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:

$n = 100, p = 0.42$

92% confidence level

So $\alpha = 0.08$ , z is the value of Z that has a pvalue of $1 - \frac{0.08}{2} = 0.96$ , so $Z = 1.75$ .

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.42 - 1.75\sqrt{\frac{0.42*0.58}{100}} = 0.3336$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.42 - 1.75\sqrt{\frac{0.42*0.58}{100}} = 0.5064$

The 92% confidence interval for the true proportion of customers who click on ads on their smartphones is (0.3336, 0.5064).

4 0

4 years ago

A merchant has coffee worth $20 a pound that she wishes to mix with 70 pounds of coffee worth $90 a pound to get a mixture that

Dvinal [7]

Let us start with assuming the amount of coffee worth $20 a pound to be "x" pounds.

Now she wants to mix "x" pounds of $20 coffee with 70 pounds of $90 coffee.

So the mixture would be (x + 70) pounds.

And the value of mixture would be = 20·x + (90)·(70) = (20x + 6300) dollars.

She want to sell this mixture at rate of $30 a pound. Her earning would be = 30·(x + 70) = (30x + 2100) dollars.

We know that the value of mixture would be equal to her earnings.

30x + 2100 = 20x + 6300

30x - 20x = 6300 - 2100

10x = 4200

x = 420

So, 420 pounds of $20 coffee would be used.

7 0

3 years ago

Read 2 more answers

Quadrilateral $ABCD$ has right angles at $B$ and $D$, and $AC=3$. If $ABCD$ has two sides with distinct integer lengths, then wh

Jet001 [13]

Answer:

34 is the answer

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers