Question

A polling agency is investigating the voter support for a ballot measure in an upcoming city election. The agency will select a random sample of 400 voters from one region, Region A, of the city. Assume that the population proportion of voters who would support the ballot measure in Region A is 0.46.

Required:
What is the probability that the proportion of voters in the sample of Region A who support the ballot measure is greater than 0.50?

Answer 1

0.0544

Explanation:

From the given information;

The population proportion of the voters that would support the ballot measures in region A = 0.46

The random sample n = 400 voters

The required probability can therefore be computed as follows:

`P[p_1>0.5] = P \bigg [ \frac{p_1-P_1}{\sqrt{(P_1(1-P_1))/(n_1)}}>\frac{0.5-0.46}{\sqrt{(0.46(1-0.46))/(400)}} \bigg]`

`P[p_1>0.5] = P \bigg [ Z>\frac{0.04}{\sqrt{(0.46(0.54))/(400)}} \bigg]`

`P[p_1>0.5] = P \bigg [ Z>\frac{0.04}{\sqrt{6.21* 10^(-4)}} \bigg]`

`P[p_1>0.5] = P \bigg [ Z>1.605 \bigg]`

`P[p_1>0.5] = 1- P [ Z<1.605 ]`

Using the Excel Function =NORMDIST(1.605)

`P[p_1>0.5] = 1- 0.9456`

`P[p_1>0.5] =0.0544`

Therefore, the required probability = 0.0544