Write an expression that will have a product greater than both factors.

Construct the 90% confidence interval for the proportion of students at the college who have completed their required English 10

igomit [66]

Answer:

The confidence interval has an lower limit of $\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = \pi - 1.645\sqrt{\frac{\pi(1-\pi)}{n}}$ and an upper limit of $\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = \pi + 1.645\sqrt{\frac{\pi(1-\pi)}{n}}$ , in which $\pi$ is the sample proportion and $n$ is the size of the sample.

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

90% confidence level

So $\alpha = 0.1$ , z is the value of Z that has a p-value of $1 - \frac{0.1}{2} = 0.95$ , so $Z = 1.645$ .

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = \pi - 1.645\sqrt{\frac{\pi(1-\pi)}{n}}$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = \pi + 1.645\sqrt{\frac{\pi(1-\pi)}{n}}$

The confidence interval has an lower limit of $\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = \pi - 1.645\sqrt{\frac{\pi(1-\pi)}{n}}$ and an upper limit of $\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = \pi + 1.645\sqrt{\frac{\pi(1-\pi)}{n}}$ , in which $\pi$ is the sample proportion and $n$ is the size of the sample.

8 0

4 years ago

2. If the sum of the first 12 terms of a geometric series is 8190 and the common ratio is 2. Find the first term and the 20th te

alexandr1967 [171]

the first term is 2 and the 20th term is 1048576 .

<u>Step-by-step explanation:</u>

Here we have , If the sum of the first 12 terms of a geometric series is 8190 and the common ratio is 2. We need to Find the first term and the 20th term. Let's find out:

We know that Sum of a GP is :

⇒ $S_n = \frac{a(r^n-1)}{r-1}$