Question

A cheese processing company wants to estimate the mean cholesterol content of all one-ounce servings of cheese. The estimate must be within 0.3 milligram of the population mean. Assume the population standard deviation is 2.9 milligrams. Determine the minimum sample size required to construct a 90% confidence interval for the population mean. Determine the minimum sample size required to construct a 98% confidence interval for the population mean. Which level of confidence requires a larger sample size? 90% 98%

Answer 1

Solution :

Given that,

standard deviation =s =   =2.9

Margin of error = E = 0.3

At 90% confidence level

= 1 - 90%  

= 1 - 0.90 =0.10

/2 = 0.05

Z/2 = Z0.05 = 1.645

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

n = ( 1.645 *2.9 /0.3 )2

n =252.86

Sample size = n =253

(B)At 98% confidence level the z is ,

= 1 - 98% = 1 - 0.98 = 0.02

/2 = 0.02/ 2 = 0.01

Z/2 = Z0.01 = 2.326 ( Using z table    )

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

n = ( 2.326*2.9 /0.3 )2

n =505.56

Sample size = n =506