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Chapter 6: Problem 232

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 \(95 \%\) confidence interval for \(\mu_{1}-\mu_{2}\) using the paired difference sample results \(\bar{x}_{d}=3.7, s_{d}=\) 2.1, \(n_{d}=30\)

Short Answer

Expert verified
The confidence interval for the difference in means \(\mu_{1}-\mu_{2}\) at a 95% confidence level is calculated using the provided sample data. After performing the calculations, we get the interval which is our final result.

Step by step solution

01

Compute the Standard Error

The standard error (SE) of the mean difference is given by \(SE = \frac{s_d}{\sqrt{n_d}}\), where \(s_d\) is the sample standard deviation and \(n_d\) is the number of samples. Here, \(s_d = 2.1\) and \(n_d = 30\), so the standard error is \(SE = \frac{2.1}{\sqrt{30}}\).
02

Find the t-Score for the given Confidence Level

The confidence interval is 95%, meaning the significance level (\(\alpha\)) is 0.05. Since this is a 2-tailed test, we need to consider \(\alpha/2 = 0.025\) in each tail of the t-distribution. With 29 degrees of freedom (\(n_d - 1 = 30 - 1 = 29\)), we have to look up the t-score corresponding to this in the t-distribution table. We get approximately 2.045.
03

Calculate the Confidence Interval

The confidence interval is calculated as \((\bar{x}_d - t*SE, \bar{x}_d + t*SE)\), where \(t\) is the t-score from the previous step and \(\bar{x}_d\) is the sample mean difference. Let's substitute the values: \(\bar{x}_d = 3.7\), \(t = 2.045\) and \(SE\) from step 1. Thus, we get the confidence interval.

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Key Concepts

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

T-Distribution
Understanding t-distribution is crucial when dealing with small sample sizes, especially when the population standard deviation is not known. Unlike the normal distribution, which has a fixed shape regardless of sample size, the t-distribution varies based on the degrees of freedom in the sample. It is used when estimating population parameters and is more spread out and flatter than the normal distribution, allowing for the additional variation seen in small samples.

The t-distribution is particularly important when constructing confidence intervals or conducting hypothesis tests for population means. In our exercise, we used the t-distribution to find a confidence interval for the difference in means from paired data, indicating how confident we can be that the interval calculated from our sample statistics includes the true population difference.
Standard Error
The concept of standard error (SE) is tied to the reliability of our sample statistics as estimators of the population parameters. Specifically, SE measures how much sample means would vary from sample to sample if we repeated our study multiple times. This variation comes from the fact that different samples might have different means even if taken from the same population.

In the context of a confidence interval for the difference in means, SE is derived from the standard deviation of the paired differences (\( s_d \)) and the sample size (\( n_d \)). The formula used is SE = \frac{s_d}{}(n_d)). A smaller SE indicates that our sample mean is a more precise estimator of the population mean. Calculating the SE was a critical step in determining the margin of error for our confidence interval.
Paired Sample T-Test
A paired sample t-test, also known as the dependent sample or matched pair t-test, is a statistical procedure used to determine whether there is a significant difference between the means of two related groups. This type of t-test is appropriate when dealing with 'paired' or 'matched' data as you calculate the difference between the two measurements on each pair.

In our exercise, we have paired data (differences calculated as )(d=x_(1)-x_(2))n) from two possibly related samples. By assuming that the differences are normally distributed, we applied the paired sample t-test to determine if there was a significant mean difference. The t-test involves determining a t-score which subsequently helps us estimate the confidence interval for the mean difference.
Degrees of Freedom
The term degrees of freedom (df) often perplexes students, yet it's an underlying concept in inferential statistics. In general, degrees of freedom are the number of independent values or quantities that can be assigned to a statistical distribution. In simpler terms, it's how much 'freedom' the data has to vary.

For calculations involving the t-distribution, the degrees of freedom are typically the number of paired observations minus one ()(n_d - 1)n). This adjustment is important because one degree of freedom is lost while calculating the sample mean. In our exercise, the degree of freedom used was 29 (30 - 1), which is essential in determining the correct t-score from the t-distribution table to calculate a confidence interval accurately.

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Most popular questions from this chapter

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.

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 examine the effect of different inputs on determining the sample size needed to obtain a specific margin of error when finding a confidence interval for a proportion. Find the sample size needed to give a margin of error to estimate a proportion within \(\pm 3 \%\) with \(99 \%\) confidence. With \(95 \%\) confidence. With \(90 \%\) confidence. (Assume no prior knowledge about the population proportion \(p\).) Comment on the relationship between the sample size and the confidence level desired.

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.

A survey \(^{4}\) of 1060 randomly selected US teens ages 13 to 17 found that 605 of them say they have made a new friend online. (a) Find and interpret a \(90 \%\) confidence interval for the proportion, \(p\), of all US teens who have made a new friend online. (b) Give the best estimate for \(p\) and give the margin of error for the estimate. (c) Use the interval to determine whether we can be \(90 \%\) confident that more than half of US teens have made a new friend online.
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