Question

The sample survey showed that 67% of internet users said the internet has generally strengthened their relationship with family and friends. Develop a 95% confidence interval for the proportion of respondents who say the internet has strengthened their relationship with family and friends. (Round your answers to four decimal places.)

Answer 1

Answer:

The 95% confidence interval for the proportion of respondents who say the internet has strengthened their relationship with family and friends is $(0.67 - 1.96\sqrt{\frac{0.67*0.33}{n}}, 0.67 + 1.96\sqrt{\frac{0.67*0.33}{n}})$, 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 $\pi$, 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

z is the z-score that has a p-value of $1 - \frac{\alpha}{2}$.

67% of internet users said the internet has generally strengthened their relationship with family and friends.

This means that $\pi = 0.67$

95% confidence level

So $\alpha = 0.05$, z is the value of Z that has a p-value of $1 - \frac{0.05}{2} = 0.975$, so $Z = 1.96$.

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.67 - 1.96\sqrt{\frac{0.67*0.33}{n}}$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.67 + 1.96\sqrt{\frac{0.67*0.33}{n}}$

The 95% confidence interval for the proportion of respondents who say the internet has strengthened their relationship with family and friends is $(0.67 - 1.96\sqrt{\frac{0.67*0.33}{n}}, 0.67 + 1.96\sqrt{\frac{0.67*0.33}{n}})$, in which n is the size of the sample.