
      
s = sample standard deviation =    3.701351105  
E = margin of error =    2  
      
Thus,      
      
n =    13.15699646  
      
Rounding up,      
      
n =    14  

Hence, as we have 5 already, we need 14 - 5 = 9 more. [ANSWER: 9]

Related Questions

Navigate

• Browse (All)
• Browse 5
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.