/*! 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 234 Use a t-distribution to find a c... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Use a t-distribution to find a confidence interval for the difference in means \(\mu_{1}-\mu_{2}\) using the relevant sample results from paired data. 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, and that differences are computed using \(d=x_{1}-x_{2}\) A \(99 \%\) confidence interval for \(\mu_{1}-\mu_{2}\) using the paired data in the following table:. $$ \begin{array}{lccccc} \hline \text { Case } & \mathbf{1} & \mathbf{2} & \mathbf{3} & \mathbf{4} & \mathbf{5} \\ \hline \text { Treatment 1 } & 22 & 28 & 31 & 25 & 28 \\ \text { Treatment 2 } & 18 & 30 & 25 & 21 & 21 \\ \hline \end{array} $$

Short Answer

Expert verified
The 99% confidence interval for \(\mu_{1}-\mu_{2}\) is given by \(\bar{d} \pm t \cdot \frac{s_{d}}{\sqrt{n}}\), where \(\bar{d}\) is the sample mean of the differences, \(t\) is the critical value from the t-distribution for a 99% confidence level, \(s_{d}\) is the sample standard deviation of the differences, and \(n\) is the number of pairs. The actual values depend on the computed mean, standard deviation, and the t-value from the t-distribution.

Step by step solution

01

Calculate Differences

Begin by calculating the difference for each pair in the data. That would mean subtracting each value of Treatment 2 from Treatment 1 to get the 'd' values.
02

Find Sample Mean and Standard Deviation

Once all the differences are calculated, compute the sample mean \(\bar{d}\) and the sample standard deviation \(s_{d}\). This will involve the formulas: \[\bar{d} = \frac{\sum{}d}{n}\] for the mean, where \(n\) is the number of pairs and \(\sum{}d\) is the sum of all 'd' values; and \[s_{d} = \sqrt{\frac{\sum{}(d - \bar{d})^2}{n-1}}\] for the standard deviation.
03

Identify Degrees of Freedom and t-value

The next step is to identify the degrees of freedom and the critical value from the t-distribution. The degrees of freedom are given by \(df = n - 1\). The t-value corresponds to the given 99% confidence level and can be retrieved from a t-distribution table or using statistical software.
04

Calculate Confidence Interval

Finally, using the t-value, calculate the confidence interval for the true mean difference using the formula: \[CI = \bar{d} \pm t \cdot \frac{s_{d}}{\sqrt{n}}\] where the term \(t \cdot \frac{s_{d}}{\sqrt{n}}\) refers to the margin of error.

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.

Confidence Interval
A confidence interval is an essential concept in statistics that estimates an unknown parameter, like the difference between two means, with a certain degree of confidence. In this context, we're looking at a 99% confidence interval.

This implies that if we were to take numerous random samples and compute a confidence interval for each one, about 99% of these intervals would actually contain the true mean difference. To achieve this, we use a t-distribution because our data involves paired differences and typically comes from a small sample size.
  • Start by calculating individual paired differences, labeled as 'd'.
  • Compute the sample mean of these differences, denoted as \(\bar{d}\).
  • Find the margin of error, which uses the standard deviation of 'd', the size of the sample, and a critical t-value.
The formula to compute the confidence interval is:\[CI = \bar{d} \pm t \cdot \frac{s_{d}}{\sqrt{n}}\]This uses:
  • \(\bar{d}\): the mean of the differences
  • \(s_{d}\): the sample standard deviation of the differences
  • \(n\): the number of paired observations
  • \(t\): the critical value from the t-distribution
A well-calculated confidence interval will offer a range, predicting where the actual difference in means \( \mu_1 - \mu_2 \) lies, while considering inherent sample variability and randomness.
Paired Data
Paired data is used when you have two sets of related observations, which often occurs in before-and-after studies or situations where each data point in one set has a direct correspondence in the other. In our example, we have two treatments applied on the same subjects, and the differences are calculated within these pairs.

Why use paired data instead of independent samples?
  • It controls for variability between subjects, focusing analysis on 'within-subject' differences.
  • It typically reduces the variance, leading to a more precise estimate of the mean difference (compared to analyzing two independent groups).
Calculating the 'd' values, or differences, involves subtracting the second treatment observation from the first for each subject, offering a direct look at how one specific condition might affect outcomes. This approach is especially useful when looking to isolate the effect of the treatment by minimizing external variability.

