Question

select the correct answer from each drop - down menu. 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? 0.23 0.04 0.90 0.45 ± 0.06 reset next end of semester test

Explanation:

Step1: Identify sample proportion and z - value

The sample proportion $\hat{p}=0.45$. For a 90% confidence interval, the z - value $z_{\alpha/2}=1.645$ (from standard normal distribution tables). The sample size $n = 200$.

Step2: Calculate the margin of error

The formula for the margin of error $E$ for a proportion is $E=z_{\alpha/2}\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}$. Substitute $\hat{p}=0.45$, $n = 200$ and $z_{\alpha/2}=1.645$ into the formula. First, calculate $(1-\hat{p})=1 - 0.45=0.55$. Then $\frac{\hat{p}(1 - \hat{p})}{n}=\frac{0.45\times0.55}{200}=\frac{0.2475}{200}=0.0012375$. $\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}=\sqrt{0.0012375}\approx0.0352$. And $E = 1.645\times0.0352\approx0.06$.

Step3: Calculate the confidence interval

The confidence interval for a proportion is $\hat{p}-E

Answer:

The first dropdown: 0.45
The second dropdown: 0.06