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Karo-lina-s [1.5K]
2 years ago
6

Sarah polled 40 randomly selected students at her high school and found that 20% ( = 0.2) are happy with the quality of the cafe

teria food. With a desired confidence level of 99%, which has a corresponding z*-score of 2.58, what is the approximate margin of error of Sarah’s poll? Remember, the margin of error, E, can be determined using the formula E = z*
Mathematics
2 answers:
Anna71 [15]2 years ago
5 0
Given:
n = 40, sample size
Confidence level = 99% => z* = 2.58
\hat{p} = 20\%=0.2, sample proportion.

By definition, the margin of error is
z^{*} \sqrt{ \frac{\hat{p}(1-\hat{p})}{n} } = 2.58 \sqrt{ \frac{(0.2)(0.8)}{40} }= 0.1632

Answer:
According to Sarah's poll, she can conclude with 99% confidence level that 20% of the high school population is happy with the quality of the cafeteria food, with a margin of error of +/- 16.3%.

solong [7]2 years ago
3 0

Answer:

The answer is D: 16%

Step-by-step explanation:

I just took the test.

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One of the properties of an isosceles trapezoid is that the angles on either side of the two bases are equal. Since line AD is equal to line BC, then angle D is equal to angle C. It also implies that angle A is equal to angle B.

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