Question

Let \(x\) denote the time (in minutes) that it takes a fifth-grade student to read a certain passage. Suppose that the mean value and standard deviation of \(x\) are \(\mu=2 \mathrm{~min}\) and \(\sigma=0.8\) min, respectively. a. If \(\bar{x}\) is the sample average time for a random sample of \(n=9\) students, where is the \(\bar{x}\) distribution centered, and how much does it spread out about the center (as described by its standard deviation)? b. Repeat Part (a) for a sample of size of \(n=20\) and again for a sample of size \(n=100\). How do the centers and spreads of the three \(\bar{x}\) distributions compare to one another? Which sample size would be most likely to result in an \(\bar{x}\) value close to \(\mu\), and why?

Short answer

The \(\bar{x}\) distribution is centered at 2 minutes for all sample sizes. As the sample size increased from 9 to 20 to 100, the standard deviation of the distribution decreased from 0.27 to 0.18 to 0.08, respectively. Therefore, a sample size of 100 would be most likely to result in a \(\bar{x}\) value close to the mean, \(\mu\), because of the smaller standard deviation or spread.

Step-by-step solution

Calculate Mean and Standard Deviation for n=9

Using the provided formulae, we get \(\mu_{\bar{x}} = \mu = 2\) minutes and \(\sigma_{\bar{x}} = \sigma / \sqrt{n} = 0.8 / \sqrt{9} = 0.27\) minutes.

Calculate Mean and Standard Deviation for n=20

Similarly, for n=20, the mean remains at \(\mu_{\bar{x}} = 2\) minutes, while the standard deviation becomes \(\sigma_{\bar{x}} = 0.8 / \sqrt{20} = 0.18\) minutes.

Calculate Mean and Standard Deviation for n=100

Finally, for n=100, the mean is still \(\mu_{\bar{x}} = 2\) minutes, and the standard deviation is now \(\sigma_{\bar{x}} = 0.8 / \sqrt{100} = 0.08\) minutes.

Comparing the Results

In all three cases, the mean of the sample means (\(\mu_{\bar{x}}\)) is the same as the population mean (\(\mu\)). As the sample size increased, the standard deviation of the sample mean (\(\sigma_{\bar{x}}\)) decreased, indicating that the sample means get closer to the population mean, i.e., they are more reliable in estimating the population mean. Therefore, the larger the sample size, the more likely the sample mean is to be close to the population mean. This suggests that the sample of size 100 would give an \(\bar{x}\) value closest to \(\mu\).

Key concepts

Central Limit Theorem

The Central Limit Theorem (CLT) is a fundamental principle in the study of statistics that applies to independent random variables with known expected value (mean) and variance. It states that when a large enough sample size is taken from a population with any distribution, the sample mean distribution approaches a normal distribution. This normal distribution will be centered around the population mean, \( \mu \), with a standard deviation equal to the population standard deviation divided by the square root of the sample size \( \sigma / \sqrt{n} \).

For the fifth-grade students' reading times, regardless of the original distribution of individual reading times, the distribution of the sample mean \( \bar{x} \) will approach normality as the sample size increases. Hence, even if the reading times are not normally distributed, the average reading times from multiple samples will display a normal distribution pattern, thanks to the CLT. This theorem underlies many statistical methods, including hypothesis testing and creating confidence intervals.

Standard Deviation of Sample Mean

The standard deviation of the sample mean, often denoted as \( \sigma_{\bar{x}} \), quantifies how much the sample mean would vary if you took multiple samples from the same population. The formula for calculating the standard deviation of the sample mean is \( \sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} \), where \( \sigma \) is the population standard deviation and \( n \) is the sample size.

As calculated in the exercise, for a sample size of 9 students, the standard deviation of the sample mean is 0.27 minutes, suggesting that if you took many samples of 9 students, their average reading times would typically vary by 0.27 minutes from the sample to sample. This measure helps us understand the reliability of the sample mean in estimating the population mean; a smaller \( \sigma_{\bar{x}} \) implies a more precise estimate.

Sample Size Effect

The effect of sample size on the distribution of the sample mean is quite significant. As seen in the exercise, increasing the sample size results in a decrease in the standard deviation of the sample mean. This effect can be observed through the given formula \( \sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} \).

In the context of the fifth-grade students, a larger sample size, such as 100 students, results in a standard deviation of 0.08 minutes, much smaller than that for a sample size of 9 or 20. This is critical because it implies that with larger samples, the probability of the sample mean lying close to the population mean increases. Hence, large samples provide more accurate estimates of population parameters, reducing the influence of random sampling errors.

Population Mean Estimation

Estimating the population mean is a standard objective in statistical analysis. The goal is to provide the best possible estimate of the population mean \( \mu \) using sample data. As suggested by the Central Limit Theorem, the distribution of the sample means can be used to estimate the population mean. With a large enough sample size, the sample mean \( \bar{x} \) is likely to be very close to the population mean due to the reduced standard deviation of the sample mean.

As demonstrated with the reading times exercise, the sample mean remains constant at 2 minutes, equal to the population mean, while the standard deviation shrinks with increasing sample size. A larger sample size, such as 100, will give the most precise estimation of the population mean. It highlights that for accurate population mean estimation, one should aim for a larger sample size to minimize the error.