Question

The most recent poll of 50 randomly-selected members of a school club with a total of 80 members showed that 38 are planning to vote for Talia as president of the club. Talia correctly determined that the margin of error, E, of the poll using a 95% confidence interval (z*-score 1.96) is approximately 12%.
What is the 95% confidence level for students who are planning to vote for Talia as president of the club?

C = + E

Answer 1

Answer with Explanation:

Total number of members in School Club = 80 members

Number of member randomly Selected,that is sample size used  (P) = 50 members

Number of members who are planning to vote for Talia = 38 members

Z Score for 95% confidence interval =1.96

it is given that Margin of Error ,E,of the poll using a 95% confidence interval=12%

$E=\frac{Z*\sigma}{\sqrt{P}}\\\\ \frac{12}{100}=\frac{1.96 * \sigma}{\sqrt{50}}\\\\ \sigma=4.32890$

Standard Deviation=4.3 (approx)

→95% confidence level for students who are planning to vote for Talia as president of the club lie between → 38 + 4.3 = 42.3 to 38 -4.3=33.7.

→So, members which will vote for Talia as president of the club lies between minimum of 33 members and maximum of 42 members.

Answer 2

Answer: D) between 64% and 88%

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