Question

Over the course of a week, an environmental activist randomly stopped 1,200 people on the street and asked whether they were willing to pay higher prices for gas in order to protect the environment. Out of the 1,200 people, 708 of them said they were willing to pay higher prices for gas. With a 99% confidence level (z*-score = 2.58), the activist wants to determine a confidence interval for the estimated proportion of people willing to pay higher prices for gas in order to protect the environment.

Answer 1

Sample Size = n = 1200
Number of people who are willing to pay higher = x = 708

Proportion of people who are willing to pay higher = p = $\frac{708}{1200}= \frac{59}{100}=0.59$

Confidence Level = 99%
Z Value =  z = 2.58

The confidence interval about a population proportion can be calculated as:

$(p-z \sqrt{ \frac{p(1-p)}{n} }, p+z \sqrt{ \frac{p(1-p)}{n} })$

Using the values, we get:

$(0.59-2.58 \sqrt{ \frac{0.59(0.41)}{1200} } ,0.59+2.58 \sqrt{ \frac{0.59(0.41)}{1200} }) \\ \\ (0.553,0.627)$

Thus, at 99% confidence level, the estimated proportion of people willing to pay higher prices for gas in order to protect the environment lies between 0.553 and 0.627.

Answer 2

Answer:

d. between 55.3% and 62.7%

Step-by-step explanation:

edgu 2020