Question

3) What sample size would be required to estimate the true proportion of American female businessexecutives who prefer the title “Ms.,” with an error of ± 0.03 and 95 percent confidence? (4 points)

Answer #1

3)

Solution :

Given that,

= 0.5

1 - = 1 - 0.5 = 0.5

margin of error = E = 0.03

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z_{/2}
= Z_{0.025} = 1.96

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

= (1.96 / 0.03)^{2} * 0.5 * 0.5

= 1067.11

sample size = 1068

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