what is the minimal sample size needed for a 95% confidence interval to have a maximal margin of error of 0.1 in the following scenarios? (round your answers up the nearest whole number.) (a) a preliminary estimate for p is 0.14 (b) there is no preliminary estimate for p

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Accepted Answer

# Explanation:
## Step1: Recall the formula for sample size
The formula for sample size $n$ when estimating a proportion is $n=\frac{z^{2}\cdot p\cdot(1 - p)}{E^{2}}$. For a 95% confidence interval, the critical value $z = 1.96$ (from standard normal distribution), and the margin of error $E=0.1$.
## Step2: Calculate sample size for part (a)
Given $p = 0.14$, then $1-p=0.86$. Substitute into the formula:
$$
\begin{align*}
n&=\frac{(1.96)^{2}\cdot0.14\cdot0.86}{(0.1)^{2}}\
&=\frac{3.8416\times0.1204}{0.01}\
&=\frac{0.4625}{0.01}\
& = 46.25\approx47
\end{align*}
$$
## Step3: Calculate sample size for part (b)
When there is no preliminary estimate for $p$, we use $p = 0.5$ (maximizes $p(1 - p)$). Then $1-p = 0.5$. Substitute into the formula:
$$
\begin{align*}
n&=\frac{(1.96)^{2}\cdot0.5\cdot0.5}{(0.1)^{2}}\
&=\frac{3.8416\times0.25}{0.01}\
&=\frac{0.9604}{0.01}\
&=96.04\approx96
\end{align*}
$$

# Answer:
(a) $47$ (b) $96$

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QUESTION IMAGE

Question

Explanation:

Step1: Recall the formula for sample size

Step2: Calculate sample size for part (a)

Step3: Calculate sample size for part (b)

Answer: