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Sampling Distributions and the t-Distribution Study Guide (page 3)

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Updated on Apr 22, 2013

Example

In a large population of high school students, the number of hours spent studying during any given week is normally distributed with mean 4.5 hours and standard deviation 18.9 hours. Consider randomly selected samples of size n = 100 students.

  1. What is the mean of the sampling distribution of the sample means?
  2. What is the standard deviation of the sampling distribution of the sample means?
  3. Use properties of the normal distribution to fill in the blanks in the following sentence: "For 68% of all randomly selected samples of size n = 100 students, the mean amount of time spent studying during a week will be between___and___hours."

Solution

  1. The mean of the sampling distribution of the sample means is 4.5 hours, the same as the population mean.
  2. The standard deviation of the sampling distribution of the sample means is = 1.89 hours.
  3. Because the population of the numbers of hours the high school students spent studying during the previous week is normally distributed, the distribution of the sample means is also normal. For a normal distribution, 68% of the population lies within one standard deviation of the mean. Thus, 68% of the sample means would lie within 1.89 hours (the standard deviation of the sampling distribution of sample means) of 4.5 hours (the mean of the sampling distribution of sample means). Thus, in repeated samples of n = 100, 68% of the samples will estimate the mean amount of time high school students spent studying during the previous week to be between 2.6 and 6.4 hours.

The t-Distribution

Again, assume that a random sample has been drawn from a normal distribution. The sampling distribution of is normal. If we know the mean and standard deviation of the population, then we can answer many questions about and the values it may assume. In practice, s is almost never known. Most statisticians have never had a real-life problem in which the standard deviation was known! (The same is true for the mean, but we will address this issue later.) Assuming that σ was known, we know that has a standard normal distribution. If we use the sample standard deviation instead of σ then a different standardized random variable, denoted by t, results in the following:

When working with z, only one quantity is varying with each sample, . For t, two quantities are varying, and s. The value of s may not be very close to σ, especially for small values of n. Consequently, the distribution of t tends to be more variable than the distribution of z, especially for small n.

The t-distribution is centered at zero. It has one parameter, called the degrees of freedom, abbreviated as df. The degrees of freedom are usually a function of the sample size n, but the exact relationship between df and n, depends on the type of problem. For this particular application we are considering The degrees of freedom are (n – 1), one less than the sample size. The t-distribution is bell shaped and looks much like the standard normal, but it has thicker tails. The thicker tails reflect the increased variability for a t-distribution compared to the standard normal distribution. As the df increase, the tails become less thick and the distribution becomes more like the normal. The normal distribution and the t-distributions with 4 and 10 degrees of freedom are displayed in Figure 12.3.

Figure 12.3

Like the normal, it is not easy to compute probabilities or to find specific values of t from the t-distribution. We must again rely on tabled values, calculators, or computers. For specified degrees of freedom, the tabulated t* in the body of the table is chosen so that P(t > t*) α . Notice that in the normal table we first used in the previous lesson, we worked with the left-tail probabilities. Here, we have the right-tail probabilities. Also, for the normal distribution, the probabilities were in the body of the table. For the t-distribution, the t* values are in the body of the table (see Table 12.1).

Example

  1. Find t* such that P(t > t*) = 0.05 when t has 8 degrees of freedom.
  2. Find t* such that P(|t| > t*) = 0.05 when t has 12 degrees of freedom.
  3. Find t* such that P(t < t*) = 0.05 when t has 16 degrees of freedom.

Solution

  1. Here, we want the right-tail probability to be 0.05. Right-tail probabilities are presented in the table. To find the proper value, look at the row with 8 df and the column headed by 0.05. The two intersect at t* = 1.86.
  2. We should first recall from algebra that P(|t| > t*) = P(t > t*) + P(t<–t*). Because the t-distribution is symmetric, P(t > t*) = P(t<–t*). Thus, we want to find t* such that P(t> t*) = = 0.025. In the t-table, the row with 12 degrees of freedom intersects with the 0.025 column at t* = 2.179. Thus, P(t > 2.179) = 0.025. By symmetry, we also have P(t<–2.179) = 0.025. This gives us that P(|t| > t*) = 0.05.

Table 12.1 Probability α of exceeding the critical value

  1. This time, we are looking for a left-tail probability. We begin by finding the t* that would provide the same size right-tail probability; that is, find t* such that P(t > t*) = 0.05. At the intersection of the row for 16 degrees of freedom and the 0.05 column, we find 1.746. By symmetry, the left-tail probability would be 0.05 for t* = –1.746. Therefore, here t* = –1.746.
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