Professor Hulbert picks 10 students at random and records their exam scores belo

Question

Professor Hulbert picks 10 students at random and records their exam scores below.

80, 80, 80, 85, 85, 85, 85, 90, 90, 90

A. Create a 90% confidence interval for the mean exam score of all his students. Assume the standard deviation of all the exam scores is 3.

B. what is the margin of error in using a sample of 10 students to estimate the population mean?

C. What size sample is needed so that the margin of error is less than .5?

D. Create another 90% confidence interval without using the assumption about the population standard deviation.

Explanation / Answer

a)

Note that              
              
Lower Bound = X - z(alpha/2) * s / sqrt(n)              
Upper Bound = X + z(alpha/2) * s / sqrt(n)              
              
where              
alpha/2 = (1 - confidence level)/2 =    0.05          
X = sample mean =    85          
z(alpha/2) = critical z for the confidence interval =    1.644853627          
s = standard deviation =    3          
n = sample size =    10          
              
Thus,              
              
Lower bound =    83.43955484          
Upper bound =    86.56044516          
              
Thus, the confidence interval is              
              
(   83.43955484   ,   86.56044516   ) [ANSWER]

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b)

Note that

Margin of error = z(alpha/2) * s / sqrt(n) = 1.560445164 [ANSWER]

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c)

Note that      
      
n = z(alpha/2)^2 s^2 / E^2      
      
where      
      
alpha/2 = (1 - confidence level)/2 =    0.05  
      
Using a table/technology,      
      
z(alpha/2) =    1.644853627  
      
Also,      
      
s = sample standard deviation =    3  
E = margin of error =    0.5  
      
Thus,      
      
n =    97.39956435  
      
Rounding up,      
      
n =    98   [ANSWER]

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d)

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 =    0.05          
X = sample mean =    85          
t(alpha/2) = critical t for the confidence interval =    1.833112933          
s = sample standard deviation =    4.082482905          
n = sample size =    10          
df = n - 1 =    9          
Thus,              
              
Lower bound =    82.63346138          
Upper bound =    87.36653862          
              
Thus, the confidence interval is              
              
(   82.63346138   ,   87.36653862   ) [ANSWER]

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