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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

write a statement that describes the relationship between these two sequence 6,7,8,9,10 and 18,21,24,27,30

Snezhnost [94]

Answer:

something something

Step-by-step explanation:

6 0

3 years ago

Mitch ordered several baseball hats for 12.65 each and paid $3.99 for shipping. If Mitch spent a total of $130.49, how many base

jek_recluse [69]

Answer: