Question

The sample survey showed that 72% 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.

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:

$0.72 \pm 1.96\sqrt{\frac{0.72*0.28}{n}}$

In which n is the sample size of the survey.

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 $1 - \frac{\alpha}{2}$.

The sample survey showed that 72% of Internet users said the Internet has generally strengthened their relationship with family and friends.

This means that $\pi = 0.72$

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 $Z = 1.96$.

Confidence interval:

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

$0.72 \pm 1.96\sqrt{\frac{0.72*0.28}{n}}$

The 95% confidence interval for the proportion of respondents who say the Internet has strengthened their relationship with family and friends is:

$0.72 \pm 1.96\sqrt{\frac{0.72*0.28}{n}}$

In which n is the sample size of the survey.