Questions: Widows A recent study indicated that 18% of the 196 women over age 55 in the study were widows. Round up your answers to the next whole number for the following questions. How large a sample must you take to be 90% confident that the estimate is within 0.05 of the true proportion of women over age 55 who are widows? n=

Widows A recent study indicated that 18% of the 196 women over age 55 in the study were widows. Round up your answers to the next whole number for the following questions.

How large a sample must you take to be 90% confident that the estimate is within 0.05 of the true proportion of women over age 55 who are widows?

n=
Transcript text: Widows A recent study indicated that $18 \%$ of the 196 women over age 55 in the study were widows. Round up your answers to the next whole number for the following questions. Part: $0 / 2$ $\square$ Part 1 of 2 How large a sample must you take to be $90 \%$ confident that the estimate is within 0.05 of the true proportion of women over age 55 who are widows? \[ n= \] $\square$
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Solution

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Solution Steps

Step 1: Given Information

We are tasked with determining the required sample size \( n \) to estimate the proportion of women over age 55 who are widows with a confidence level of \( 90\% \). The following values are provided:

  • Estimated proportion \( p = 0.18 \)
  • Margin of error \( E = 0.05 \)
  • Z-score for \( 90\% \) confidence level \( Z \approx 1.645 \)
Step 2: Sample Size Formula

The formula for calculating the required sample size \( n \) for estimating a population proportion is given by:

\[ n = \frac{Z^2 \cdot p \cdot (1 - p)}{E^2} \]

Step 3: Substitute Values

Substituting the known values into the formula:

\[ n = \frac{(1.645)^2 \cdot 0.18 \cdot (1 - 0.18)}{(0.05)^2} \]

Calculating each component:

  • \( (1.645)^2 \approx 2.706025 \)
  • \( 1 - 0.18 = 0.82 \)
  • \( (0.05)^2 = 0.0025 \)

Thus, we have:

\[ n = \frac{2.706025 \cdot 0.18 \cdot 0.82}{0.0025} \]

Step 4: Calculate Sample Size

Calculating the numerator:

\[ 2.706025 \cdot 0.18 \cdot 0.82 \approx 0.4001 \]

Now, substituting back into the equation for \( n \):

\[ n = \frac{0.4001}{0.0025} \approx 160.04 \]

Step 5: Round Up

Since the sample size must be a whole number, we round up \( 160.04 \) to the next whole number:

\[ n = 161 \]

Final Answer

The required sample size to be \( 90\% \) confident that the estimate is within \( 0.05 \) of the true proportion is:

\[ \boxed{n = 161} \]

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