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A) Banking fees have received much attention during the recent economic recessio

ID: 3183381 • Letter: A

Question

A) Banking fees have received much attention during the recent economic recession as banks look for ways to recover from the crisis. A sample of 39 customers paid an average fee of $12.16 per month on their interest-bearing checking accounts. Assume the population standard deviation is $1.86. Complete parts a and b below.

a.1 Construct a 90% confidence interval to estimate the average fee for the population.

The 90% confidence interval has a lower limit of ___ and an upper limit of ___.

b.1 What is the margin of error for this interval? (Round to the nearest cent as needed.)

B) Construct an 80% confidence interval to estimate the population mean using the data below. What assumptions need to be made about this population?

The 80% confidence interval for the population mean is from a lower limit of ___ to an upper limit of ___.

What assumptions need to be made about this population?

A. The population follows the Student's t-distribution.

B.The population distribution is skewed to one side.

C.The population follows the normal probability distribution.

D.The population size is larger than 30.

C) Construct a 95% confidence interval to estimate the population mean when x=128 and s =28 for the sample sizes below.

The 95% confidence interval for the population mean when n=20 is from a lower limit of ___ to to an upper limit of ___.

The 95% confidence interval for the population mean when n=50 is from a lower limit of ___ to to an upper limit of ___.

The 95% confidence interval for the population mean when n=80 is from a lower limit of ___ to to an upper limit of ___.

Explanation / Answer

A)as std error =std deviation/(n)1/2 =0.2978

for 90% CI, z=1.6449

hence confidence interval =sample mean -/+ z*Std error =11.6701 ; 12.6499

b1) margin of error = z*Std error =0.4899

B) for 80% CI, z=1.2815

confidence interval =sample mean -/+ z*Std error =11.7783 ; 12.5417

C.The population follows the normal probability distribution.

C) 1) confidence interval =sample mean -/+ z*Std error =115.7287 ; 140.2713

2) when n=50 ;  confidence interval =sample mean -/+ z*Std error = 120.2389 ; 135.7611

3)  when n=80 ;  confidence interval =sample mean -/+ z*Std error = 121.8643 ; 134.1357

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