Question

A survey of 750 likely voters in Ohio was conducted by the Rasmussen Poll just prior to the general election, The state of the economy was thought to be an important determinant of how people would vote. Among other things, the survey found that 165 of the respondents rated the economy as good or excellent, 315 rated the economy as poor (the rest of the respondents were in between these two categories).

a. (10p) Develop a point estimate of the proportion of likely voters in Ohio who rated the economy as good or excellent.

b. (20p) Construct a 90% confidence interval for the proportion of likely voters in Ohio who rated the economy as good or excellent.

c. (20p) Construct a 90% confidence interval for the proportion of likely voters in Ohio who rated the economy as poor.

Answer 1

Solution :

Given that,

n = 750

x = 165

a) Point estimate = sample proportion = = x / n = 165 / 750 = 0.22

1 - = 1 - 0.22 = 0.78

b) At 90% confidence level

= 1 - 90%

=1 - 0.90 =0.10

/2 = 0.05

Z/2 = Z0.05  = 1.645

Margin of error = E = Z / 2 * (( * (1 - )) / n)

= 1.645 (((0.22 * 0.78) / 750)

= 0.025

A 90% confidence interval for population proportion p is ,

± E

= 0.22 ± 0.025

= ( 0.195, 0.245 )

c) x = 315

Point estimate = sample proportion = = x / n = 315 / 750 = 0.42

1 - = 1 - 0.42 = 0.58

At 90% confidence level

= 1 - 90%

=1 - 0.90 =0.10

/2 = 0.05

Z/2 = Z0.05  = 1.645

Margin of error = E = Z / 2 * (( * (1 - )) / n)

= 1.645 (((0.42 * 0.58) / 750)

= 0.030

A 90% confidence interval for population proportion p is ,

± E

= 0.42 ± 0.030

= ( 0.390, 0.450 )