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Sagot :
Answer:
The 95% confidence intervals is (0.109, 0.125).
Step-by-step explanation:
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 zscore that has a pvalue of [tex]1 - \frac{\alpha}{2}[/tex].
Out of a sample of 6,458, a total of 754 reported that they have experienced physical bullying.
This means that [tex]n = 6458, \pi = \frac{754}{6458} = 0.1168[/tex]
95% confidence level
So [tex]\alpha = 0.05[/tex], z is the value of Z that has a pvalue of [tex]1 - \frac{0.05}{2} = 0.975[/tex], so [tex]Z = 1.96[/tex].
The lower limit of this interval is:
[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.1168 - 1.96\sqrt{\frac{0.1168*0.8832}{6458}} = 0.109[/tex]
The upper limit of this interval is:
[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.1168 + 1.96\sqrt{\frac{0.1168*0.8832}{6458}} = 0.125[/tex]
The 95% confidence intervals is (0.109, 0.125).
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