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mafiozo [28]
3 years ago
14

A medical researcher wants to investigate the amount of time it takes for patients' headache pain to be relieved after taking a

new prescription painkiller. She plans to use statistical methods to estimate the mean of the population of relief times. She believes that the population is normally distributed with a standard deviation of 20 minutes. How large a sample should she take to estimate the mean time to within 3 minutes with 91% confidence?
Mathematics
1 answer:
Mariana [72]3 years ago
3 0
<span>A medical researcher wants to investigate the amount of time it takes for patients' headache pain to be relieved after taking a new prescription painkiller. She plans to use statistical methods to estimate the mean of the population of relief times. She believes that the population is normally distributed with a standard deviation of 18 minutes. How large a sample should she take to estimate the mean time to within 4 minutes with 97% confidence?
-----
n = [z*s/E]^2
---

n = [2.17*18/4]^2 = 96 when rounded up

</span>
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6 0
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Dwayne's family has many pets. at the house they have both cats and birds. between all the cats and birds, they have 16 eyes and
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Sandy spent 4% of x hours at her part-time job.what is x if 4% of x is about 22 hours? Explain how you estimated and which prope
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5 0
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Which fraction (in simplest form) is equivalent to 0.03?<br> A.3/100<br> B.3/10<br> C.1/3<br> D.1/30
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3 0
3 years ago
A random sample is selected from a normally distributed population. The following sample statistics are obtained: n = 20, = 30,
Flura [38]

Answer:

b) The margin of error is approximately 3.24

e) The critical value is 1.7921.

Step-by-step explanation:

<u>Step(i</u>):-

Given sample size 'n' =20

Given sample standard deviation 's' = 10

<u><em> Margin of error </em></u>

<u><em>The margin of error is determined by</em></u>

<u><em /></u>M.E = \frac{t_{\alpha } S.D }{\sqrt{n} }<u><em /></u>

<em>The level of significance ∝ =0.95</em>

<em>The degrees of freedom = n-1 = 20-1=19</em>

t₀.₉₅ = 1.729

M.E = \frac{1.729 X10 }{\sqrt{20} }

Margin of error = 3.866

Step(ii):-

<u><em> Margin of error </em></u>

<u><em>The margin of error is determined by</em></u>

<u><em /></u>M.E = \frac{t_{\alpha } S.D }{\sqrt{n} }<u><em /></u>

<u><em>Given another sample size n =30</em></u>

<em>The level of significance ∝ =0.95</em>

<em>The degrees of freedom = n-1 = 30-1=29</em>

t₀.₉₅ = 1.70

M.E = \frac{1.7021 X10 }{\sqrt{30} } =3.24

<u><em>Margin of error = 3.24</em></u>

5 0
3 years ago
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