A sample of n10 observations from Normal distribution yields

A sample of n=10 observations from Normal distribution yields a sample mean =2.4.

Assuming a population standard deviation =0.8,

(a) Construct a 95% confidence interval for the population mean.

(b) What is the minimum sample size required for estimating the population mean with a margin of error not exceeding 0.4?

(a) Enter the confidence interval in the form [...,...]: [ , ].

(b) The minimum required sample size is ... observations.

Solution

a)

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 =    2.4          
t(alpha/2) = critical t for the confidence interval =    2.262157163          
s = sample standard deviation =    0.8          
n = sample size =    10          
df = n - 1 =    9          
Thus,              
Margin of Error E =    0.572285525          
Lower bound =    1.827714475          
Upper bound =    2.972285525          
              
Thus, the confidence interval is              
              
(   1.827714475   ,   2.972285525   ) [ANSWER]

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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 =    0.8  
E = margin of error =    0.4  
      
Thus,      
      
n =    15.36583528  
      
Rounding up,      
      
n =    16   [ANSWER]

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