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Gennadij [26K]
3 years ago
6

I need the answer to 35-38 please

Mathematics
1 answer:
Vadim26 [7]3 years ago
3 0
Good night. How ru?

35) x => 20
36) x =< 8 
37) x > 36
38) x < 9
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The sample size to obtain the desired margin of error is 160.

Step-by-step explanation:

The Margin of Error is given as

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n_2=\left[z_{crit}\times \dfrac{\sigma}{M_2}\right]^2\\n_2=\left[z_{crit}\times \dfrac{\sigma}{M/2}\right]^2\\n_2=\left[z_{crit}\times \dfrac{2\sigma}{M}\right]^2\\n_2=2^2\left[z_{crit}\times \dfrac{\sigma}{M}\right]^2\\n_2=4\left[z_{crit}\times \dfrac{\sigma}{M}\right]^2\\n_2=4n

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n_2=4n\\n_2=4*40\\n_2=160

So the sample size to obtain the desired margin of error is 160.

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The dwarves of the Grey Mountains wish to conduct a survey of their pick-axes in order to construct a 99% confidence interval ab
Dmitry_Shevchenko [17]

Answer:

The minimum sample size needed is 125.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of \pi, and a confidence level of 1-\alpha, we have the following confidence interval of proportions.

\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}

In which

z is the zscore that has a pvalue of 1 - \frac{\alpha}{2}.

The margin of error is:

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For this problem, we have that:

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99% confidence level

So \alpha = 0.01, z is the value of Z that has a pvalue of 1 - \frac{0.01}{2} = 0.995, so Z = 2.575.

What minimum sample size would be necessary in order ensure a margin of error of 10 percentage points (or less) if they use the prior estimate that 25 percent of the pick-axes are in need of repair?

This minimum sample size is n.

n is found when M = 0.1

So

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

0.1 = 2.575\sqrt{\frac{0.25*0.75}{n}}

0.1\sqrt{n} = 2.575{0.25*0.75}

\sqrt{n} = \frac{2.575{0.25*0.75}}{0.1}

(\sqrt{n})^{2} = (\frac{2.575{0.25*0.75}}{0.1})^{2}

n = 124.32

Rounding up

The minimum sample size needed is 125.

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3 years ago
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