Question

On September 25, 2011 Michael Vick, the quarterback for the Philadelphia Eagles broke his non-throwing hand in a football game against the New York Giants. ESPN then posted a poll on their website.
The poll asked viewers to predict which team would win the NFC East Division. The Eagles play in the NFC East Division. By 4:15 pm, 914 fans had voted. 25% of them thought the Eagles would still win the division.


What can we conclude from a 95% confidence interval about the opinions of ESPN viewers?


A. We are 95% confident that between about 22% and 28% of fans thought the Eagles would still win the division

B. nothing because the count of successes is too small

C. nothing because the sample was not randomly selected

D. We are 95% confident that between about 22% and 28% of ESPN Viewers thought the Eagles would still win the division.

Answer 1

Answer:

D. We are 95% confident that between about 22% and 28% of ESPN Viewers thought the Eagles would still win the division.

Step-by-step explanation:

We have to calculate a 95% confidence interval for the population proportion.  The population is made up of ESPN viewers.

The sample proportion is p=0.25.

The standard error of the proportion is:

$\sigma_p=\sqrt{\dfrac{p(1-p)}{n}}=\sqrt{\dfrac{0.25*0.75}{914}}\\\\\\ \sigma_p=\sqrt{0.00021}=0.014$

The critical z-value for a 95% confidence interval is z=1.96.

The margin of error (MOE) can be calculated as:

$MOE=z\cdot \sigma_p=1.96 \cdot 0.014=0.028$

Then, the lower and upper bounds of the confidence interval are:

$LL=p-z \cdot \sigma_p = 0.25-0.028=0.222\\\\UL=p+z \cdot \sigma_p = 0.25+0.028=0.278$

The 95% confidence interval for the proportion of ESPN viewers  is (0.222, 0.278) or (22.2%, 27.8%).