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?

The 95% confidence interval indicates that the proportion of voters who thought Eagles would still win the division was between 22% and 29%.

Step-by-step explanation:

Let X = number of ESPN viewers who thought the Eagles would win the division.

The probability that a viewer thought the Eagles would win is, p = 0.25.

A random sample of, n = 914 voters voted for the poll asking the viewers of ESPN to predict which team would win the NFC East Division.

A random viewer's opinion is independent from others.

The random variable X follows a Binomial distribution with parameters n = 914 and p = 0.25.

Now, as the sample size is quite large, i.e. n = 914 > 30, according to the Central limit theorem the sampling distribution of sample proportion can be described by the Normal distribution.

The mean of this sampling distribution is same as the population proportion, i.e. .

And the standard deviation of the sampling distribution of sample proportion is:

A (1 - α)% confidence interval for population proportion is given by:

The critical value of z for 95% confidence level is:

Compute the 95% confidence interval for the proportion of voters who thought Eagles would still win the division as follows:

Thus, the 95% confidence interval indicates that the proportion of voters who thought Eagles would still win the division was between 22% and 29%.

answer: 10 meter

step-by-step explanation:

need details

16 is the absolute value

step-by-step explanation: