Question

Suppose you want to estimate one of four parameters- \(\mu, \mu_{1}-\mu_{2}, p,\) or \(p_{1}-p_{2}-\) to within a given bound with a certain amount of confidence. Use the information given to find the appropriate sample size(s). Estimating the difference between two means with a margin of error equal to ±5 . Assume that the sample sizes will be equal and that \(\sigma_{1} \approx \sigma_{2} \approx 24.5\).

Short answer

Answer: The appropriate sample size for the given criteria is 49.

Step-by-step solution

Identify the formula for the margin of error of the difference between two means

The formula for the margin of error \( E \) of the difference between two means, assuming equal variances, is: \[ E = z \sqrt{ \frac{ (\sigma_{1}^2 + \sigma_{2}^2)}{n} } \] Here, \(z\) indicates Z-score that corresponds to the desired level of confidence, and \(n\) is the sample size. Since the sample sizes are equal and the standard deviations are approximately the same, we can simplify the formula to: \[ E = z \sqrt{ \frac{ 2 \sigma^{2}}{n} } \] In our case, \(E = 5\), and \(\sigma \approx 24.5\).

Find the Z-score corresponding to the desired confidence level

The Z-score depends on the desired confidence level, which is not explicitly given in the problem. However, we can discuss common confidence levels and their corresponding Z-scores: - For a 90% confidence level, \(z = 1.645\) - For a 95% confidence level, \(z = 1.96\) - For a 99% confidence level, \(z = 2.576\) We will use the 95% confidence level (with a Z-score of 1.96) as it is the most commonly used one. If you need to work with a different confidence level, substitute the corresponding Z-score in the calculations.

Solve for the sample size \(n\)

Now, we can substitute the values provided in the problem into the simplified margin of error formula: \[ 5 = 1.96 \sqrt{ \frac{ 2 (24.5)^{2}}{n} } \] To solve for the sample size \(n\), first, square both sides of the equation: \[25 = \left(1.96\right)^2 \frac{2 \cdot (24.5)^2}{n} \] Now, isolate the sample size \(n\): \[n = \frac{\left(1.96\right)^2 \cdot 2 \cdot (24.5)^2}{25} \] Finally, calculate the value of \(n\): \[n \approx 48.47\] Since the sample size must be a whole number, round up to the nearest whole number, and thus the appropriate sample size is \(n = 49\).

Key concepts

Margin of Error

Understanding the margin of error is crucial for conducting precise statistical estimates and surveys. It represents the extent of random sampling error in the results. In simpler terms, it’s the radius of the confidence interval for a given statistic that tells us how much we can expect a sample’s estimate to fluctuate if we were to repeat the survey multiple times.

To conceptualize this better, imagine measuring the average height of plants in a large greenhouse. The margin of error tells you how close the measured average height from your sample is likely to be to the true average height of all plants in the greenhouse. It’s typically expressed as a plus-or-minus figure, showing the range within which the true value lies with a specified level of certainty.

To calculate the margin of error for estimating the difference between two means, as in our exercise, we used the formula \( E = z \sqrt{ \frac{ 2 \sigma^{2}}{n} } \), where \( E \) is the margin of error, \( z \) is the Z-score, \( \sigma \) is the standard deviation, and \( n \) is the sample size.

Confidence Level

The confidence level is a measure of certainty regarding how well a sample reflects the population as a whole. Essentially, it’s the percentage of all possible samples that can be expected to include the true population parameter.

For example, if we say we have a 95% confidence level, it means that if we took 100 random samples from the population, we would expect about 95 of those samples to contain the true population mean within the margin of error. The selected confidence level directly affects the width of the confidence interval: a higher confidence level results in a wider interval, reflecting greater uncertainty.

Our exercise doesn't specify the confidence level, so we assumed a 95% confidence level, a common choice in statistical analysis. Adjusting the confidence level affects the Z-score, which is an integral part of the margin of error calculation.

Z-score

A Z-score, also known as a standard score, indicates how many standard deviations an element is from the mean. In the context of confidence intervals, the Z-score correlates to the confidence level and tells us how 'far out' from the mean our tolerance for error extends.

The higher the Z-score we choose, the more confident we can be that the population parameter lies within our calculated margin of error. However, a higher Z-score also means a wider margin of error. Z-scores are derived from the standard normal distribution, and common values for typical confidence levels are 1.645 for 90%, 1.96 for 95%, and 2.576 for 99% confidence levels.

In the given problem, a 95% confidence level corresponds to a Z-score of 1.96. This Z-score, when plugged into the margin of error formula, influences the required sample size.

Standard Deviation

Standard deviation is a measure of the amount of variation or dispersion in a set of values. A low standard deviation indicates that the values tend to be close to the mean, while a high standard deviation indicates that the values are spread out over a wider range.

In practical terms, if you measured the heights of students in a classroom, a small standard deviation would mean most students are about the same height, whereas a large standard deviation would indicate a wide variety of student heights. The symbol \( \sigma \) stands in for standard deviation in formulas.

In our exercise example, the standard deviation of each group was approximately 24.5, and this number was squared within the margin of error formula. This standard deviation is vital because it quantifies the expected variability of the means we are comparing, which, along with the Z-score and the margin of error, helps us determine the necessary sample size for our study.