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Find the tabled value of \(t\left(t_{a}\right)\) corresponding to a right-tail area a and degrees of freedom given. $$ a=.10, d f=7 $$

Short Answer

Expert verified
Answer: The tabled value of \(t\left(t_{a}\right)\) for a right-tail area of 0.10 and degrees of freedom equal to 7 is approximately 1.415.

Step by step solution

01

Look up the t-distribution table

To find the tabled value, we need to locate a t-distribution table or a t-distribution calculator. A t-distribution table can be found in most statistics textbooks or online. For this problem, let's use a t-distribution table.
02

Identify the degrees of freedom row

Refer to the t-distribution table and find the row that corresponds to the given degrees of freedom. In our case, we're looking for a row with degrees of freedom (\(df\)) equal to 7.
03

Identify the right-tail area column

Now, find the column in the t-distribution table that corresponds to the given right-tail area (\(a = 0.10\)). This value can be found in the top row of the table, often labeled with "One Tail" to indicate the tail areas.
04

Find the intersecting cell

Locate the cell that intersects the row with degrees of freedom equal to 7 and the column with the right-tail area equal to 0.10. The value in this cell is the tabled value of \(t\left(t_{a}\right)\) for the given degrees of freedom and right-tail area.
05

Report the tabled value of t

After identifying the intersecting cell, we find that the tabled value of \(t\left(t_{a}\right)\) for \(df=7\) and \(a=0.10\) is approximately 1.415. This is our answer.

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Key Concepts

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

Degrees of Freedom
In the context of probability and statistics, particularly when dealing with the t-distribution, the term 'degrees of freedom' (often abbreviated as df) plays a crucial role. Fundamentally, the degrees of freedom represent the number of values in the final calculation of a statistic that are free to vary.

Consider a simple analogy: if you know the average of a set of numbers and all but one of the numbers in this set, you can determine the last number. In this scenario, the 'last number' would be the degree of freedom.

In the specific case of the t-distribution, the degrees of freedom are associated with sample size. When you draw a sample from a population and calculate the sample mean, the number of degrees of freedom is usually the sample size minus one (- 1). This is because the calculation of the sample mean fixes one value to set the average, leaving the rest of the data points free to vary.

The concept of degrees of freedom is essential when looking up values in the t-distribution table, as it determines the shape of the t-distribution curve. A higher degree of freedom corresponds to a t-distribution that more closely resembles the normal distribution, with less variability and a narrower spread.
Right-tail Area
In probability and statistics, the 'right-tail area' refers to the probability that a t-distribution random variable will be greater than a certain value. Visually, this is represented by the area under the curve to the right of the specified value, hence the name.

Different from the left-tail area, which looks at values smaller than the specified value, the right-tail area can be crucial in hypothesis testing, particularly in one-tailed tests. In a one-tailed test, you may be interested in testing if a parameter is greater (or sometimes less) than a hypothesized value. The right-tail area can provide the probability of observing a test statistic as extreme or more extreme than the one calculated from your sample data, assuming the null hypothesis is true.

Using the right-tail area is straightforward with the help of a t-distribution table. After determining the degrees of freedom, you locate the column corresponding to the desired right-tail area, which gives you the critical t-value. This t-value can then be used to make decisions about the null hypothesis in a statistical test.
Probability and Statistics
The field of 'probability and statistics' is a vast and fundamental area of mathematics that deals with the study of uncertainty and data analysis. Probability theory is used to predict the likelihood of events based on a model, while statistics help us analyze data and make inferences about populations based on samples.

Key concepts within this field include population characteristics (parameters), sample characteristics (statistics), the normal distribution, and various other probability distributions like the t-distribution. The t-distribution is particularly important when working with small sample sizes or unknown population variances.

Probability and statistics are widely applied across various disciplines such as business, medicine, social sciences, and engineering. They offer the tools needed to make informed decisions based on quantitative data and to test hypotheses using significance tests such as the t-test, where the t-distribution table is frequently used to determine critical values for decision-making.

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Most popular questions from this chapter

A pharmaceutical manufacturer is concerned about the variability of the impurities from shipment to shipment from two different suppliers. To compare the variation in percentage impurities, the manufacturer selects 10 shipments from each of the two suppliers and measures the percentage of impurities in the raw material for each shipment. The sample means and variances are shown in the table. $$\begin{array}{ll}\hline \text { Supplier A } & \text { Supplier B } \\\\\hline \bar{x}_{1}=1.89 & \bar{x}_{2}=1.85 \\\s_{1}^{2}=.273 & s_{2}^{2}=.094 \\\n_{1}=10 & n_{2}=10\end{array}$$ a. Do the data provide sufficient evidence to indicate a difference in the variability of the shipment impurity levels for the two suppliers? Test using \(\alpha=.01 .\) Based on the results of your test, what recommendation would you make to the pharmaceutical manufacturer? b. Find a \(99 \%\) confidence interval for \(\sigma_{2}^{2}\) and interpret your results.

In an experiment to study an oral rinse designed to prevent plaque buildup, subjects were divided into two groups: One group used a rinse with an antiplaque ingredient, and the control group used a rinse containing inactive ingredients. Suppose that the plaque growth on each person's teeth was measured after using the rinse after 4 hours and then again after 8 hours. If you wish to estimate the difference in plaque growth from 4 to 8 hours, should you use a confidence interval based on a paired or an unpaired analysis? Explain.

Find the critical value(s) of t that specify the rejection region for the situations $$\text { A left-tailed test with } \alpha=.01 \text { and } 7 d f$$

Why use paired observations to estimate the difference between two population means rather than estimation based on independent random samples selected from the two populations? Is a paired experiment always preferable? Explain.

Cerebral blood flow (CBF) in the brains of healthy people is normally distributed with a mean of 74 . A random sample of 25 stroke patients resulted in an average \(\mathrm{CBF}\) of 69.7 with a standard deviation of \(16.0 .\) a. Test the hypothesis that the standard deviation of CBF measurements for stroke patients is greater than \(\sigma=10\) at the \(\alpha=.05\) level of significance. b. Find bounds on the \(p\) -value for the test. Does this support your conclusion in part a? c. Find a \(95 \%\) confidence interval for \(\sigma^{2}\). Does it include the value \(\sigma^{2}=100 ?\) Does this validate your conclusion in part a?

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