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:

$n = 18, \pi = \frac{5}{18} = 0.2778$

99.5% confidence level

So $\alpha = 0.005$ , z is the value of Z that has a pvalue of $1 - \frac{0.005}{2} = 0.9975$ , so $Z = 2.81$ .

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.2778 - 2.81\sqrt{\frac{0.2778*0.7222}{18}} = -0.01 = 0$

We cannot have a negative proportion, so we use 0.

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.2778 + 2.81\sqrt{\frac{0.2778*0.7222}{18}} = 0.5745$

The 99.5% confidence interval for the proportion of all shrubs of this species that will resprout after fire is (0, 0.5745).

3 0

3 years ago

Help Me Please. Explain if possible, I just don't understand.

Kazeer [188]

Honestly i don’t even know how to answer this myself

7 0

3 years ago

Describe the pattern in the table

Anton [14]

5 is added each time.

5 0

4 years ago