Question

Zarina wants to determine a confidence interval based on a random poll she conducted about the percent of attendees at a conference who are vegetarian. She determined the margin of error for her poll results using the equation below.
E = 1.96 /.3(1-.3)/540





Based on Zarina’s work, what can be concluded?







With 90% confidence, between 4% and 30% of the attendees are vegetarian.




With 90% confidence, between 26% and 34% of the attendees are vegetarian.




With 95% confidence, between 4% and 30% of the attendees are vegetarian.




With 95% confidence, between 26% and 34% of the attendees are vegetarian.

Answer 1

Answer:

With 95% confidence, between 26% and 34% of the attendees are vegetarian.

Step-by-step explanation:

Given:

Zarina used 1.96 for confidence interval for the proportion

We know that a sample proportion will have standard error as

square root of pq/n

So from the information given,E = 1.96 /.3(1-.3)/540

we find that since 1.96 is used, 95% confidence level was done.

Sample size n =540

p = 0.3 and q = 0.7

Std error =$\sqrt{\frac{0.3(0.7)}{540} } =0.0197$

Margin of error = 1.96(std error) = 0.038

=0.040 (after rounding off)

=4%

Based on Zarina’s work, it can be concluded that: 

suitable option would be which as 4 as margin of error and 95% Conf level.

With 95% confidence, between 26% and 34% of the attendees are vegetarian.

is right answer.

Answer 2

Based on Zarina’s work, it can be concluded that: 

With 95% confidence, between 26% and 34% of the attendees are vegetarian.

She did a binomial experiment. The experiment consists of a number of repeated trials and each trial can result in just two possible outcomes. That can be either a success or a failure.