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?

Explanation:

Step1: Identify sample proportion and sample size

Let $\hat{p}$ be sample - proportion and $n$ be sample size. $\hat{p}=0.45$, $n = 200$.

Step2: Find critical value for 90% confidence interval

For a 90% confidence interval, the significance level $\alpha=1 - 0.90=0.10$, and $\alpha/2=0.05$. The critical value $z_{\alpha/2}=z_{0.05}\approx1.645$.

Step3: 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:
\[

$$\begin{align*} E&=1.645\sqrt{\frac{0.45\times(1 - 0.45)}{200}}\\ &=1.645\sqrt{\frac{0.45\times0.55}{200}}\\ &=1.645\sqrt{\frac{0.2475}{200}}\\ &=1.645\sqrt{0.0012375}\\ &=1.645\times0.0352\\ &\approx0.06 \end{align*}$$

\]

Step4: Calculate the confidence interval

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

Answer:

The 90% confidence interval for the true proportion of club members who use compost is $(0.39,0.51)$. The value to fill in the blank for the margin - of - error is approximately $0.06$.