/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 8 Find the tabled value of \(t\lef... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Find the tabled value of \(t\left(t_{a}\right)\) corresponding to a left-tail area a and degrees of freedom given. $$ a=.05, d f=8 $$

Short Answer

Expert verified
Answer: The tabled value of the t-distribution with a left-tail area of 0.05 and 8 degrees of freedom is approximately 1.860.

Step by step solution

01

Identify the row and column in the t-distribution table

To find the tabled value, we need to first locate the row with the correct degrees of freedom and the column with the given left-tail area. Since the degrees of freedom are \(8\), we will look for the row with \(df=8\) and the column with \(a=0.05\).
02

Check the t-distribution table

After identifying the row and column that correspond to \(df=8\) and \(a=0.05\), look at the t-distribution table to find the tabled value. The table can be found in most statistics textbooks or through an online search.
03

Find the tabled value

Now that we know which row and column to look at in the t-distribution table, simply find the intersection of the row for \(df = 8\) and the column for \(a=0.05\). In this case, the tabled value of \(t\left(t_{a}\right)\) is approximately \(1.860\). So, the tabled value of \(t\left(t_{a}\right)\) for the given degrees of freedom \(8\) and left-tail area \(0.05\) is approximately \(1.860\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Degrees of Freedom
When performing statistical tests, the concept of degrees of freedom (df) is crucial in accurately estimating population parameters from sample data. Imagine it as a measure of how much 'freedom' your data has to vary while still estimating the same statistic. In the context of a t-distribution, which is used when the population standard deviation is unknown and the sample size is small, degrees of freedom are typically calculated as the sample size minus one, that is, (df = n - 1).For example, if a sample size is 9, the degrees of freedom will be 8. This value determines which row we will use in the t-distribution table. As the degrees of freedom increase, meaning more data or a larger sample size, the shape of the t-distribution gradually resembles the standard normal distribution. Understanding this concept helps in interpreting the t-distribution table correctly, ensuring that your statistical test results are accurate.
Left-Tail Area
The left-tail area under a t-distribution curve represents the probability that a t-statistic will fall below a certain threshold. It quantifies the area to the left of a specified value on the t-distribution curve. When calculating the left-tail area, statisticians can determine the critical value for hypothesis tests, including the likes of one-tailed t-tests.In the context of our given exercise, the left-tail area of 0.05 corresponds to a 5% probability of falling below the critical value in the t-distribution. This is a common cutoff used for establishing statistical significance, which can also indicate the likelihood of obtaining a test statistic at least as extreme as the one observed, assuming the null hypothesis is true. Selecting the correct left-tail area is pivotal as it directly impacts the critical value, which ultimately influences decision-making in hypothesis testing.
Statistical Significance
The term statistical significance is used to make a judgment about a null hypothesis, typically regarding the relationship or difference between groups in an experiment. A statistically significant result implies that the observed effect in your study is unlikely to be due to chance alone and may reflect a true effect in the population.Statistical significance is often determined by a p-value which is compared against a pre-determined significance level, typically 0.05 or 5%. If the p-value is less than the chosen significance level, the result is deemed statistically significant, and the null hypothesis is rejected. It's important to note that statistical significance does not necessarily mean practical significance or that the findings are of substantial real-world impact; rather, it's a way to convey the confidence on the reliability of the observed effect . Understanding statistical significance aids in interpreting the results of hypothesis tests and in making decisions based on statistical evidence.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Calculate the number of degrees of freedom for \(s^{2}\), the pooled estimator of \(\sigma^{2}\). $$ n_{1}=10, n_{2}=12 $$

Suppose you wish to compare the mean amount of oil required to produce 1 acre of corn versus 1 acre of cauliflower. The readings (in barrels of oil per acre), based on 20 -acre plots, seven for each crop, are shown in the table. $$ \begin{array}{lc} \hline \text { Corn } & \text { Cauliflower } \\ \hline 5.6 & 15.9 \\ 7.1 & 13.4 \\ 4.5 & 17.6 \\ 6.0 & 16.8 \\ 7.9 & 15.8 \\ 4.8 & 16.3 \\ 5.7 & 17.1 \\ \hline \end{array} $$ a. Use these data to find a \(90 \%\) confidence interval for the difference between the mean amounts of oil required to produce these two crops. b. Based on the interval in part a, is there evidence of a difference in the average amount of oil required to produce these two crops? Explain.

Use Table 4 in Appendix I to approximate the \(p\) -value for the \(t\) statistic in the situations given $$\text { A two-tailed test with } t=-1.19 \text { and } 25 d f$$

What assumptions are made about the populations from which random samples are drawn when the \(t\) distribution is used to make small-sample inferences about the difference in population means?

In a study on the effect of an oral rinse on plaque buildup on teeth, 14 people whose teeth were thoroughly cleaned and polished were randomly assigned to two groups of seven subjects each. \(^{7}\) Both groups were assigned to use oral rinses (no brushing) for a 2 -week period. Group 1 used a rinse that contained an antiplaque agent. Group \(2,\) the control group, received a similar rinse except that the rinse contained no antiplaque agent. A measure of plaque buildup was recorded at 14 days with means and standard deviations for the two groups shown in the table. $$ \begin{array}{lcc} \hline & \text { Control Group } & \text { Antiplaque Group } \\ \hline \text { Sample Size } & 7 & 7 \\ \text { Mean } & 1.26 & 0.78 \\ \text { Standard Deviation } & 0.32 & 0.32 \\ \hline \end{array} $$ a. State the null and alternative hypotheses that should be used to test the effectiveness of the antiplaque oral rinse. b. Do the data provide sufficient evidence to indicate that the oral antiplaque rinse is effective? Test using \(\alpha=.05\) c. Find the approximate \(p\) -value for the test.

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.