Question

A recent poll of 380 randomly selected Americans showed that 24% ( = 0.24) are happy with their cell phone carriers. The country’s largest cell phone carrier would like to know, within a 90% confidence level, the margin of error for this poll. (90% confidence level z*-score of 1.645) Remember, the margin of error, E, can be determined using the formula E = z*. To the nearest tenth of a percent, the margin of error for the poll is %.

Answer 1

THE ANSWER IS 3.6% BECAUSE I JUST TOOK THE TEST AND GOT A 100%

Answer 2

3.6%

Explanation:

Given that A recent poll of 380 randomly selected Americans showed that 24% ( = 0.24) are happy with their cell phone carriers.

Sample size n =380

Sample proportion p = 0.24

q = 1-0.24 =0.76

Std error of p =
`\sqrt{(pq)/(n) } =0.0219`

Z critical value for 90% would be 1.645

Margin of error = ±1.645*std error

=±1.645(0.0219)

=±0.0361

i.e. in percent margin of error

3.61% = 3.6%