Question

In: Math

An advertising executive wants to estimate the mean weekly amount of time consumers spend watching traditional...

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 21 minutes. The executive wants to​ estimate, with​ 99% confidence, the mean weekly amount of time to within ± 6 minutes.

a. What sample size is​ needed?

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

Solutions

Expert Solution

Solution :

Given that,

standard deviation = = 21

margin of error = E = 6

a ) At 99% confidence level the z is ,

  = 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z/2 = Z0.005 = 2.576

Sample size = n = ((Z/2 * ) / E)2

= ((2.576 * 21) / 6)2

= 81.288

= 81

Sample size = 81

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 = 1.960

Sample size = n = ((Z/2 * ) / E)2

= ((1.960 * 21) / 6)2

= 47.0596

= 47

Sample size = 47


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