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 zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

For this problem, we have that:

$n = 186, \pi = 0.11$

95% confidence level

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

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.11 - 1.96\sqrt{\frac{0.11*0.89}{186}} = 0.065$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.11 + 1.96\sqrt{\frac{0.11*0.89}{186}} = 0.155$

His 95% confidence interval is (0.065, 0.155).

4 0

2 years ago

PLEASE HELP ITS URGENT!

Rudik [331]

Answer:

Step-by-step explanation:

The two equations are similar, because if you multiply the first one by 3 it will be equal to the second one, so when that happens we say that there are infinite solutions.

8 0

2 years ago

On a quiz, Tom received 2 points for each correct spelling word, and 4 points for each correct definition, Tom received 80 point

Lesechka [4]

Answer:

Tom got 20 words correct and 30 definitions correct

Step-by-step explanation:

6 0

2 years ago