/*! 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 69 Use a t-distribution to answer t... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Use a t-distribution to answer the question. Assume the sample is a random sample from a distribution that is reason ably normally distributed and we are doing inference for a sample mean. Find endpoints of a t-distribution with 0.025 beyond them in each tail if the sample has size \(n=25\)

Short Answer

Expert verified
The endpoints of the t-distribution with an area of 0.025 in each tail for a sample of size 25 are approximately +2.064 and -2.064.

Step by step solution

01

Understanding the problem

We are tasked with finding the endpoints of a t-distribution which leaves an area of 0.025 in each tail. This is also known as finding the critical t-values. A T-distribution is a type of probability distribution that is symmetric and similar to the standard Normal distribution but has heavier tails, which can come in handy when the sample size is small.
02

Calculate degrees of freedom

The number of degrees of freedom for a sample is typically defined as the number of values in the final calculation of a statistic that are free to vary. In our case, since the sample size \(n=25\), the degrees of freedom will be \(n-1 = 24\). The degrees of freedom are used to adjust the sample variance by the same estimate and it is an important component of many statistical calculations.
03

Find the t-values

The question is to find t-scores when the area beyond them in each tail is 0.025 and the degrees of freedom is 24. These t-scores are often listed on a t-distribution table or can be found using a calculator. Using a t-table or calculator, we find that the t-score is approximately ±2.064.

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.

Understanding Degrees of Freedom
To truly grasp the concept of degrees of freedom, think of it as a way to define how much freedom you have when choosing data points that can vary in your sample set. When you calculate a statistic, like the sample mean, there needs to be a certain number of data points that are free to vary without affecting the overall structure of your data.
For example, in a set of 25 data points (a sample size of 25), each data point could theoretically change while keeping the mean constant, except for the final one. This is because the mean is a fixed number based on all 25 points, so if 24 points are set, the 25th must bring the mean to the value stated. Hence, the degrees of freedom in this case is 24, calculated as \(n - 1\).
Degrees of freedom are crucial in various statistical formulas since they adjust the precision of statistical estimates, especially when assessing sample variance. A higher number of degrees of freedom gives a more reliable estimate of the population parameters.
Deciphering Critical t-values
Critical t-values play an essential role in hypothesis testing, helping you determine the cut-off points where you decide whether to reject or not reject the null hypothesis. These values are specific points on the t-distribution where only a small percentage (like 2.5% in each tail in our problem) of the data lies beyond these points.
When you need to find critical t-values, you'll rely on the t-distribution table, which varies depending on the degrees of freedom. In our case, with 24 degrees of freedom, you're looking for values which account for 0.025 in each tail, typically marked as \(t_{0.025, 24}\).
The critical t-values were found to be approximately \(±2.064\). If your sample's t-score falls beyond these values, it provides evidence to reject your null hypothesis, implying a significant difference or effect. These values give a sense of how extreme your sample results have to be to consider them statistically significant.
Sample Mean Inference Using t-distribution
Conducting inference about a sample mean when the population standard deviation is unknown involves using the t-distribution. This is particularly important when dealing with smaller sample sizes, typically under 30, where normal approximation through the z-distribution isn't suitable.
Sample mean inference steps usually start by calculating the sample mean and the sample standard deviation. Then, the standard error is determined to account for sample size. With these elements, you compute the test statistic, which is a t-score here, showing how many standard errors your sample mean is from the hypothesized population mean.
Inference about the sample mean allows us to make predictions or decisions about the population. By comparing this t-statistic to critical t-values, you can infer whether the observed sample mean significantly deviates from a proposed value under the null hypothesis. Essentially, it helps in understanding the likelihood of the sample results assuming the null hypothesis is true.

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

Find a \(95 \%\) confidence interval for the proportion two ways: using StatKey or other technology and percentiles from a bootstrap distribution, and using the normal distribution and the formula for standard error. Compare the results. Proportion of Reese's Pieces that are orange, using \(\hat{p}=0.48\) with \(n=150\)

In Exercises 6.150 and \(6.151,\) use StatKey or other technology to generate a bootstrap distribution of sample differences in proportions and find the standard error for that distribution. Compare the result to the value obtained using the formula for the standard error of a difference in proportions from this section. Sample A has a count of 90 successes with \(n=120\) and Sample \(\mathrm{B}\) has a count of 180 successes with \(n=300\).

Use a t-distribution and the given matched pair sample results to complete the test of the given hypotheses. Assume the results come from random samples, and if the sample sizes are small, assume the underlying distribution of the differences is relatively normal. Assume that differences are computed using \(d=x_{1}-x_{2}\). Test \(H_{0}: \mu_{1}=\mu_{2}\) vs \(H_{a}: \mu_{1}>\mu_{2}\) using the paired data in the following table: $$ \begin{array}{lllllllllll} \hline \text { Situation } & 1 & 125 & 156 & 132 & 175 & 153 & 148 & 180 & 135 & 168 & 157 \\ \text { Situation } & 2 & 120 & 145 & 142 & 150 & 160 & 148 & 160 & 142 & 162 & 150 \\ \hline \end{array} $$

Gender Bias In a study \(^{52}\) examining gender bias, a nationwide sample of 127 science professors evaluated the application materials of an undergraduate student who had ostensibly applied for a laboratory manager position. All participants received the same materials, which were randomly assigned either the name of a male \(\left(n_{m}=63\right)\) or the name of a female \(\left(n_{f}=64\right) .\) Participants believed that they were giving feedback to the applicant, including what salary could be expected. The average salary recommended for the male applicant was \(\$ 30,238\) with a standard deviation of \(\$ 5152\) while the average salary recommended for the (identical) female applicant was \(\$ 26,508\) with a standard deviation of \(\$ 7348\). Does this provide evidence of a gender bias, in which applicants with male names are given higher recommended salaries than applicants with female names? Show all details of the test.

Survival Status and Heart Rate in the ICU The dataset ICUAdmissions contains information on a sample of 200 patients being admitted to the Intensive Care Unit (ICU) at a hospital. One of the variables is HeartRate and another is Status which indicates whether the patient lived (Status \(=0)\) or died (Status \(=1\) ). Use the computer output to give the details of a test to determine whether mean heart rate is different between patients who lived and died. (You should not do any computations. State the hypotheses based on the output, read the p-value off the output, and state the conclusion in context.) Two-sample \(\mathrm{T}\) for HeartRate \(\begin{array}{lrrrr}\text { Status } & \text { N } & \text { Mean } & \text { StDev } & \text { SE Mean } \\ 0 & 160 & 98.5 & 27.0 & 2.1 \\ 1 & 40 & 100.6 & 26.5 & 4.2 \\ \text { Difference } & = & m u(0) & -\operatorname{mu}(1) & \end{array}\) Estimate for difference: -2.13 \(95 \% \mathrm{Cl}\) for difference: (-11.53,7.28) T-Test of difference \(=0\) (vs not \(=\) ): T-Value \(=-0.45\) P-Value \(=0.653\) DF \(=60\)

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.