Question

- Find the minimum sample size required to estimate a population proportion with a margin of error = 0.05 a confidence level of 90%, and from a prior study, p is estimated to be .25

(a) 203 (b) 329 (c) 247 (d) 396 (e) 289

Answer #1

Solution :

Given that,

= 0.25

1 - = 1 - 0.25 = 0.75

margin of error = E = 0.05

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} *
* (1 -
)

= (1.645 * 0.05)^{2} * 0.25 * 0.75

= 202.95

Sample size = 203 (rounded)

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