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:

560 random tests conducted, 26 employees failed the test. This means that $n = 560, \pi = \frac{26}{560} = 0.046$

98% confidence level

So $\alpha = 0.02$ , z is the value of Z that has a pvalue of $1 - \frac{0.02}{2} = 0.99$ , so $Z = 2.33$ .

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.046 - 2.33\sqrt{\frac{0.046*0.954}{560}} = 0.025$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.046 + 2.33\sqrt{\frac{0.046*0.954}{560}} = 0.067$

The 98% confidence interval for the proportion of applicants that fail the test is (0.025, 0.067).

5 0

4 years ago

Y=2/3x-1 ×=-6,-3,0,3

Y_Kistochka [10]

Answer:

y=-5,-3,-1,1 are value for y

4 0

3 years ago

Please answer this question :)

sveticcg [70]

Answer:

a) 49.41

b) not quite sure

Step-by-step explanation:

sorry

6 0

3 years ago