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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 \(1 \%\) beyond them in each tail if the sample has size \(n=18 .\)

Short Answer

Expert verified
To find the endpoints of a t-distribution with 1% beyond them in each tail for a sample size of 18, first work out the degrees of freedom to be 17. Then look up the t-values \(t_{0.01}\) and \(t_{0.99}\) using a t-distribution table or a calculator set to a degree of freedom of 17. These t-values are the endpoints of the t-distribution.

Step by step solution

01

Finding degrees of freedom

Degrees of freedom, often denoted by \(df\), is calculated as the sample size \(n\) minus 1. In this case, the sample size is 18, therefore the degrees of freedom is \(df = n - 1 = 18 - 1 = 17.\)
02

Finding the t-values

Since we want 1% beyond the t-values in each tail, we are looking for t-values that leave a total area of 0.02 (0.01 in each tail) between them, leaving 0.98 in between. These t-values are usually found using a table for the t-distribution or by using a calculator. The t-values are denoted as \(t_{0.01}\) and \(t_{0.99}\) to reflect the cumulative areas from negative infinity.
03

Look up t-values

Using a t distribution table or a calculator set with a degree of freedom of 17, find \(t_{0.01}\) and \(t_{0.99}\). These are the t-values that leave a cumulative area of 0.01 and 0.99 from negative infinity respectively. These are the endpoints of the t-distribution.

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Key Concepts

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

Degrees of Freedom
When it comes to statistical analyses, the concept of 'degrees of freedom' is a crucial one. It refers to the number of independent values or quantities which can be assigned to a statistical distribution. In most cases, the degrees of freedom in a statistical calculation are equal to the number of values that are free to vary. Imagine having a set of numbers with a fixed mean; if you already know all but one of those numbers, the last one can be determined to maintain the mean. Therefore, it only has one degree of freedom. Similarly, when working with a sample of size n, the degrees of freedom for calculations related to the sample mean is usually n-1, because the total of the sample values is constrained by the sample mean.
To make it relatable, consider a real-life scenario where a teacher has ten apples to distribute among nine students. She can give out apples to eight students in any manner she wishes, but the ninth student's number of apples is no longer a free choice—it’s determined by the remaining apples. This simple example mirrors the rationale behind degrees of freedom in statistics, which explains why, as in our exercise, the sample size of 18 leads to 17 degrees of freedom (\(df = n - 1 = 18 - 1 = 17\)).
Inference for Sample Mean
The process of making decisions or predictions about a population based on sample data is known as statistical inference. When it comes to the inference for the sample mean, this involves using sample data to estimate the population mean. Since we don't have access to the whole population, we rely on our sample to provide insight. However, we must account for variability in our estimate; this is where the t-distribution comes into play when the sample size is small or the population standard deviation is unknown.
By using the t-distribution, we can determine confidence intervals or conduct hypothesis tests regarding the population mean. In our exercise, the t-distribution is employed to find the endpoints of the confidence interval. This tells us within what range we can expect to find the population mean, based on our sample, with a certain level of confidence. A key factor in such inference is the aforementioned degrees of freedom, which affects the shape of the t-distribution and the accuracy of our estimates.
So, in essence, inference for the sample mean is like piecing together a puzzle with some missing pieces. We use the pieces we do have (our sample) to estimate what the full picture (population mean) is likely to be. In our problem, we're determining how far out the puzzle's edges go, with \(1\%\) uncertainty on each end.
Normal Distribution
The normal distribution, often called the bell curve due to its shape, is a common probability distribution that's symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean. In fact, the farther away from the mean, the lower the frequency of occurrence. Notably, in a normal distribution, the mean, median, and mode are all equal.
A standard normal distribution is a normal distribution with a mean of zero and a standard deviation of one. It is a foundational concept in statistics and underpins many statistical techniques, including the t-distribution. When the sample size is sufficiently large, usually more than 30, the central limit theorem tells us that the sampling distribution of the sample mean will be approximately normally distributed, even if the underlying population distribution is not.
Returning to our exercise, we assume the sample is from a distribution that is 'reasonably normally distributed’, which means it's similar enough to a normal distribution for purposes of the statistical methods being used. When we have smaller sample sizes or an unknown population standard deviation, the normal distribution isn't a proper fit anymore. This is where the t-distribution steps in, aptly adjusting for the increased uncertainty and providing a more accurate tool for our inference about the sample mean.

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Most popular questions from this chapter

In Exercises 6.188 to 6.191 , use the t-distribution to find a confidence interval for a difference in means \(\mu_{1}-\mu_{2}\) given the relevant sample results. Give the best estimate for \(\mu_{1}-\mu_{2},\) the margin of error, and the confidence interval. Assume the results come from random samples from populations that are approximately normally distributed. A \(95 \%\) confidence interval for \(\mu_{1}-\mu_{2}\) using the sample results \(\bar{x}_{1}=75.2, s_{1}=10.7, n_{1}=30\) and \(\bar{x}_{2}=69.0, s_{2}=8.3, n_{2}=20 .\)

Survival in the ICU and Infection In the dataset ICUAdmissions, the variable Status indicates whether the ICU (Intensive Care Unit) patient lived (0) or died \((1),\) while the variable Infection indicates whether the patient had an infection ( 1 for yes, 0 for no) at the time of admission to the ICU. Use technology to find a \(95 \%\) confidence interval for the difference in the proportion who die between those with an infection and those without.

Close Confidants and Social Networking Sites Exercise 6.93 introduces a study \(^{48}\) in which 2006 randomly selected US adults (age 18 or older) were asked to give the number of people in the last six months "with whom you discussed matters that are important to you." The average number of close confidants for the full sample was \(2.2 .\) In addition, the study asked participants whether or not they had a profile on a social networking site. For the 947 participants using a social networking site, the average number of close confidants was 2.5 with a standard deviation of 1.4 , and for the other 1059 participants who do not use a social networking site, the average was 1.9 with a standard deviation of \(1.3 .\) Find and interpret a \(90 \%\) confidence interval for the difference in means between the two groups.

Restaurant Bill by Gender In the study described in Exercise 6.224 the diners were also chosen so that half the people at each table were female and half were male. Thus we can also test for a difference in mean meal cost between females \(\left(n_{f}=24, \bar{x}_{f}=44.46, s_{f}=15.48\right)\) and males \(\left(n_{m}=24, \bar{x}_{m}=43.75, s_{m}=14.81\right) .\) Show all details for doing this test.

Use StatKey or other technology to generate a bootstrap distribution of sample proportions and find the standard error for that distribution. Compare the result to the standard error given by the Central Limit Theorem, using the sample proportion as an estimate of the population proportion \(p\). Proportion of home team wins in soccer, with \(n=120\) and \(\hat{p}=0.583\)

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