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Find the sample size required to estimate the population mean. Assume that all grade-point averages are to be standardized on a scale between 0 and \(4 .\) How many grade-point averages must be obtained so that the sample mean is within 0.01 of the population mean? Assume that a \(95 \%\) confidence level is desired. If we use the range rule of thumb, we can estimate \(\sigma\) to be range/4 \(=(4-0) / 4=1 .\) Does the sample size seem practical?

Short Answer

Expert verified
38416 grade-point averages are required for the desired accuracy at 95% confidence level.

Step by step solution

01

- Identify Variables

Note the given information: - Confidence level: 95%, which gives a Z-value of 1.96 - Standard deviation, \(\sigma\), estimated as 1 (from the range rule of thumb) - Margin of error, \(\text{E}\text{, is 0.01}\)
02

- Use the Sample Size Formula

The formula for calculating sample size is: \[n = \left(\frac{Z \sigma}{E}\right)^2\]Substitute the values: - \(\text{Z} = 1.96\) - \(\sigma = 1\) - \(\text{E} = 0.01\)
03

- Plug in the Values

Substitute and compute: \[n = \left(\frac{1.96 \times 1}{0.01}\right)^2\] \[n = \left(196\right)^2\] \[n = 38416\text{..} \]
04

- Interpret the Result

The sample size required is 38,416. To achieve a mean within 0.01 of the population mean with a 95% confidence level, a very large sample size is needed.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Population Mean
The population mean is the average value of a set of measurements or values of the entire population. In our problem, it represents the average grade-point average (GPA) of all students. It is denoted by the symbol \(\mu\).

Understanding the population mean is critical when drawing conclusions from sample data. It gives us a measure to compare individual scores and sample statistics. In real-world applications, we often don't have access to the entire population, so we estimate this mean using a sample mean \(\bar{x}\).
Confidence Level
The confidence level indicates the probability that the value of a parameter falls within a specified range of values. In our example, a 95% confidence level means we are 95% certain that the sample mean will be within the margin of error from the population mean.

Confidence levels are represented typically as percentages (like 90%, 95%, and 99%). They are linked to Z-values in the context of the normal distribution. For a 95% confidence level, the corresponding Z-value is approximately 1.96. This Z-value is what we use in our sample size formula to ensure that our estimate has the desired confidence.
Margin of Error
The margin of error is the maximum amount by which the sample mean is expected to differ from the actual population mean. In our exercise, we desired a margin of error of 0.01 for the GPA. This means we want our sample mean to be within 0.01 points of the actual population mean.

A smaller margin of error requires a larger sample size because we want to be more precise in our estimation. The margin of error is represented by \(E\) and directly influences the sample size calculation:

= \left( \frac{Z \sigma}{E} \right) ^2
Standard Deviation
Standard deviation, denoted by \(\sigma\), measures the amount of variability or dispersion of a set of values. In our case, it represents the standard deviation of GPAs. A higher standard deviation means the values are more spread out.

Using the range rule of thumb, we can estimate the standard deviation as the range divided by four. For GPAs that range from 0 to 4, the standard deviation is: \[ \sigma = \frac{4-0}{4} = 1 \].

In our sample size calculation, the standard deviation helps determine how spread out our data are and affects how large our sample size needs to be to achieve the desired margin of error and confidence level.

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Most popular questions from this chapter

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