Question

1. Crest toothpaste is reviewing plans for its annual survey of toothpaste purchasers. Assume the market share of Crest toothpaste last year was 25%, please calculate the sample size needed for this year’s survey if acceptable error is 4% and desired confidence level is 99%.

2. A recent survey conducted by Wall Street Journal indicates that the average reading time of 100 its readers/participants was 45 minutes and the standard deviation was 20 minutes. The journal is now planning another survey study. If the acceptable error is 3 minutes and a confidence level of 99% is desired, what would be the minimum sample size needed for the upcoming survey?

Answer 1

Solution :

Given that,

= 0.25

1 - = 1 - 0.25 = 0.75

margin of error = E = 4% = 0.04

At 99% confidence level the z is,

= 1 - 99%

= 1 - 0.99 = 0.01

/2 = 0.005

Z/2 = 2.58    ( Using z table )

Sample size = n = (Z/2 / E)2 * * (1 - )

= (2.58 / 0.04)2 * 0.25 * 0.75

=780.046

Sample size = 780

(B)

Solution :

Given that,

standard deviation =   =20

Margin of error = E = 3

At 99% confidence level the z is,

= 1 - 99%

= 1 - 0.99 = 0.01

/2 = 0.005

Z/2 = 2.58 ( Using z table ( see the 0.005 value in standard normal (z) table corresponding z value is 2.58 )  

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

n = ( 2.58* 20 /3 )2

n =295.84

Sample size = n =296