Question

A random sample of \(n=100\) measurements has been selected from a population with unknown mean \(\mu\) and known standard deviation \(\sigma=10 .\) Calculate the width of the confidence interval for \(\mu\) for the confidence levels given. What effect do the changing confidence levels have on the width of the interval? \(99 \%(\alpha=.01)\)

Short answer

Answer: The width of the 99% confidence interval for the mean is 5.152.

Step-by-step solution

Identify the given information

In this problem, we are given: - A random sample of \(n=100\) - A population with unknown mean \(\mu\) - A known standard deviation \(\sigma=10\) - Confidence level of 99% (\(\alpha=0.01\))

Determine the appropriate method to use

Since the sample size is large (\(n\geq 30\)) and the standard deviation is known, we should use a Z-distribution to calculate the width of the confidence interval.

Calculate the standard error of the mean

The standard error of the mean (SEM) is calculated using the following formula: $$SEM = \frac{\sigma}{\sqrt{n}}$$ where \(\sigma\) is the standard deviation and \(n\) is the sample size. Plug in the given values: $$SEM = \frac{10}{\sqrt{100}}$$ $$SEM = \frac{10}{10}$$ $$SEM = 1$$

Determine the critical Z-value

For a 99% confidence interval, we will use a Z-table, which is a table of values showing the area under a standard normal distribution curve, to find the critical Z-value (Z\(\alpha/2\)). Since \(\alpha = 0.01\), we have \(\alpha/2 = 0.005\). The Z-value that corresponds with this area in the tails of the distribution is \(2.576\).

Calculate the margin of error

Multiply the critical Z-value by the standard error of the mean to find the margin of error (ME): $$ME = Z_{\alpha/2} \times SEM$$ $$ME = 2.576 \times 1$$ $$ME = 2.576$$

Find the width of the confidence interval

The width of the confidence interval is twice the margin of error, as it accounts for both the lower and upper bounds of the interval: $$Width = 2 \times ME$$ $$Width = 2 \times 2.576$$ $$Width = 5.152$$ The width of the 99% confidence interval for the mean is 5.152.

Discuss the effect of changing confidence levels on the width of the interval

As the confidence level increases, the width of the confidence interval also increases. This means that if we were to calculate a confidence interval with a higher confidence level (e.g., 99.9%), the width would be larger than the one we calculated for the 99% confidence level. Conversely, if we were to calculate a confidence interval with a lower confidence level (e.g., 95%), the width would be smaller. The reason for this is that increasing the confidence level requires a larger interval to ensure that the true population mean has a higher likelihood of being included within the interval.

Key concepts

Standard Deviation

Understanding standard deviation is essential for analyzing the variation or spread among numbers in a data set. It measures how much the numbers differ from the mean, or average, value. The more spread out the numbers, the higher the standard deviation. For instance, a standard deviation of \(\sigma=10\) suggests that, typically, the data points deviate from the mean by 10 units.

When we discuss sample standard deviation, we take into account only a subset of the entire population. Therefore, in our exercise, a known standard deviation of \(\sigma=10\) aids in calculating other statistics, such as the standard error of the mean, and consequently broadens our understanding of the data's variability linked to the average value.

Standard Error of the Mean

The standard error of the mean (SEM) tells us how precise our sample mean is as an estimate of the true population mean. It's an indication of the uncertainty around the mean. The formula \( SEM = \frac{\sigma}{\sqrt{n}} \) combines the sample size \(n\) and the population standard deviation \(\sigma\) to estimate this uncertainty.

In our exercise, the SEM is calculated as \(\frac{10}{\sqrt{100}}\), resulting in an SEM of 1. This suggests that, based on our sample, our estimate of the population mean is likely to be within 1 unit of the actual population mean. As the sample size increases, the standard error decreases, meaning our estimate becomes more precise.

Z-distribution

The Z-distribution, or standard normal distribution, is a bell-shaped curve that characterizes a dataset and its values relative to the mean, standardized to units of the standard deviation. Critical Z-values, obtained from a Z-table, represent standard deviations away from the mean and correspond to desired confidence levels.

For our 99% confidence level, we find a Z-value of 2.576. This means that the confidence interval for the mean—within which we are 99% sure the true population mean lies—extends 2.576 standard deviations from the sample mean in either direction.

Margin of Error

The margin of error (ME) quantifies the range around the sample mean where the true population mean is expected to fall. It's found by multiplying the Z-value by the SEM. A higher Z-value or a larger SEM increases the margin of error, while a lower Z-value or smaller SEM decreases it.

With a margin of error of 2.576 for our 99% confidence interval, we interpret that there is a 99% chance that the population mean is within \(\pm2.576\) units of our sample mean. This margin reflects our level of certainty regarding where the true mean lies compared to our sample estimate.

Sample Size

The sample size (\(n\)) is a crucial element in determining the reliability and precision of the statistical estimates we make about a population. Larger samples generally lead to more precise estimates of the population parameter, which means lower standard error and tighter confidence intervals.

In our exercise, the use of a sample size of 100 ensures that the standard error is low (\(SEM = 1\)), creating a confidence interval that's relatively precise. However, if we were to use a smaller sample size, the SEM would increase, thereby widening the confidence interval and making our estimate less precise.