Question

Almost all employees working for financial companies in New York City receive large bonuses at the end of the year. A sample of 63 employees selected from financial companies in New York City showed that they received an average bonus of $56,000 last year with a standard deviation of $16,000. Construct a 98% confidence interval for the average bonus that all employees working for financial companies in New York City received last year.

Round your answers to cents.

$_____ to $_____

Answer 1

Solution :

Given that,

Point estimate = sample mean = = $56000

Population standard deviation =    = $16000

Sample size = n =63

At 98% confidence level the z is ,

= 1 - 98% = 1 - 0.98 = 0.02

/ 2 = 0.02/ 2 = 0.01

Z/2 = Z0.01 = 2.326 ( Using z table    )

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

= 2.326 * (16000 /  63 )

E= 4688.8
At 98% confidence interval estimate of the population mean
is,

- E < < + E

56000 - 4688.8 <   < 56000 + 4688.8

51311.2 <   < 60688.8

( $ 51311.2 ,$ 60688.8)