After obtaining these paired differences, we treat them as a single sample to perform further statistical analysis. This method allows us to directly assess the impact of changing treatments, assuming any remaining variability is narrower than it would be if we simply compared average outcomes from two independent groups.
Degrees of Freedom
Degrees of freedom (df) is an important statistical concept that refers to the number of values in the final calculation of a statistic that are free to vary. In the context of calculating confidence intervals for paired data, the degrees of freedom influence the shape of the t-distribution used.

For paired data, the degrees of freedom are given by \(df = n - 1\), where \(n\) is the number of pairs. It helps us determine the t-value needed to calculate the confidence interval.
  • As the sample size increases, degrees of freedom increases, making the t-distribution resemble a normal distribution more closely.
  • The degrees of freedom factor into how 'stretched' the t-distribution looks; fewer degrees of freedom typically result in wider confidence intervals reflecting greater uncertainty.
When referencing a t-table or using statistical software, the degrees of freedom help locate the right t-value that corresponds with our desired confidence level (e.g., 99%).

Understanding degrees of freedom enhances the clarity of statistical procedures, as they play a key role in quantifying the uncertainty in our estimate interval, leading to more accurate interpretations of data.

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

Home Field Advantage in Baseball 2009 There were 2430 Major League Baseball (MLB) games played in \(2009,\) and the home team won the game in \(54.9 \%\) of the games. \({ }^{15}\) If we consider the games played in 2009 as a sample of all MLB games, test to see if there is evidence, at the \(1 \%\) level, that the home team wins more than half the games. Show all details of the test.

In a survey of 2255 randomly selected US adults (age 18 or older), 1787 of them use the Internet regularly. Of the Internet users, 1054 use a social networking site. \({ }^{7}\) Find and interpret a \(95 \%\) confidence interval for each of the following proportions: (a) Proportion of US adults who use the Internet regularly. (b) Proportion of US adult Internet users who use a social networking site. (c) Proportion of all US adults who use a social networking site. Use the confidence interval to estimate whether it is plausible that \(50 \%\) of all US adults use a social networking site.

Statistical Inference in Babies Is statistical inference intuitive to babies? In other words, are babies able to generalize from sample to population? In this study, \(1 \quad 8\) -month-old infants watched someone draw a sample of five balls from an opaque box. Each sample consisted of four balls of one color (red or white) and one ball of the other color. After observing the sample, the side of the box was lifted so the infants could see all of the balls inside (the population). Some boxes had an "expected" population, with balls in the same color proportions as the sample, while other boxes had an "unexpected" population, with balls in the opposite color proportion from the sample. Babies looked at the unexpected populations for an average of 9.9 seconds \((\mathrm{sd}=4.5\) seconds) and the expected populations for an average of 7.5 seconds \((\mathrm{sd}=4.2\) seconds). The sample size in each group was \(20,\) and you may assume the data in each group are reasonably normally distributed. Is this convincing evidence that babies look longer at the unexpected population, suggesting that they make inferences about the population from the sample? (a) State the null and alternative hypotheses. (b) Calculate the relevant sample statistic. (c) Calculate the t-statistic.

Is B a Good Choice on a Multiple-Choice Exam? Multiple-choice questions on Advanced Placement exams have five options: \(\mathrm{A}, \mathrm{B}, \mathrm{C}, \mathrm{D}\) and \(\mathrm{E}\). A random sample of the correct choice on 400 multiple-choice questions on a variety of \(\mathrm{AP}\) exams \(^{18}\) shows that \(\mathrm{B}\) was the most common correct choice, with 90 of the 400 questions having \(\underline{B}\) as the answer. Does this provide evidence that \(\mathrm{B}\) is more likely to be the correct choice than would be expected if all five options were equally likely? Show all details of the test. The data are available in APMultipleChoice.

We saw in Exercise 6.221 on page 466 that drinking tea appears to offer a strong boost to the immune system. In a study extending the results of the study described in that exercise, \(^{58}\) blood samples were taken on five participants before and after one week of drinking about five cups of tea a day (the participants did not drink tea before the study started). The before and after blood samples were exposed to e. coli bacteria, and production of interferon gamma, a molecule that fights bacteria, viruses, and tumors, was measured. Mean production went from \(155 \mathrm{pg} / \mathrm{mL}\) before tea drinking to \(448 \mathrm{pg} / \mathrm{mL}\) after tea drinking. The mean difference for the five subjects is \(293 \mathrm{pg} / \mathrm{mL}\) with a standard deviation in the differences of \(242 .\) The paper implies that the use of the t-distribution is appropriate. (a) Why is it appropriate to use paired data in this analysis? (b) Find and interpret a \(90 \%\) confidence interval for the mean increase in production of interferon gamma after drinking tea for one week.

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.