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Chapter 8: Problem 74

High school students who take the SAT Math exam a second time generally score higher than on their first try. Past data suggest that the score increase has a standard deviation of about 50 points. How large a sample of high school students would be needed to estimate the mean change in SAT score to within 2 points with \(95 \%\) confidence? Show your work.

Short Answer

Expert verified
A sample of 2401 students is required.

Step by step solution

01

Identify Known Values

The standard deviation of the score increase is given as \( \sigma = 50 \). We want to estimate the mean change in SAT score to within 2 points, which is the margin of error \( E = 2 \), with a confidence level of \( 95\% \).
02

Determine Z-Score for Confidence Level

The confidence level of \( 95\% \) corresponds to a Z-score of approximately \( 1.96 \). This Z-score is found using a standard normal distribution table or a calculator.
03

Use Formula for Sample Size

The formula to determine the required sample size \( n \) is: \[ n = \left( \frac{Z \times \sigma}{E} \right)^2 \]Substitute the known values into this formula: \[n = \left( \frac{1.96 \times 50}{2} \right)^2 \]
04

Calculate Sample Size

Calculate the expression:\[n = \left( \frac{98}{2} \right)^2 = \left( 49 \right)^2 = 2401 \]Thus, the required sample size \( n \) is 2401 students.

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Key Concepts

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

Confidence Interval
A confidence interval gives us a range of values, which is likely to contain the population parameter of interest, such as a mean or proportion. This range is associated with a specified confidence level, such as 95%, indicating how sure we are that the true parameter lies within this interval.

In the context of the SAT score problem, we want to estimate the mean change in scores for the entire population of students. The 95% confidence interval tells us that if we were to take many samples and construct an interval from each one, approximately 95% of these intervals would contain the true mean change in scores.

Confidence intervals are crucial in statistics as they help provide insight into the precision and reliability of sample estimates.
Standard Deviation
Standard deviation, denoted as \( \sigma \), measures the amount of variation or dispersion in a set of values. In simpler terms, it tells us how much the values in a data set differ from the average (mean) of the data set. A low standard deviation indicates that the values tend to be close to the mean, while a high standard deviation indicates larger variance.

For the SAT scores, the standard deviation of 50 points implies that most students' score changes fall within 50 points of the average change. Knowing this helps in understanding the spread of score improvements, which is critical for accurately estimating the mean score change with a certain level of precision.
Z-score
A Z-score is a numerical measurement that describes a value's relation to the mean of a group of values, in terms of standard deviations. If a Z-score is 0, it represents a value equal to the mean. A Z-score indicates how many standard deviations an element is from the mean.

For example, a Z-score of 1.96 relates to a confidence level of 95%. This means if we take many samples and calculate the mean of each, roughly 95% of those means will fall within this Z-score above or below the original population mean.

In our SAT score problem, the Z-score of 1.96 is used to assure us that the mean score change estimate is within our chosen confidence interval, which is essential for determining how many students we need to reliably estimate this mean change.
Margin of Error
Margin of error represents the range or amount of random sampling error in a survey's results. It's often expressed alongside a confidence interval, giving us a range in which the true answer likely falls. Margin of error quantifies the uncertainty in estimating a population parameter from a sample.

In the SAT score context, the margin of error is 2 points, meaning we want our estimate of the mean score change to be within 2 points of the actual mean score change for the entire population.

Understanding margin of error is critical because it helps determine the required sample size to achieve a desired confidence level. In this exercise, it directly affects the calculation of the sample size, as seen in the formula applied in the solution.

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Most popular questions from this chapter

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