Question

In 2010, an online security firm estimated that 65% of computer users don't change their passwords very often. Because this estimate may be outdated, suppose that you want to carry out a new survey to estimate the proportion of students at your school who do not change their password. You would like to determine the sample size required to estimate this proportion to within a margin of error of 0.05 with 95% confidence.

(a)

Using 0.65 as a preliminary estimate for the proportion, what is the required sample size if you want to estimate this proportion with a margin of error of 0.05? (Give your answer as a whole number.)

n ≥

(b)

Calculate the required sample size if no preliminary estimate for the proportion is used. (Give your answer as a whole number.)

n ≥

  

Answer 1

Solution :

a)

Given that,

$\hat p$ = 0.65

1 - $\hat p$ = 0.65

margin of error = E = 0.05

At 95% confidence level the z is ,

$\alpha$ = 1 - 95% = 1 - 0.95 = 0.05

$\alpha$ / 2 = 0.05 / 2 = 0.025

Z$\alpha$/2 = Z0.025 = 1.96

sample size = n = (Z$\alpha$ / 2 / E )2 * $\hat p$ * (1 - $\hat p$ )

= (1.96 / 0.05)2 * 0.65 * 0.35

= 349.59

sample size = 350

b)

Given that,

$\hat p$ = 0.5

1 - $\hat p$ = 0.5

margin of error = E = 0.05

At 95% confidence level the z is ,

$\alpha$ = 1 - 95% = 1 - 0.95 = 0.05

$\alpha$ / 2 = 0.05 / 2 = 0.025

Z$\alpha$/2 = Z0.025 = 1.96

sample size = n = (Z$\alpha$ / 2 / E )2 * $\hat p$ * (1 - $\hat p$ )

= (1.96 / 0.05)2 * 0.5 * 0.5

= 384.16

sample size = 385