Question

2. A random sample of 144 checking accounts at a bank showed an average daily balance of $290. The standard deviation of the population is known to be $66. 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 90% confidence interval estimates for the mean.

Answer 1

2)

Solution :

Given that,

(b)

Point estimate = sample mean = = 290

Population standard deviation = = 66

Sample size = n = 144

At 90% confidence level the z is ,

= 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 0.10 / 2 = 0.05

Z_/2 = Z _0.05 = 1.645

(a)

( /n) = (66 / 144) = 66 / 12 = 5.5

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

= 1.645 * (66 / 144)

= 9

(c)

At 90% confidence interval estimate of the population mean is,

- E < < + E

290 - 9 < < 290 + 9

281 < < 299

(281 , 299 )