Question

Question 1:

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 $ 61 , 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.

Question 2:

The high price of medicines is a source of major expense for those seniors in the United States who have to pay for these medicines themselves. A random sample of 1800 seniors who pay for their medicines showed that they spent an average of $ 4600 last year on medicines with a standard deviation of $ 600 . Make a 95 % confidence interval for the corresponding population mean.

Round your answers to cents.

Answer 1

Solution :

1) degrees of freedom = n - 1 = 63 - 1 = 62

t/2,df = t_0.01,62 = 2.388

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

= 2.388 * (16000 / 63)

Margin of error = E = 4813.76

The 98% confidence interval estimate of the population mean is,

  ± E  

= 61000  ± 4813.76

= ( $ 56186.24, $ 65813.76 )

2) Z/2 = Z_0.025 = 1.96

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

= 1.96 * ( 600 /  1800 )

= 27.72

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

  ± E   

= 4600  ± 27.72

= ($ 4572.28, $ 4627.72 )