Question

Using the sample information given in Exercises \(20-21\), give the best point estimate for the population mean \(\mu\) and calculate the margin of error: A random sample of \(n=50\) observations from a quantitative population produced \(\bar{x}=56.4\) and \(s^{2}=2.6\).

Short answer

Answer: The best point estimate for the population mean is 56.4, and the margin of error is approximately 0.46.

Step-by-step solution

Find the sample standard deviation

To find the sample standard deviation, we need to find the square root of the sample variance: $$s = \sqrt{s^{2}}$$ In our case, the sample variance is \(2.6\). Therefore, the sample standard deviation is: $$s = \sqrt{2.6} \approx 1.61$$

Calculate the standard error

The standard error is given by the formula: $$SE=\frac{s}{\sqrt{n}}$$ We are given the sample size \(n=50\) and sample standard deviation \(s\approx 1.61\). So, the standard error is: $$SE = \frac{1.61}{\sqrt{50}} \approx 0.23$$

Find the degrees of freedom

The degrees of freedom for this problem is given by \(df = n-1\). In our case, \(n=50\), so the degrees of freedom is: $$df = 50-1 = 49$$

Find the t-critical value

We will use the t-distribution table to find the t-critical value that corresponds to a 95% confidence level (most commonly used level). For a two-tailed test, we find the critical value that leaves 2.5% of the area in each tail. For \(df = 49\) and \(\alpha=0.025\), the t-critical value (\(t_{\alpha/2,df}\)) is approximately: $$t_{\alpha/2,df} = 2.009$$

Calculate the margin of error

The margin of error is calculated using the formula: $$ME = t_{\alpha/2,df} \cdot SE$$ Where: \(ME\) - Margin of Error \(t_{\alpha/2,df}\) - t-critical value \(SE\) - Standard error In our case, the t-critical value is \(2.009\) and the standard error is \(0.23\). Therefore, the margin of error is: $$ME = 2.009 \cdot 0.23 \approx 0.46$$ The best point estimate for the population mean \(\mu\) is the sample mean \(\bar{x}=56.4\), and the margin of error is approximately \(0.46\).

Key concepts

Point Estimate

A point estimate stands as the single best guess for a population parameter based on sample data. In statistical terms, it is a single value estimate of a parameter. For example, the sample mean \bar{x} is a point estimate of the population mean \(\mu\). In the context of our exercise, the sample mean calculated from the data, \(\bar{x}=56.4\), acts as the point estimate for the population mean \(\mu\).

It is important to note that a point estimate is not perfect and usually has some error associated with it, which leads us to quantify the uncertainty of our estimate with a margin of error and construct a confidence interval around the point estimate for better analysis.

Population Mean

The population mean, denoted as \(\mu\), represents the average value of a characteristic in the entire population. As it's practically challenging to collect data from every individual in the population, a sample mean (\bar{x}), which is the average of data points in a sample, is used to estimate \(\mu\).

In statistical practice, the true population mean is unknown, and hence the sample mean becomes crucial for making inferences about the population. Our exercise involved estimating this unknown \(\mu\) using the sample mean obtained from 50 observations, which turned out to be 56.4.

Standard Error

The standard error (SE) measures the variability of the sampling distribution of a statistic, most commonly the mean. It is essentially the estimated standard deviation of the sample mean and reflects how much the sample mean \(\bar{x}\) varies from sample to sample.

In our exercise, to calculate the SE, we first found the sample standard deviation \(s\) to be about 1.61, then divided it by the square root of the sample size, \(n\). Thus, with \(n=50\) observations, the standard error was calculated to be approximately 0.23. The SE is pivotal in determining the margin of error and hence the precision of the sample mean as an estimate of the population mean.

Degrees of Freedom

Degrees of freedom (df) typically represent the number of values in a calculation that are free to vary. In the context of our sample, it is related to the number of independent pieces of information left after estimating a parameter, like the population mean. The formula for degrees of freedom is \(df = n - 1\), where \(n\) is the sample size.

In the step-by-step solution provided, with our sample size of 50, the degrees of freedom were calculated as \(df = 50 - 1 = 49\). This value is crucial when referring to statistical tables such as the t-distribution table to find critical values for hypothesis testing or constructing confidence intervals around our point estimate.

T-Distribution

The t-distribution, also known as Student’s t-distribution, is a type of probability distribution that is symmetric and bell-shaped, similar to the normal distribution, but has heavier tails. The t-distribution is used instead of the normal distribution when the sample size is small and the population standard deviation is unknown.

It is indexed by the degrees of freedom (df), with the shape of the distribution becoming closer to the normal distribution as the df increases. In our example, the critical value from the t-distribution was used for calculating the margin of error. With 49 degrees of freedom, the t-critical value at a 95% confidence level was 2.009, which is essential to determine the reliability of our interval estimate that surrounds the point estimate of the population mean.