1A bank manager wants to know the mean amount of mortgage pa

1.A bank manager wants to know the mean amount of mortgage paid per month by homeowners in an area. A random sample of 117 homeowners selected from this area showed that they pay an average of $1578 per month for their mortgages. The population standard deviation of such mortgages is $219.

_____ to ______ dollars

b. Suppose the confidence interval obtained in part a is too wide. Select all of the ways the width of this interval can be reduced.

which one?

Solution

a)

Note that              
Margin of Error E = z(alpha/2) * s / sqrt(n)              
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 =    1578          
z(alpha/2) = critical z for the confidence interval =    1.959963985          
s = sample standard deviation =    219          
n = sample size =    117          
              
Thus,              
Margin of Error E =    39.68252285          
Lower bound =    1538.317477          
Upper bound =    1617.682523          
              
Thus, the confidence interval is              
              
(   1538.317477   ,   1617.682523   ) [ANSWER]

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b)

Higher sample sizes have narrower confidence intervals.

Lower confidence levels have narrower confidence intervals.

Thus,

OPTION B: Increasing the sample size AND
OPTION D: Lowering the confidence level [ANSWER]