The 95% confidence interval is (8.3%,17.7%), and the correct option is C.
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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].
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13% of a sample of 200 students do not like ice cream.
This means that [tex]\pi = 0.13, n = 200[/tex]
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95% confidence level
So [tex]\alpha = 0.05[/tex], z is the value of Z that has a p-value of [tex]1 - \frac{0.05}{2} = 0.975[/tex], so [tex]Z = 1.96[/tex].
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The lower limit of this interval is:
[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.13 - 1.96\sqrt{\frac{0.13*0.87}{200}} = 0.083[/tex]
The upper limit of this interval is:
[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.13 + 1.96\sqrt{\frac{0.13*0.87}{200}} = 0.177[/tex]
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As a percentage:
- 0.083x100% = 8.3%
- 0.177x100% = 17.7%
Thus, the 95% confidence interval is (8.3%,17.7%), and the correct option is C.
A similar problem is given at https://brainly.com/question/22223066
