A random sample of 23 people at Burger King revealed a mean

A random sample of 23 people at Burger King revealed a mean amount spent of $5.60 and a standard deviation of amount spent $1.10. Assume the amount spent per person at McDonald\'s is approximately a normally distributed random variable. Find the 95% confidence interval for p. How large a sample size is needed if you wanted to reduce the above bound by 15%?

Solution

2.

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 =    5.6          
t(alpha/2) = critical t for the confidence interval =    2.073873068          
s = sample standard deviation =    1.1          
n = sample size =    23          
df = n - 1 =    22          
Thus,              
Margin of Error E =    0.47567567          
Lower bound =    5.12432433          
Upper bound =    6.07567567          
              
Thus, the confidence interval is              
              
(   5.12432433   ,   6.07567567   ) [ANSWER]

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For the next part, the problem is unclear. Does he mean to reduce the margin of error by 15%?

If so, then the new margin of error is

E = 0.47567567*0.85 = 0.40432432

Hence,

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 =    1.1  
E = margin of error =    0.40432432  
      
Thus,      
      
n =    28.43294354  
      
Rounding up,      
      
n =    29   [ANSWER]

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