Question

Cameron is a member of a national gardening club. she asked 200 of her fellow members whether they use compost to fertilize their plants, and 45% responded favorably. what is the 90% confidence interval for the true proportion of club members who use compost?

Answer 1

Using the z-distribution, it is found that the 90% confidence interval for the true proportion of club members who use compost is (0.392, 0.508).

What is a confidence interval of proportions?

A confidence interval of proportions is given by:

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

In which:

• $\pi$ is the sample proportion.

• z is the critical value.

• n is the sample size.

In this problem, we have a 90% confidence level, hence$\alpha = 0.9$, z is the value of Z that has a p-value of $\frac{1+0.9}{2} = 0.95$, so the critical value is z = 1.645.

The estimate and the sample size are given by:

$\pi = 0.45, n = 200$

Hence, the bounds of the interval are:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.45 - 1.645\sqrt{\frac{0.45(0.55)}{200}} = 0.392$

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.45 + 1.645\sqrt{\frac{0.45(0.55)}{200}} = 0.508$

More can be learned about the z-distribution at  https://brainly.com/question/25890103

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Answer 2

Answer:

The answers are 0.45 and 0.06

Step-by-step explanation:

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