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The 99% confidence interval for the fraction of U.S. adult Twitter users who get some news on Twitter is (0.46, 0.58). This means that we are 99% sure that the true proportion of all U.S. adult Twitter users who get some news on Twitter is between these two values.

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

A poll found that 52% of U.S. adult Twitter users get at least some news on Twitter. The standard error for this estimate was 2.4%

This means that:

$\pi = 0.52, \sqrt{\frac{\pi(1-\pi)}{n}} = 0.024$

99% confidence level

So $\alpha = 0.01$ , z is the value of Z that has a p-value of $1 - \frac{0.01}{2} = 0.995$ , so $Z = 2.575$ .

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.52 - 2.575(0.024) = 0.46$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.52 + 2.575(0.024) = 0.58$

The 99% confidence interval for the fraction of U.S. adult Twitter users who get some news on Twitter is (0.46, 0.58). This means that we are 99% sure that the true proportion of all U.S. adult Twitter users who get some news on Twitter is between these two values.

8 0

3 years ago

Charlie bought a 2and 1 half bag of oranges for $3.75.What is the cost of one pound of oranges?

Nikolay [14]

Charlie buys 2.5 pounds of oranges for 3.75.

To find out how much he will pay for 1 pound of oranges, use the proportion

2.5/3.75 = 1/x
Cross multiply to get x = $1.50 per pound of oranges.

8 0

3 years ago

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