Questions: Question 14 What is the smallest sample size that guarantees that the margin of error is less than 10 when constructing a 98% confidence interval for a population mean with the assumed population standard deviation of 335.22? (Round UP to the nearest integer)

Question 14 What is the smallest sample size that guarantees that the margin of error is less than 10 when constructing a 98% confidence interval for a population mean with the assumed population standard deviation of 335.22? (Round UP to the nearest integer)
Transcript text: Question 14 What is the smallest sample size that guarantees that the margin of error is less than 10 when constructing a $98 \%$ confidence interval for a population mean with the assumed population standard deviation of 335.22 ? (Round UP to the nearest integer) Question Help: Video Written Example Post to forum Submit Question
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Solution

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Solution Steps

Step 1: Determine the Z-Score

To find the Z-score for a \(98\%\) confidence level, we use the formula:

\[ Z = \text{PPF}\left(1 - \frac{1 - 0.98}{2}\right) = \text{PPF}(0.99) = 2.3263 \]

Step 2: Calculate the Sample Size

The formula for calculating the sample size \(n\) is given by:

\[ n = \left(\frac{Z \cdot \sigma}{\text{Margin of Error}}\right)^2 \]

Substituting the values:

  • \(Z = 2.3263\)
  • \(\sigma = 335.22\)
  • \(\text{Margin of Error} = 10\)

We have:

\[ n = \left(\frac{2.3263 \cdot 335.22}{10}\right)^2 = (2.3263 \cdot 33.522)^2 = (77.96)^2 = 6081.4783 \]

Rounding up to the nearest integer, we find:

\[ n \approx 6082 \]

Final Answer

The smallest sample size that guarantees a margin of error less than \(10\) is:

\[ \boxed{6082} \]

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