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?
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Explanation:

Step1: Identify the sample proportion and size

The sample proportion $\hat{p}$ is 45% or 0.45, and the sample size $n = 200$.

Step2: Find the z-score for 90% confidence

For a 90% confidence interval, the z-score ($z_{\alpha/2}$) is 1.645 (from standard normal distribution tables).

Step3: Calculate the standard error

The formula for the standard error (SE) of a proportion is $SE=\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}$.
Substituting $\hat{p}=0.45$ 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 (ME)

The margin of error is $ME = z_{\alpha/2}\times SE$.
Substituting $z_{\alpha/2}=1.645$ and $SE\approx0.0352$:
$$ME = 1.645\times0.0352\approx0.0579$$

Step5: Form the confidence interval

The confidence interval is $\hat{p}\pm ME$, so it is $0.45\pm0.0579$ (or 45% $\pm$ 5.79%).

Answer:

$0.45 \pm 0.0579$ (or in percentage terms, $45\% \pm 5.79\%$)