Question

Studies show that massage therapy has a variety of health benefits and it is not too expensive. A sample of 11 typical one-hour massage therapy sessions showed an average charge of 59. The population standard deviation for a one-hour session is 5. What assumptions about the population should we be willing to make if a margin of error is desired? Using 95% confidence, what is the margin of error ( 2 decimals)? Using 99% confidence, what is the margin of error (2 decimals)?

Answer 1

Solution :

Given that,

= 59

= 8.8

n = 11

A ) At 95% confidence level the z is ,

  = 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z_/2 = Z_0.025 = 1.960

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

= 1.960 * ( 5 / 11 )

= 2.95

Margin of error = E = 2.95

B ) At 99% confidence level the z is ,

  = 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z_/2 = Z_0.005 = 2.576

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

= 2.576 * ( 5 / 11 )

= 3.88

Margin of error = E = 3.88