Question

Question 4

A statistician employed by a consumer testing organization reports that at 90% confidence he has determined that the true average content of the Uncola soft drinks is between 11.8 to 12.2 ounces. He further reports that his sample revealed an average content of 12 ounces, but he forgot to report the size of the sample he had selected. Assuming the standard deviation of the population is 0.72, determine the size of the sample.

Answer 1

Solution:

Given that,

Confidence interval = ( 11.8 , 12.2 )

The sample mean is ,

= ( Lower confidence interval + Upper confidence interval ) / 2

= (11.8 + 12.2) / 2

=24 / 2

= 12

The margin of error is,

E = Upper confidence interval -

= 12.2 - 12

= 0.2

Population standard deviation = = 0.72

Margin of error = E = 0.2

At 90% confidence level the t is ,

= 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 0.10 / 2 = 0.05

Z_/2 = Z_0.05 = 1.645

sample size = n = (Z_/2* / E) ²

n = (1.645 * 0.72/ 0.2)²

n = 35.07

n = 36

Sample size = 36