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4 years ago

Some one help me idk what to do

Mathematics

1 answer:

Vinvika [58]4 years ago

6 0

Hey this is your answer that you are looking for

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-19+31 gets you 12.

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3 years ago

A college student organization wants to start a nightclub for students under the age of 21. To assess support for this proposal,

eduard

Answer:

a) Sample size = 822

b) Margin of error = 0.03407

Step-by-step explanation:

We are given the following in the question:

p = 74% = 0.74

a) Sample size is required to obtain margin of error of 0.03

Formula:

$\text{Margin of error} = z_{\text{statistic}}\times \sqrt{\dfrac{p(1-p)}{n}}}$

$z_{critical}\text{ at}~\alpha_{0.05} = 1.96$

Putting values, we get,

$0.03 = 1.96\times \sqrt{\dfrac{0.74(1-0.74)}{n}}\\\\n = (\dfrac{1.96}{0.03})^2(0.74)(1-0.74)\\\\n = 821.24 \approx 822$

Thus, the sample size must be approximately 822 to obtain a 95% confidence interval with an approximate margin of error of 0.03

b) Margin of error for the 95% confidence interval

p = 54% = 0.54

Formula:

$\text{Margin of error} = z_{\text{statistic}}\times \sqrt{\dfrac{p(1-p)}{n}}}$

$z_{critical}\text{ at}~\alpha_{0.05} = 1.96$

Putting values, we get,

$\text{Margin of error} = 1.96\times \sqrt{\dfrac{0.54(1-0.54)}{822}}\\\\=0.03407$

The margin of error now will be 0.03407.

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3 years ago