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Find \(t_{.975}(5)\)

Short Answer

Expert verified
Using a t-distribution table or calculator, we find that \(t_{.975}(5) \approx 2.571\), which is the critical t-value for a 97.5% confidence level with 5 degrees of freedom.

Step by step solution

01

Understand the t-distribution and degrees of freedom

A t-distribution, also known as the Student's t-distribution, is a probability distribution used to estimate population parameters when the sample size is small and/or the population standard deviation is unknown. Degrees of freedom (df) is a parameter in the t-distribution that determines the shape of the distribution. It is related to the sample size (n) as df = n-1. In this exercise, we have the degrees of freedom (5), and we need to find the t-value for a 97.5% confidence level.
02

Identify the appropriate t-distribution table or calculator

To find the t-value, we need a t-distribution table or calculator. A t-distribution table lists t-values corresponding to various degrees of freedom and confidence levels (usually denoted by the proportion of area to the right of t-value, i.e., the right tail area). These tables are usually found in statistics textbooks or can be found online. Alternatively, you can use a t-distribution calculator (available online) to find the value directly.
03

Find the t-value from the table or calculator

Using a t-distribution table, locate the row corresponding to 5 degrees of freedom and the column corresponding to a right tail area of 0.025 (1-0.975). The intersection of this row and column will give you the t-value, or use an online calculator that requires you to input the degrees of freedom (5) and the desired confidence level (0.975) to obtain the t-value.
04

Interpret the result

The t-value that you find using the table or calculator is the critical value, \(t_{.975}(5)\), for a 97.5% confidence level with 5 degrees of freedom. This t-value can be used for various statistical tests and calculations, such as constructing confidence intervals for the population mean. After using the table or calculator, we find that \(t_{.975}(5) \approx 2.571\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Degrees of Freedom
The concept of 'degrees of freedom' might sound a bit technical, but it's quite straightforward once you break it down. Degrees of freedom (often abbreviated as df) are a measure of how much independent information remains after certain statistics have been calculated. Essentially, it tells you how many values in the final calculation are free to vary.
Generally, degrees of freedom are related to the sample size in statistics, where:
  • Degrees of Freedom = Sample Size - 1
This simple equation helps determine the shape of the t-distribution you're working with. A smaller number of degrees of freedom results in a broader and more variable distribution, while more degrees of freedom make the distribution more similar to the normal distribution.
In the context of hypothesis testing or constructing confidence intervals using a t-distribution, knowing the degrees of freedom is essential. It ensures you're using the correct distribution for your data, particularly when the sample size is small and the population standard deviation is unknown.
Confidence Level
The confidence level is a key concept when you're interpreting statistical data. It's a way to express how certain we are about our estimates or results. A confidence level is often expressed as a percentage, such as 95% or 97.5%, and it represents the proportion of times you expect your interval estimate to include the true population parameter if you were to repeat the study multiple times.
Here's what you need to know about confidence levels:
  • A higher confidence level means a wider interval, but it gives more certainty that the interval contains the true parameter.
  • Conversely, a lower confidence level means a narrower interval, providing less certainty that the interval contains the true parameter.
We often choose confidence levels like 90%, 95%, or 99%, but any level is possible depending on the requirements and context of your analysis. A 97.5% confidence level means you can be 97.5% certain that the true population parameter lies within your calculated confidence interval. This is particularly useful in fields such as scientific research or quality control, where understanding the precision of estimates is crucial.
T-value
The t-value is a specific critical value you obtain from a t-distribution. It's an essential component when calculating confidence intervals or conducting hypothesis tests with a t-distribution. This is especially useful when dealing with small sample sizes or when the population standard deviation is unknown.
Here’s how to understand and use t-values effectively:
  • The t-value helps you determine the cut-off point on the distribution's tail for your chosen confidence level.
  • It varies depending on the degrees of freedom and the specified confidence level you are working with.
When all parameters are defined, a t-distribution table or an online calculator can be used to find the precise t-value. Once found, this critical t-value can be used to form the range within which the true population mean is likely to exist.
In our exercise, we needed to find the t-value for 5 degrees of freedom and a 97.5% confidence level, resulting in a t-value of approximately 2.571. This means that if you were to construct a confidence interval for the population mean, taking these parameters into account, you would be able to determine a reliable range where the true mean is likely to fall, with 97.5% certainty.

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Most popular questions from this chapter

Each of the following sampling procedures is to be classified as producing a random sample or as producing a biased sample. Consider each case and decide whether the procedure is random or biased. a.) The population about which inference is to be made is a population of scores is dart throwing, and the question to be decide is whether men achieve higher scores than women. The investigator selects 10 men and 10 women at random; he then obtain five score from each, making a sample of 100 scores . Is this a random sample? b.) In attacking the same problem, the investigator takes 100 averages of five scores, each average coming from a different randomly selected person. Is this sample of average scores random sample? c.) The population of interest is population of attitude scores at Alpha College; assume that it is an infinite population. The experimenter whishes to obtain a sample of about 100 scores. He administers the attitude test to four existing groups. In this case they are four classes selected at random whose total enrollment is 100 . Is this a random sample? d.) The same investigator interested in attitude scores sends out the attitude test by mail to a sample of 100 students whose names he has selected by taking every seventy-fifth name in the student directory after randomly choosing a starting point. He receives 71 completed tests; the other 29 students fail to respond. Is the original list a random sample? Is the sample of 71 tests random?

The chi-square density function is the special case of a gamma density with parameters \(\alpha=\mathrm{K} / 2\) and \(\lambda=1 / 2\). Find the mean, variance and moment-generating function of a chi-square random variable.

Suppose that light bulbs made by a standard process have an average life of 2000 hours, with a standard deviation of 250 hours, and suppose that it is worthwhile to change the process if the mean life can be increased by at least 10 percent. An engineer wishes to test a proposed new process, and he is willing to assume that the standard deviation of the distribution of lives is about the same as for the standard process. How large a sample should he examine if he wishes the probability to be about \(.01\) that he will fail to adopt the new process if in fact it produces bulbs with a mean life of 2250 hours?

If \(\mathrm{S}^{2}=[1 /(\mathrm{n}-1)]^{\mathrm{n}} \sum_{\mathrm{i}=1}\left(\mathrm{X}_{\mathrm{i}}-\mathrm{X}\right)^{2}\) is the unbiased sample variance of a random sample from a normal distribution with mean \(\mu\) and variance \(\sigma^{2}\), then show \(\mathrm{U}=\left[\left\\{(\mathrm{n}-1) \mathrm{S}^{2}\right\\} / \sigma^{2}\right]\) has a chi-square distribution with \(\mathrm{n}-1\) degrees of freedom.

For a large sample, the distribution of \(\underline{X}\) is always approximately normal. Find the probability that a random sample mean lies within a) one standard error of the mean. b) two standard errors.

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