Questions: Assume that all grade point averages are to be standardized on a scale between 0 and 5. How many grade point averages must be obtained so that the sample mean is within 0.011 of the population mean? Assume that a 98% confidence level is desired. If using the range rule of thumb, σ can be estimated as (range/4)=(5-0)/4=1.25. Does the sample size seem practical? The required sample size is 69.886? (Round up to the nearest whole number as needed.) Does the sample size seem practical? A. Yes, because the required sample size is a fairly small number. B. No, because the required sample size is a fairly large number C. No, because the required sample size is a fairly small number D. Yes, because the required sample size is a fairly large number.

Solution Steps

To determine the required sample size for a 98% confidence level, we first calculate the Z-score using the formula:

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

Using the range rule of thumb, the standard deviation \( \sigma \) is estimated as:

\[ \sigma = \frac{\text{range}}{4} = \frac{5 - 0}{4} = 1.25 \]

The sample size \( n \) can be calculated using the formula:

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

Substituting the values:

\[ n = \left(\frac{2.3263 \cdot 1.25}{0.011}\right)^2 = (2.3263 \cdot 113.6364)^2 \approx 69885.0004 \]

Rounding up, we find:

\[ n \approx 69886 \]

The calculated sample size of \( 69886 \) is considered fairly large. Therefore, the conclusion regarding the practicality of the sample size is:

B. No, because the required sample size is a fairly large number.

Final Answer