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sydneip2394
14.07.2019 •
Mathematics
Arecent 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 %.
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Ответ:
THE ANSWER IS 3.6% BECAUSE I JUST TOOK THE TEST AND GOT A 100%
Ответ:
3.6%
Step-by-step 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{\frac{pq}{n} } =0.0219](/tpl/images/0089/0006/e4d47.png)
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%
Ответ: