Question

For budgetary purposes, a manufacturing company needs to estimate the proportion of defective parts that a production process will create. What (minimum) sample size is needed to form a 98% confidence interval to estimate the true proportion of defective parts to within ± 0.04?

**Do not round intermediate calculations. Round up your
final answer to the next whole number.**

Sample size = parts

Answer #1

Solution :

Given that,

= 1 - = 0.5

margin of error = E = 0.04

At 98% confidence level the z is ,

= 1 - 98% = 1 - 0.98 = 0.02

/ 2 = 0.02 / 2 = 0.01

Z_{/2}
= Z _{0.01} = 2.326

sample size = n = (Z_{
/ 2} / E )^{2} *
* (1 -
)

= (2.326 / 0.04)^{2} * 0.5 * 0.5

= 845

**sample size** = **845**

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