Question

An advertising executive wants to estimate the mean weekly amount of time consumers spend watching traditional television daily. Based on previous​ studies, the standard deviation is assumed to be 20 minutes. The executive wants to​ estimate, with ​99% confidence, the mean weekly amount of time to within ±5 minutes.

a. What sample size is​ needed?

b. If​ 95% confidence is​ desired, how many consumers need to be​ selected?

Answer 1

Solution :

Given that,

standard deviation =s =   =20

Margin of error = E = 5

At 99% confidence level the z is,

= 1 - 99%

= 1 - 0.99 = 0.01

/2 = 0.005

Z/2 = 2.576

sample size = n = [Z/2* / E] 2

n = ( 2.576* 20/ 5)2

n =106.17

Sample size = n =106

b.

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z/2 = Z0.025 = 1.96

sample size = n = [Z/2* / E] 2

n = ( 1.96* 20/ 5 )2

n =61.4656

Sample size = n =61

some time sample size 62 this is also right