The average selling price of a smartphone purchased by a ran

The average selling price of a smartphone purchased by a random sample of 34 customers was $324 Assume the population standard deviation was $31.

a. Construct a 95% confidence interval to estimate the average selling price in the population with this sample.

b. What is the margin of error for this interval?

a. The 95% confidence interval has a lower limit of $? and an upper limit of $?

(Round to the nearest cent as needed.)

b. The margin of error is $?

(round to nearest cent as needed)

Solution

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 =    0.025          
X = sample mean =    324          
z(alpha/2) = critical z for the confidence interval =    1.959963985          
s = sample standard deviation =    31          
n = sample size =    34          
              
Thus,              
              
Lower bound =    313.5799374          
Upper bound =    334.4200626          
              
Thus, the confidence interval is              
              
(   313.5799374   ,   334.4200626   ) [ANSWER]

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B)

Margin of error = (upper bound - lower bound)/2 = 10.42006256 [ANSWER]