Question

The lowest and highest observations in a population are 28 and 62, respectively. What is the minimum sample size n required to estimate μ with 90% confidence if the desired margin of error is E = 1.6? What happens to n if you decide to estimate μ with 99% confidence? Use Table 1. (Round intermediate calculations to 4 decimal places and "z-value" to 3 decimal places. Round up your answers to the nearest whole number.)

  

Confidence Level	n
90%
99%

Answer 1

Solution :

Given that,

Maximum = 62

Minimum = 28

Range = maximum - minimum = 62 - 28 = 34

Using Range rule of thumb

Standard deviation = Range / 4 = 34 / 4 = 8.5

Population standard deviation = = 8.5

Margin of error = E = 1.6

At 90% confidence level the z 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 * 8.5/ 1.6)²

n = 76.37

n = 76

Sample size = 76

At 99% confidence level the z is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z_/2 = Z_0.005 = 2.576

sample size = n = (Z_/2* / E) ²

n = (2.576 * 8.5/ 1.6)²

n = 187.27

n = 187