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 \(\mu\) to within 1.6 with probability .95. Prior experience suggests that \(\sigma=12.7\).

Short answer

Answer: 241

Step-by-step solution

Understand the Margin of Error Formula

The margin of error formula used for estimation of the population mean with a known population standard deviation is given by: Margin of Error = \(Z_{\frac{\alpha}{2}} * \frac{\sigma}{\sqrt{n}}\) Where \(Z_{\frac{\alpha}{2}}\) is the z-score corresponding to the desired confidence level, \(\sigma\) is the population standard deviation, and n is the sample size.

Find the Z-score for the Confidence Level

For a confidence level of 0.95, we need to find the z-score corresponding to a 95% confidence interval which leaves 2.5% in the tails on both sides. That is 0.975 in the standard normal distribution. Using a standard normal table: \(Z_{\frac{\alpha}{2}}=Z_{0.975}=1.96\)

Determine the Margin of Error

In this exercise, we need to estimate the population mean to within 1.6. Therefore, the margin of error is 1.6. Margin of Error = 1.6

Plug the Values into the Formula and Solve for n

Now, plug in the values for the margin of error, z-score, and population standard deviation into the margin of error formula: 1.6 = 1.96 * \(\frac{12.7}{\sqrt{n}}\)

Solve for Sample Size (n)

Now, we need to solve the equation for the sample size (n): 1. Divide both sides by 1.96: \(\frac{1.6}{1.96}=\frac{12.7}{\sqrt{n}}\) 2. Square both sides: \((\frac{1.6}{1.96})^{2}=(\frac{12.7}{\sqrt{n}})^{2}\) 3. Rearrange the equation to get n: \(n=(\frac{12.7}{\frac{1.6}{1.96}})^{2}\) 4. Calculate n: \(n=(\frac{12.7}{0.8163})^{2}\) \(n=240.829\) Since we can't have a fraction of a sample, round up to the nearest whole number as we need at least that many samples to achieve the desired margin of error. \(n = 241\)

Conclusion

The appropriate sample size needed to estimate the population mean (\(\mu\)) to within 1.6 with a 95% confidence level is 241.

Key concepts

Margin of Error

Understanding the margin of error is crucial for anyone conducting a statistical estimation. The margin of error represents a buffer around a population mean estimate, signifying the range within which we can expect our estimate to fall with a certain level of confidence. It essentially tells us how 'close' we are to the true mean of the population. For instance, if we say the margin of error is 1.6, we imply that the true population mean is likely within 1.6 units of our sample mean.

The formula for margin of error, when the population standard deviation is known, looks like this:
Margin of Error = \(Z_{\frac{\alpha}{2}}\times\frac{\sigma}{\sqrt{n}}\).

In this formula, \(Z_{\frac{\alpha}{2}}\) is the z-score that aligns with the desired confidence level, \(\sigma\) is known as the population standard deviation, and \(n\) represents the number of samples. Adjusting this margin helps determine an adequate sample size in research, ensuring accurate and reliable estimations.

Confidence Interval

A confidence interval is a range of values, derived from the sample data, that likely contains the value of an unknown population parameter. When we express a 95% confidence interval for a population mean, we assert that if we were to take 100 different samples and compute 100 confidence intervals, we would expect about 95 of them to contain the true population mean.

This concept is intrinsically linked to the margin of error, as a narrower margin represents a more precise estimate, subsequently leading to a narrower confidence interval. When we interpret the confidence level in relation to the z-score, which was 1.96 for our 95% confidence level, we can understand that this z-score corresponds to the upper limit where 95% of the area under the curve lies within. Therefore, this value is critical for constructing our confidence interval.

Population Mean Estimation

Estimating the population mean is a fundamental aspect of statistical analysis. The population mean, denoted by \(\mu\), is the average of a set of values representing the entire population. Since it's not typically feasible to survey an entire population, we estimate \(\mu\) by taking a sample and calculating the sample mean. This estimate is not perfect and will have some degree of error, which is where the margin of error and confidence interval come into play.

To ensure our estimation is reasonable, we determine the sample size needed to get a margin of error that reflects the level of precision we aim to achieve. The smaller the margin of error, the more closely our sample mean will approximate the population mean—at the cost of requiring a larger sample size for this increased precision.

Z-Score

The z-score is a statistical metric that tells us how many standard deviations an element is from the mean. In the realm of sample size determination, the z-score serves a specific role. It translates confidence levels into a quantifiable number that we can utilize in our margin of error calculation.

When we wanted to find a sample size with a 95% confidence level in our exercise, we used a z-score of 1.96, which reflects the standard deviation cut-off for a 95% confidence level in a standard normal distribution. This score is essential in determining how large a sample is required for a specific level of confidence, as higher confidence levels will generally increase the z-score and, subsequently, the required sample size.

Population Standard Deviation

The population standard deviation, symbolized by \(\sigma\), plays a pivotal role in determining the variability or spread of a population's values. In terms of sample size calculation, the exact value of \(\sigma\) is directly proportional to the required sample size. The higher the population standard deviation, the more spread out the population data is, which means that we'll need a larger sample size to accurately estimate the population mean within a given margin of error.

Knowing the population standard deviation allows for more precise calculations when determining the necessary sample size, ensuring that the sample adequately represents the population. In the problem exercise, the population standard deviation was given as 12.7, which is a crucial component in calculating the margin of error and, ultimately, the proper sample size for the estimation.