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Answer:
The 95% confidence interval for the proportion of respondents who say the internet has strengthened their relationship with family and friends is [tex](0.67 - 1.96\sqrt{\frac{0.67*0.33}{n}}, 0.67 + 1.96\sqrt{\frac{0.67*0.33}{n}})[/tex], in which n is the size of the sample.
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 z-score that has a p-value of [tex]1 - \frac{\alpha}{2}[/tex].
67% of internet users said the internet has generally strengthened their relationship with family and friends.
This means that [tex]\pi = 0.67[/tex]
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].
The lower limit of this interval is:
[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.67 - 1.96\sqrt{\frac{0.67*0.33}{n}}[/tex]
The upper limit of this interval is:
[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.67 + 1.96\sqrt{\frac{0.67*0.33}{n}}[/tex]
The 95% confidence interval for the proportion of respondents who say the internet has strengthened their relationship with family and friends is [tex](0.67 - 1.96\sqrt{\frac{0.67*0.33}{n}}, 0.67 + 1.96\sqrt{\frac{0.67*0.33}{n}})[/tex], in which n is the size of the sample.
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