Questions: Assume that all grade-point averages are to be standardized on a scale b of the population mean? Assume that a 95% confidence level is desired. seem practical? The required sample size is □ (Round up to the nearest whole number as needed.)

Solution Steps

We need to determine the required sample size \( n \) for estimating a population mean with a 95% confidence level. The formula for the sample size is given by:

\[ n = \left( \frac{Z \cdot \sigma}{E} \right)^2 \]

where:

• \( Z \) is the Z-score corresponding to the desired confidence level,
• \( \sigma \) is the population standard deviation,
• \( E \) is the margin of error.

For this calculation, we assume the following values:

• \( Z = 1.96 \) (for a 95% confidence level),
• \( \sigma = 1.0 \) (assumed population standard deviation),
• \( E = 0.1 \) (assumed margin of error).

Substituting the values into the formula:

\[ n = \left( \frac{1.96 \cdot 1.0}{0.1} \right)^2 \]

Calculating the expression inside the parentheses:

\[ \frac{1.96 \cdot 1.0}{0.1} = 19.6 \]

Now squaring this value:

\[ n = (19.6)^2 = 384.16 \]

Since the sample size must be a whole number, we round up \( 384.16 \) to the nearest whole number:

\[ n = 385 \]

Final Answer