What is the value of the sample mean x ? One CI has a confidence level of 95%, a

Question

What is the value of the sample mean x ? One CI has a confidence level of 95%, and the other has a confidence level of 99.9%. Match the confidence level with the confidence interval, and justify your answer. In each of the following problems, the population standard deviation, the bound on the error of estimation, and the confidence level are given. Find a value for the sample size n necessary to satisfy these requirements. sigma = 7.9, B = 2.5, 95% sigma = 10.77, B = 5, 99% sigma = 0.55, B = 0.001, 98% sigma = 35.97, B = 3.5, 95% sigma = 55, B = 2, 99.9% Applications Physical Science The number of active coal mines in Utah has steadily decreased since 1960. However, peak

Explanation / Answer

a)

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 =    7.9  
E = margin of error =    2.5  
      
Thus,      
      
n =    38.3592712  
      
Rounding up,      
      
n =    39   [ANSWER]

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

b)

Note that      
      
n = z(alpha/2)^2 s^2 / E^2      
      
where      
      
alpha/2 = (1 - confidence level)/2 =    0.005  
      
Using a table/technology,      
      
z(alpha/2) =    2.575829304  
      
Also,      
      
s = sample standard deviation =    10.77  
E = margin of error =    5  
      
Thus,      
      
n =    30.78403592  
      
Rounding up,      
      
n =    31   [ANSWER]

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

c)

Note that      
      
n = z(alpha/2)^2 s^2 / E^2      
      
where      
      
alpha/2 = (1 - confidence level)/2 =    0.01  
      
Using a table/technology,      
      
z(alpha/2) =    2.326347874  
      
Also,      
      
s = sample standard deviation =    0.55  
E = margin of error =    0.001  
      
Thus,      
      
n =    1637098.065  
      
Rounding up,      
      
n =    1637099   [ANSWER]

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

d)

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 =    35.97  
E = margin of error =    3.5  
      
Thus,      
      
n =    405.7335949  
      
Rounding up,      
      
n =    406   [ANSWER]

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

e)

Note that      
      
n = z(alpha/2)^2 s^2 / E^2      
      
where      
      
alpha/2 = (1 - confidence level)/2 =    0.0005  
      
Using a table/technology,      
      
z(alpha/2) =    3.290526731  
      
Also,      
      
s = sample standard deviation =    55  
E = margin of error =    2  
      
Thus,      
      
n =    8188.346917  
      
Rounding up,      
      
n =    8189   [ANSWER]

Related Questions

Navigate

• Browse (All)
• Browse W
• Subjects

---

---

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