Question

Almost all employees working for financial companies in New York City receive large bonuses at the end of the year. A sample of 65 employees selected from financial companies in New York City showed that they received an average bonus of $51,000 last year with a standard deviation of $17,000. Construct a 99% confidence intervale 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 = = $51000

sample standard deviation = s = $17000

sample size = n = 65

Degrees of freedom = df = n - 1 = 65 - 1 = 64

At 99% confidence level the t is ,

= 1 - 99% = 1 - 0.90 = 0.01

/ 2 = 0.01 / 2 = 0.005

t _/2,df = t_0.005,64 = 2.655

Margin of error = E = t_/2,df * (s /n)

= 2.655 * ( 17000/ 65)

= 5598.31

The 95% confidence interval estimate of the population mean is,

- E < < + E

51000 - 5598.31 < < 51000 + 5598.31

45401.69 < < 56598.31

($45401.69 to  $56598.31)