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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 0005 of the Assume that a \( 98 \% \) confidence level is desired. If using the range rule of thumb, \( \sigma \) can be estimated as \( \frac{\text { range }}{4}=\frac{5-0}{4}=1.25 \). Does the sample size seem practical? The required sample size is \( \square \). (Round up to the nearest whole number as needed.)

Ask by Montgomery Floyd. in the United States
Mar 15,2025

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To determine the required sample size for a 98% confidence level with a margin of error of 0.005, using a standard deviation of 1.25, the calculation yields a sample size of approximately 339,307. This sample size is impractical due to its large scale.

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To determine the required sample size for a sample mean of grade-point averages within \(0.0005\) of the true population mean at a \(98\%\) confidence level, we can use the formula for the sample size \(n\): \[ n = \left( \frac{Z \cdot \sigma}{E} \right)^2 \] Where: - \(Z\) is the Z-value corresponding to the desired confidence level (for \(98\%\), \(Z \approx 2.33\)). - \(\sigma\) is the estimated standard deviation (\(1.25\) from the range rule of thumb). - \(E\) is the margin of error (\(0.0005\)). Now, plug in the values and calculate \(n\): \[ n = \left( \frac{2.33 \cdot 1.25}{0.0005} \right)^2 \] \[ = \left( \frac{2.9125}{0.0005} \right)^2 \] \[ = (5825)^2 \] \[ = 33819225 \] Rounding up gives us \(n\), which is \(33819225\). So, the required sample size is \(33819225\). Considering practicality, such a large sample size is quite unrealistic, as it would be cumbersome and resource-intensive to collect and analyze that many grade-point averages. In real-world applications, you typically aim for smaller, feasible samples while balancing confidence and precision.

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