You may need to use the appropriate appendix table or technology to answer this question.
In the past, of all homes with a stay-at-home parent, the father is the stay-at-home parent. An independent research
firm has been charged with conducting a sample survey to obtain more current information.
(a) What sample size is needed if the research firm’s goal is to estimate the current proportion of homes with a stay-at-
home parent in which the father is the stay-at-home parent with a margin of error of 0.04 ? Use a confidence
level. (Round your answer up to the nearest whole number.)
(b) Repeat part (a) using a confidence level. (Round your answer up to the nearest whole number.)
(b)

Ask by Mathis Reyes. in the United States

Mar 31,2025

Question

You may need to use the appropriate appendix table or technology to answer this question.
In the past, of all homes with a stay-at-home parent, the father is the stay-at-home parent. An independent research
firm has been charged with conducting a sample survey to obtain more current information.
(a) What sample size is needed if the research firm’s goal is to estimate the current proportion of homes with a stay-at-
home parent in which the father is the stay-at-home parent with a margin of error of 0.04 ? Use a confidence
level. (Round your answer up to the nearest whole number.)
(b) Repeat part (a) using a confidence level. (Round your answer up to the nearest whole number.)
(b)

Ask by Mathis Reyes. in the United States

Mar 31,2025

UpStudy AI · Accepted Answer

To determine the sample size needed for estimating the proportion of homes with a stay-at-home parent in which the father is the stay-at-home parent with a given margin of error and confidence level, we can use the formula for calculating the sample size for estimating a population proportion:

$$ n = \left( \frac{Z_{\alpha/2}}{E} \right)^2 \times p \times (1-p) $$

where:
- $$ n $$ is the sample size needed,
- $$ Z_{\alpha/2} $$ is the Z-score corresponding to the desired confidence level,
- $$ E $$ is the margin of error,
- $$ p $$ is the estimated proportion of homes with a stay-at-home parent in which the father is the stay-at-home parent.

Given:
- Margin of error ($$ E $$) = 0.04
- Confidence level = 95% (a) and 99% (b)

(a) For a 95% confidence level:
- $$ Z_{\alpha/2} $$ for 95% confidence level is approximately 1.96

(b) For a 99% confidence level:
- $$ Z_{\alpha/2} $$ for 99% confidence level is approximately 2.576

Let's calculate the sample size needed for both confidence levels.
Calculate the value by following steps:
- step0: Calculate:
  $$\left(\frac{1.96}{0.04}\right)^{2}\times 0.26\left(1-0.26\right)$$
- step1: Divide the terms:
  $$49^{2}\times 0.26\left(1-0.26\right)$$
- step2: Subtract the numbers:
  $$49^{2}\times 0.26\times 0.74$$
- step3: Multiply the terms:
  $$49^{2}\times 0.1924$$
- step4: Rewrite the expression:
  $$2401\times 0.1924$$
- step5: Multiply the numbers:
  $$461.9524$$
The sample size needed for estimating the proportion of homes with a stay-at-home parent in which the father is the stay-at-home parent with a margin of error of 0.04 and a 95% confidence level is approximately 462.
Calculate the value by following steps:
- step0: Calculate:
  $$\left(\frac{2.576}{0.04}\right)^{2}\times 0.26\left(1-0.26\right)$$
- step1: Divide the terms:
  $$\left(\frac{322}{5}\right)^{2}\times 0.26\left(1-0.26\right)$$
- step2: Subtract the numbers:
  $$\left(\frac{322}{5}\right)^{2}\times 0.26\times 0.74$$
- step3: Multiply the terms:
  $$\left(\frac{322}{5}\right)^{2}\times 0.1924$$
- step4: Convert the expressions:
  $$\left(\frac{322}{5}\right)^{2}\times \frac{481}{2500}$$
- step5: Multiply the numbers:
  $$\frac{12468001}{5^{6}}$$
- step6: Evaluate the power:
  $$\frac{12468001}{15625}$$
The sample size needed for estimating the proportion of homes with a stay-at-home parent in which the father is the stay-at-home parent with a margin of error of 0.04 and a 99% confidence level is approximately 798.
For a 95% confidence level, the sample size needed is approximately 462. For a 99% confidence level, the sample size needed is approximately 798.

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