Question

A population has standard deviation 17.7.

Part 1 of 2
(a) How large a sample must be drawn so that a 90% confidence interval for u will have a margin of error equal to 3.2?
Round the critical value to no less than three decimal places. Round the sample size up to the nearest integer.

A sample size of ___ is needed to be drawn in order to obtain a 90% confidence interval with a margin of error equal to 3.2.

Part 2 of 2
(b) If the required confidence level were 99.5, would the necessary sample size be larger or smaller

(larger/smaller), because the confidence level is (higher/lower).

Answer 1

Solution :

Given that,

standard deviation = = 17.7

margin of error = E = 3.2

At 90% confidence level the z is ,

= 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 0.10 / 2 = 0.05

Z/2 = Z0.05 = 1.645

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

= ((1.645 * 17.7) / 3.2)2

= 837

Sample size = 83

b )At 99.5% confidence level the z is ,

  = 1 - 99.5% = 1 - 0.995 = 0.005

/ 2 = 0.005 / 2 = 0.0025

Z/2 = Z0.0025 = 2.807

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

= ((2.807 * 17.7) / 3.2)2

= 241

Sample size = 241

Sample size is larger  the confidence level is higher