Respuesta :
Using confidence interval of proportions, it is found that:
a) The margin of error is of 8.3%.
b) The margin of error is of 8.8%.
c) The interval is between 20% and 40%.
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In a sample with a number n of people surveyed with a probability of a success of [tex]\pi[/tex], and a confidence level of [tex]1-\alpha[/tex], we have the following confidence interval of proportions.
[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
In which
z is the z-score that has a p-value of [tex]1 - \frac{\alpha}{2}[/tex].
The margin of error is of:
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
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Item a:
- 110 students, thus, [tex]n = 110[/tex]
- 30 sophomores, thus, [tex]\pi = \frac{30}{110} = 0.2727[/tex]
- Standard confidence level of 95%, thus [tex]z = 1.96[/tex]
The margin of error is of:
[tex]M = 1.96\sqrt{\frac{0.2727(0.7273)}{110}} = 0.083[/tex]
As a percentage, 0.083*100% = 8.3%.
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Item b:
- 65 students, thus, [tex]n = 65[/tex]
- 10 sophomores, thus, [tex]\pi = \frac{10}{65} = 0.1538[/tex]
The margin of error is of:
[tex]M = 1.96\sqrt{\frac{0.1538(0.8462)}{65}} = 0.088[/tex]
As a percentage, 0.088*100% = 8.8%.
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Item c:
- The interval is the sample proportion plus/minus the margin of error.
[tex]0.3 - 0.1 = 0.2[/tex]
[tex]0.3 + 0.1 = 0.4[/tex]
As a percentage, 20% and 40%.
A similar problem is given at https://brainly.com/question/16807970.
