Question

A research center conducted a telephone survey of 2,000 adults to learn about the major economic concerns for the future. The survey results showed that 1,500 of the respondents think the future health of Social Security is a major economic concern.

(a)

What is the point estimate of the population proportion of adults who think the future health of Social Security is a major economic concern?

(b)

At 90% confidence, what is the margin of error?

(c)

Develop a 90% confidence interval for the population proportion of adults who think the future health of Social Security is a major economic concern.

(d)

Develop a 95% confidence interval for this population proportion.

Answer #1

Solution :

Given that,

n = 2000

x = 1500

(a)

Point estimate = sample proportion = = x / n = 1500 / 2000 = 0.750

1 - = 1 - 0.750 = 0.250

At 90% confidence level the z is ,

= 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 0.10 / 2 = 0.05

Z_{/2}
= Z_{0.05} = 1.645

(b)

Margin of error = E = Z_{ / 2} * (( * (1 - )) / n)

= 1.645 * (((0.750 * 0.250) / 2000)

= 0.016

(c)

A 90% confidence interval for population proportion p is ,

- E < p < + E

0.750 - 0.016 < p < 0.750 + 0.016

0.734 < p < 0.766

The 90% confidence interval for the population proportion p is : 0.734 to 0.766

(d)

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

Margin of error = E = Z_{ / 2} * (( * (1 - )) / n)

= 1.96 * (((0.750 * 0.250) / 2000)

= 0.019

A 95% confidence interval for population proportion p is ,

- E < p < + E

0.750 - 0.019 < p < 0.750 + 0.019

0.731 < p < 0.769

The 95% confidence interval for the population proportion p is : 0.731 to 0.769

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