Question

Banking fees have received much attention during the recent economic recession as banks look for ways to recover from the crisis. A sample of 32 customers paid an average fee of ​$12.11 per month on their​ interest-bearing checking accounts. Assume the population standard deviation is ​$1.65

Complete parts a and b below.

a. Construct a 90​% confidence interval to estimate the average fee for the population.The 90​% confidence interval has a lower limit of blank and an upper limit of.

​(Round to the nearest cent as​ needed.)

Answer 1

Solution :

Given that,

Point estimate = sample mean = $\bar x$ = $12.11

Population standard deviation = $\sigma$ = $1.65

Sample size = n = 32

At 90% confidence level the z is ,

  = 1 - 90% = 1 - 0.90$\alpha$ = 0.10

$\alpha$ / 2 = 0.10 / 2 = 0.05

Z_$\alpha$/2 = Z_0.05 = 1.645

Margin of error = E = Z_$\alpha$/2* ($\sigma$ /$\sqrt$n)

= 1.645 * (1.65 / $\sqrt$ 32)

= 0.48

At 99% confidence interval estimate of the population mean is,

$\bar x$ - E < $\mu$ < $\bar x$ + E

12.11 - 0.48 < $\mu$ < 12.11 + 0.48

11.63 < $\mu$ < 12.59

lower limit = 11.63

upper limi = 12.59

Answer 2

Banking fees have received much attention during the recent economic recession as banks look for ways to recover from the crisis. A sample of

45

customers paid an average fee of

$12.21

per month on their interest-bearing checking accounts. Assume the population standard deviation is

$1.55.

Complete parts a and b below