1 7pts The average loans they owed in a big city were 110000

1. [7pts The average loans they owed in a big city were $110,000 of a random sample of 1600 households. Given the standard deviation of household loan amount in this city sigma = $40,000. An approximate 95% confidence interval for the average amount of loans that all households owned is a)$110,000-$120,000 b)$30,000-$190,000 c)$108,000-$112,000 d)$70,000-$150,000 2. (7pts)Suppose that scores on the mathematics part of the National Assessment of Educational progress (NAEP) test for high school seniors follow a Normal distribution with standard deviation sigma = 30. You want to estimate the mean score within + 10 with 90% confidence. How large an SRS of scores must you choose? a)10 b)30 c)25 d) 24.4 3.[5pts] Determine whether a 95% confidence Internal involves a t-procedure, z-procedure, or neither. Sample data: n = 50, x = 984, s = 25. a) T-procedure B) Neither b) Z-procedure

Solution

1.

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 =    110000          
z(alpha/2) = critical z for the confidence interval =    1.959963985          
s = sample standard deviation =    40000          
n = sample size =    1600          
              
Thus,              
Margin of Error E =    1959.963985          
Lower bound =    108040.036          
Upper bound =    111959.964          
              
Thus, the confidence interval is              
              
(   108040.036   ,   111959.964   )

or

OPTION C: 108000-112000 [ANSWER]

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