Question

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 :

degrees of freedom = n - 1 = 1800 - 1 = 1799

t/2,df =  t_{0.025 , 1799 = 1.961}

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

,= 1.961* ( 600 / 1800)

Margin of error = E =14.14

The 99% confidence interval estimate of the population mean is,

  ± E  

4600 ± 14.14

(4585.86 , 4614.14)