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prohojiy [21]
3 years ago
8

Describe how a company that produce earphones can design a survey

Mathematics
1 answer:
DiKsa [7]3 years ago
3 0
First of all this survey need to be <span>anonymous. The company need to know the population whom the survey is addressed. So, I think the following questions can be a good example:

1. </span><span>Where do you usually use earphones? (Please select all that apply)

</span><span>library 
home
transportation
kitchen
driving/riding
Other

2. What are the purposes you use earphones? (Please select all that apply)</span><span>  learn language
listen to music
watch video
privacy concerns(show ‘don’t bother me’)
for beauty
Other
</span>
<span>3. What kind of earphones you like the most?
<span> 
circumaural
supra-aural
earbuds
in-ear 

<span>4. What are the reasons make you like the earphone you chose in question 3? (Please select all that apply)
<span> 
appearance
comfortability
function(plays music well) 
brand 
price 
portability 
Other

<span>5. Have you ever felt vulnerable or unsafe while wearing headphones and listening to music because you are not as aware of your surroundings?
<span><span> 
yes</span><span> 
no</span></span></span></span></span></span></span>
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Kim wants to determine a 90 percent confidence interval for the true proportion of high school students in the area who attend t
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Answer:

We need a sample of at least 752 students.

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

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The margin of error of the interval is:

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So \alpha = 0.1, z is the value of Z that has a pvalue of 1 - \frac{0.1}{2} = 0.95, so Z = 1.645.

How large of a sample must she have to get a margin of error less than 0.03

We need a sample of at least n students.

n is found when M = 0.03.

We have no information about the true proportion, so we use \pi = 0.5.

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

0.03 = 1.645\sqrt{\frac{0.5*0.5}{n}}

0.03\sqrt{n} = 1.645*0.5

\sqrt{n} = \frac{1.645*0.5}{0.03}

(\sqrt{n})^{2} = (\frac{1.645*0.5}{0.03})^{2}

n = 751.67

Rounding up

We need a sample of at least 752 students.

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