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?
(there are two drop - down boxes with a ± sign in between, and also reset and next buttons at the bottom left.)

Explanation:

Step1: Identify the sample proportion and sample size

The sample proportion \(\hat{p}\) is \(0.45\) (since \(45\% = 0.45\)) and the sample size \(n = 200\).

Step2: Find the critical value for 90% confidence

For a 90% confidence interval, the critical value \(z^*\) (from the standard normal distribution) is approximately \(1.645\).

Step3: Calculate the standard error

The formula for the standard error \(SE\) of a proportion is \(SE=\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}\).
Substitute \(\hat{p}=0.45\) and \(n = 200\):

$$ SE=\sqrt{\frac{0.45\times(1 - 0.45)}{200}}=\sqrt{\frac{0.45\times0.55}{200}}=\sqrt{\frac{0.2475}{200}}\approx\sqrt{0.0012375}\approx0.0352 $$

Step4: Calculate the margin of error

The margin of error \(ME = z^*\times SE\).
Substitute \(z^* = 1.645\) and \(SE\approx0.0352\):

$$ ME=1.645\times0.0352\approx0.0579 $$

Step5: Construct the confidence interval

The confidence interval is \(\hat{p}\pm ME\), so \(0.45\pm0.0579\) (or \(45\%\pm5.79\%\)).

Answer:

The 90% confidence interval is \(0.45\pm0.0579\) (or \(45\%\pm5.8\%\) approximately). The first drop - down should be \(0.45\) and the second drop - down should be approximately \(0.058\) (or \(5.8\%\) if in percentage terms).