In the past, zest of al homes whth a stapatheme parent. the father is the stay atheme parent An mdependent researeh
(a) What sample she is needed if the researd firmis goat is to estimate the curfent prepertion of hemes with a stay atheme parett in which the father is the stajatheme parent with a margin of effer of 0.04B Use a cenfldence leveh (hound your anstrer up to the nearest whele number.)

(b) Repeat part (a) using a 890 confidence leveh (Round your answer up to the nearest whole number)

Ask by Osborne Maxwell. in the United States

Mar 31,2025

Question

In the past, zest of al homes whth a stapatheme parent. the father is the stay atheme parent An mdependent researeh
(a) What sample she is needed if the researd firmis goat is to estimate the curfent prepertion of hemes with a stay atheme parett in which the father is the stajatheme parent with a margin of effer of 0.04B Use a cenfldence leveh (hound your anstrer up to the nearest whele number.)

(b) Repeat part (a) using a 890 confidence leveh (Round your answer up to the nearest whole number)

Ask by Osborne Maxwell. in the United States

Mar 31,2025

UpStudy AI · Accepted Answer

We wish to determine the sample size needed to estimate a proportion to within a margin of error $$E=0.04$$. In the worst‐case scenario (where the variability is maximized) we take the proportion as $$p=0.5$$. The formula for the sample size $$n$$ for estimating a proportion is

$$
n=\frac{z_{\alpha/2}^2 \, p(1-p)}{E^2}.
$$

In both parts we use $$p(1-p) = 0.5(1-0.5)=0.25$$ and $$E^2=0.04^2=0.0016$$.

**(a) For an 85% Confidence Level**

When the confidence level is 85%, the total area in both tails is $$15\%$$ (since $$100\%-85\%=15\%$$). Thus, the area in each tail is

$$
\frac{15\%}{2} = 7.5\%.
$$

We need the $$z$$-value for which

$$
P(Z\leq z_{\alpha/2})=1-0.075=0.925.
$$

This $$z$$-value is approximately

$$
z_{0.075}\approx 1.44.
$$

Now substitute into the formula:

$$
n=\frac{(1.44)^2 (0.25)}{0.0016}.
$$

Calculate step by step:

1. Compute $$ (1.44)^2 $$:
   $$
   (1.44)^2 = 2.0736.
   $$

2. Multiply by 0.25:
   $$
   2.0736\times 0.25=0.5184.
   $$

3. Divide by 0.0016:
   $$
   n=\frac{0.5184}{0.0016}=324.
   $$

Since we round up to the nearest whole number, the sample size needed is

$$
\boxed{324}.
$$

---

**(b) For a 90% Confidence Level**

In part (b) we are asked to repeat the calculation using a 90% confidence level. (Here we assume that “890 confidence level” is a typographical error and that the intended confidence level is 90%.) For a 90% confidence level, the total area in the tails is $$10\%$$ (since $$100\%-90\%=10\%$$); thus, each tail has an area of

$$
\frac{10\%}{2} = 5\%.
$$

The $$z$$-value satisfies

$$
P(Z\leq z_{\alpha/2}) = 1-0.05=0.95,
$$

so

$$
z_{0.05}\approx 1.645.
$$

Now, substitute into the sample size formula:

$$
n=\frac{(1.645)^2(0.25)}{0.0016}.
$$

Calculate step by step:

1. Compute $$ (1.645)^2 $$:
   $$
   (1.645)^2\approx 2.706.
   $$

2. Multiply by 0.25:
   $$
   2.706\times 0.25\approx 0.6765.
   $$

3. Divide by 0.0016:
   $$
   n=\frac{0.6765}{0.0016}\approx 422.81.
   $$

Rounding up to the nearest whole number gives

$$
\boxed{422}.
$$

---

Thus, the answers are:

(a) Sample size needed is $$324$$.

(b) Sample size needed is $$422$$.
(a) A sample size of 324 is needed. (b) A sample size of 422 is needed.

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