Question

Calculate the margin of error in estimating a population mean \(\mu\) for the values given in Exercises \(7-10 .\) Comment on how an increased sample size affects the margin of error: \(n=100, s^{2}=4\)

Short answer

Answer: As the sample size increases, the margin of error decreases, resulting in a more precise estimate of the population mean. This happens because the denominator in the margin of error formula \(\frac{s}{\sqrt{n}}\) increases with the sample size, causing the margin of error to become smaller.

Step-by-step solution

Identify the given values

In this exercise, we are given the following values: Sample size, \(n = 100\) Sample variance, \(s^{2} = 4\)

Calculate the standard deviation from the variance

To calculate the standard deviation \(s\), we take the square root of the variance: \(s = \sqrt{s^{2}} = \sqrt{4} =2\)

Determine the degrees of freedom and critical value

The degrees of freedom for this problem are equal to the sample size minus 1: \(df = n - 1 = 100 - 1 = 99\) Now, we need to find the critical value from a t-distribution for a 95% confidence interval. This value is denoted as \(t_{\frac{0.05}{2}}\): For 99 degrees of freedom and a 95% confidence interval, the critical value is approximately equal to 1.984.

Calculate the margin of error

Now that we have the critical value, we can calculate the margin of error using the following formula: Margin of error = \(t_{\frac{0.05}{2}} \times \frac{s}{\sqrt{n}}\) Substitute the values: Margin of error = \(1.984 \times \frac{2}{\sqrt{100}} = 1.984 \times \frac{2}{10} = 1.984 \times 0.2 = 0.3968\)

Comment on the effect of increased sample size on margin of error

To provide a general comment on the effect of increased sample size on the margin of error, let's consider the relationship between the sample size and margin of error: Margin of error = \(t_{\frac{0.05}{2}} \times \frac{s}{\sqrt{n}}\) As \(n\) increases, the term \(\frac{s}{\sqrt{n}}\) will decrease because the denominator is increasing. This implies that the margin of error will decrease as the sample size increases. In other words, as the sample size gets larger, our estimate of the population mean becomes more precise, and the margin of error becomes smaller.

Key concepts

Population Mean Estimation

Estimating the population mean involves using sample data to infer the average value of an entire population. This is particularly useful because often one does not have access to the entire population data due to constraints like time, money, or logistics. To estimate the population mean \( \mu \), we often rely on the sample mean along with some measure of reliability, known as the confidence interval.

In this case, a margin of error is calculated to provide a range within which the true population mean is likely to fall. The margin of error accounts for variations within the sample and helps in constructing a confidence interval around the sample mean that, based on the central limit theorem, is assumed to be normally distributed if the sample size is sufficiently large or the population is normally distributed.

For accurate estimation, one must carefully consider the sample size, the standard deviation, and the level of confidence desired (often 95% or 99%) which impacts the critical value used in calculations. These factors are interconnected, and understanding their role is crucial for valid inferences about the population mean.

Sample Size Effect

The sample size is pivotal in statistical analysis, especially when estimating a population parameter such as the mean. It’s a common mistake to overlook the impact that the sample size can have on the precision of an estimation. As seen in the provided exercise, an increase in the sample size results in a reduced margin of error, enhancing the precision of the population mean estimation.

Why does this happen? The mathematical formula for margin of error includes the expression \( \frac{s}{\sqrt{n}} \) where \( s \) is the sample standard deviation and \( n \) is the sample size. As the sample size increases, the denominator of this fraction, which is the square root of the sample size \( \sqrt{n} \) gets larger, thus making the entire fraction smaller. This smaller fraction, when multiplied by the critical value (which remains constant for a given confidence level), leads to a smaller margin of error.

In practical terms, a larger sample size means that the sample is likely to more closely reflect the true characteristics of the population, assuming the sample is randomly selected and adequately represents the population.

Standard Deviation Calculation

The standard deviation is a measure of the amount of variation or dispersion of a set of values. When it comes to estimating population parameters, like the mean, understanding this variability within a sample is essential. The formula provided in the exercise demonstrates the initial step in calculating the standard deviation from the variance \( (s^2) \) in a sample - by taking the square root of the variance.

The formula \( s = \sqrt{s^2} \) simplifies the calculations, as it converts the variance into a more interpretable metric, which represents the average distance between each data point in the sample and the sample mean. In most statistical tests and confidence intervals, the standard deviation is used rather than the variance due to its direct interpretation in terms of the original units of measurement.

It's important to note that the standard deviation is a crucial element in calculating the margin of error. It directly contributes to the variability component of the equation. Hence, a larger standard deviation indicates greater spread in the data, which can result in a larger margin of error for the population mean estimation.