Question

1. A newsgroup is interested in constructing a 90% confidence interval for the proportion of all Americans who are in favor of a new Green initiative. Of the 526 randomly selected Americans surveyed, 353 were in favor of the initiative. Round answers to 4 decimal places where possible.

a. With 90% confidence the proportion of all Americans who favor the new Green initiative is between____ and ____ .

b. If many groups of 526 randomly selected Americans were surveyed, then a different confidence interval would be produced from each group. About _____ percent of these confidence intervals will contain the true population proportion of Americans who favor the Green initiative and about ____ percent will not contain the true population proportion.

2. 121 students at a college were asked whether they had completed their required English 101 course, and 88 students said "yes". Find the best point estimate for the proportion of students at the college who have completed their required English 101 course. Round to four decimal places.

3. In a recent poll, 330 people were asked if they liked dogs, and 69% said they did. Find the Margin of Error for this poll, at the 95% confidence level. Give your answer to four decimal places if possible.

4. Assume that a sample is used to estimate a population proportion p. Find the 90% confidence interval for a sample of size 116 with 85% successes. Enter your answer as a tri-linear inequality using decimals (not percents) accurate to three decimal places.
_____ < p < ______

5. Express the confidence interval 85%±7.5%85%±7.5% in the form of a trilinear inequality.
______ % <p<<p< _____ %

6. You recently sent out a survey to determine if the percentage of adults who use social media has changed from 61%, which was the percentage of adults who used social media five years ago. Of the 2357 people who responded to survey, 1713 stated that they currently use social media.
Use the data from this survey to construct a 93% confidence interval estimate of the proportion of adults who use social media. Record the result below in the form of (#,#)(#,#). Round your final answer to four decimal places.

Answer 1

1. n = 526

x = 353

p̄ = x/n = 0.6711

a) 90% Confidence interval :

At α = 0.1, two tailed critical value, z_c = NORM.S.INV(0.1/2) = 1.645

Lower Bound = p̄ - z_c*√( p̄ *(1- p̄ )/n) = 0.6711 - 1.645 *√(0.6711*0.3289/526) = 0.6374

Upper Bound = p̄ + z_c*√( p̄ *(1- p̄ )/n) = 0.6711 + 1.645 *√(0.6711*0.3289/526) = 0.7048

0.6374 < p < 0.7048

b) If many groups of 526 randomly selected Americans were surveyed, then a different confidence interval would be produced from each group. About 90 percent of these confidence intervals will contain the true population proportion of Americans who favor the Green initiative and about 10 percent will not contain the true population proportion.

-----------------------------------

2. n = 121

x = 88

Point estimate, p̄ = x/n = 0.7273

-----------------------------------

3. n = 330

p̄ = x/n = 0.69

At α = 0.05, two tailed critical value, z = NORM.S.INV(0.05/2) = 1.960

Margin of error, E = z*√( p̄ *(1- p̄ )/n) =1.96 *√(0.69*0.31/330) = 0.0499

95% Confidence interval :

Lower Bound = p̄ - E = 0.69 - 0.0499 = 0.6401

Upper Bound = p̄ + E = 0.69 + 0.0499 = 0.7399

-----------------------------------

4. n = 116

p̄ = x/n = 0.85

90% Confidence interval :

At α = 0.1, two tailed critical value, z_c = NORM.S.INV(0.1/2) = 1.645

Lower Bound = p̄ - z_c*√( p̄ *(1- p̄ )/n) = 0.85 - 1.645 *√(0.85*0.15/116) = 0.795

Upper Bound = p̄ + z_c*√( p̄ *(1- p̄ )/n) = 0.85 + 1.645 *√(0.85*0.15/116) = 0.905

0.795 < p < 0.905

------------------------ -----------

5. 85%-7.5% < p < 85%+7.5%

77.5% < p < 92.5%

-----------------------------------

6. n = 2357

x = 1713

p̄ = x/n = 0.7268

93% Confidence interval :

At α = 0.07, two tailed critical value, z_c = NORM.S.INV(0.07/2) = 1.812

Lower Bound = p̄ - z_c*√( p̄ *(1- p̄ )/n) = 0.7268 - 1.812 *√(0.7268*0.2732/2357) = 0.7101

Upper Bound = p̄ + z_c*√( p̄ *(1- p̄ )/n) = 0.7268 + 1.812 *√(0.7268*0.2732/2357) = 0.7434

0.7101 < p < 0.7434