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The sample survey showed that 67% of internet users said the internet has generally strengthened their relationship with family

SpyIntel [72]

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.

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A recipe for Potato Soup is listed below:

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That’s a pretty good recipe

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What is 25 divided by 2350

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