Question

1.Express the confidence interval (45.3%,61.9%) in the form of ˆp ± E

2.Assume that a sample is used to estimate a population proportion p. Find the 80% confidence interval for a sample of size 260 with 24% successes. Enter your answer as a tri-linear inequality using decimals (not percents) accurate to three decimal places.

3.Out of 400 people sampled, 240 preferred Candidate A. Based on this, estimate what proportion of the entire voting population (p) prefers Candidate A.

Use a 90% confidence level, and give your answers as decimals, to three places.

Answer 1

Solution :

1)  confidence interval (45.3%,61.9%)

confidence interval (0.453,0.619)

Point estimate = = (Lower confidence interval + Upper confidence interval ) / 2

Point estimate = = (0.453 + 0.619) / 2

Point estimate = = 0.536

Point estimate = = 53.6%

Margin of error = E = Upper confidence interval -

Margin of error = E = 0.619 - 0.536

Margin of error = E = 0.083

Margin of error = E = 8.3%

In the form of ˆp ± E

= 53.6% ± 8.3%

2) Given that,

n = 260

Point estimate = sample proportion = = 0.24

1 - = 1 - 0.24 = 0.76

At 80% confidence level

= 1 - 80%

=1 - 0.80 =0.20

/2 = 0.10

Z/2 = Z_{0.10 = 1.282}

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

= 1.282 (((0.24 * 0.76) / 260)

= 0.034

A 80% confidence interval for population proportion p is ,  

- E < p < + E

0.24 - 0.034 < p < 0.24 + 0.034

( 0.206 < p < 0.274 )

3) Given that,

n = 400

x = 240

Point estimate = sample proportion = = x / n = 240 / 400 = 0.600

1 - = 1 - 0.600 = 0.400

At 90% confidence level

= 1 - 90%

=1 - 0.90 =0.10

/2 = 0.05

Z/2 = Z_{0.05 = 1.645}

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

= 1.645 (((0.600 * 0.400) / 400 )

= 0.040

A 90% confidence interval for population proportion p is ,

± E

= 0.600 ± 0.040

( 0.560, 0.640 )