Question

A random sample of 144 checking accounts at a bank showed an average daily balance of $395. The standard deviation of the population is known to be $72. Please answer the following questions:

(a) Find the standard error of the mean.

(b) Give a point estimate of the population mean.

(c) Construct a 95% confidence interval estimates for the mean.

Answer 1

Solution :

Given that,

Point estimate = sample mean =     =$395

Population standard deviation =    =$72

Sample size n =144

(A)Point estimate =$395

(B)standard error=( /n) =(72 / 144 ) =6

(C)At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z/2 = Z0.025 = 1.96   ( Using z table )

Margin of error = E = Z/2 * ( /n)

= 1.96 * (72 / 144 )

= 11.76
At 95% confidence interval estimate of the population mean
is,

- E < < + E

395 -11.76 <   < 395+ 11.76

383.24 <   < 406.76

( 383.24 , 406.76 )