Question

An admissions director wants to estimate the mean age of all students enrolled at a college. The estimate must be within 1 years of the population mean. Assume the population of ages is normally distributed and the population standard deviation is 9.5 years. Determine the minimum sample size required to construct a 80% confidence interval for the population mean age. Determine the minimum sample size required to construct a 95% confidence interval for the population mean age. Which level of confidence requires a larger sample size? 80% 95%

Answer 1

Solution :

Given that,

standard deviation = = 9.5

margin of error = E = 1

a ) At 80% confidence level the z is ,

= 1 - 80% = 1 - 0.80 = 0.20

/ 2 = 0.20 / 2 = 0.10

Z/2 = Z 0.10 = 1.280

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

= ((1.280 * 9.5 ) /1 )2

= 148

Sample size = 148

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 * 9.5 ) /1 )2

= 347

Sample size = 347

The Increasing confidence level is increasing sample size