To estimate the average monthly rent for 1 bedroom apartment

To estimate the average monthly rent for 1 bedroom apartments, 14 complexes were randomly selected in Orlando. The mean cost of the selected complexes is $995 with a standard devaition of $97.

1. If the confidence level was decreased to 90% of 1 bedroom apartments, what would happen to the width of the interval? (What is the relationship between the confidence level and interval width)

Solution

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 =    995          
t(alpha/2) = critical t for the confidence interval =    1.770933396          
s = sample standard deviation =    97          
n = sample size =    14          
df = n - 1 =    13          
Thus,              
              
Lower bound =    949.0897197          
Upper bound =    1040.91028          
              
Thus, the confidence interval is              
              
(   949.0897197   ,   1040.91028   ) [ANSWER]

which is narrower than the 98% confidence interval.

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As confidence level increases, the width increases. [ANSWER]