It is estimated that the proportion of Californians in favor of a Proposition 5 is between 0.4 and 0.6. The sponsors of the proposition would like to survey Californians and get a 95 percent confidence interval for the true proportion of people in favor of Proposition 5. They want to restrict the error level to 4 percentage points (4%). Given the likelihood of the survey being delivered to a qualified Californian is 0.8 (incidence rate), and the response rate is 0.6, how many surveys should they send out?

Answer :

Answer:

The number of surveys must be 600

Explanation:

Proposition a is between 0.4 and 0.6

Confidence 95%

Error level 4%

Incident rate 0.8

Response rate 0.6

Surveys should they send out are:

E = 0.04 (this is the margin error of 4%)

P = (0.4+0.6)/2=0.5

Alpha = 1 - 0.95 = 0.05

So, the z value that we need to search in the table of z distribution is alpha/2 = 0.025

Z-value = 1.96

P (z>1.96) = 0.025 nearly

The number of surveys is

n = p*q (2 * half-alpha) ^2

=0.5 * 0.5 * (1.96/0.04)^2

=(0.5^2) * (49^2)

=0.25 * 2.401

= 600.25

The number of surveys must be 600

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