/*! 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 271 A \(90 \%\) confidence interval ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A \(90 \%\) confidence interval for \(\mu_{1}-\mu_{2}\) using the paired difference sample results \(\bar{x}_{d}=556.9\), \(s_{d}=143.6, n_{d}=100\)

Short Answer

Expert verified
The 90% confidence interval for \(\mu_{1}-\mu_{2}\) is \(\bar{x}_{d} \pm E\) where \(\bar{x}_{d}\) is the sample mean difference and E is the margin of error calculated in the above steps. Substitute the calculated values to get the final confidence interval.

Step by step solution

01

Calculate the Margin of Error

First, the standard error needs to be calculated. The standard error is given by the equation:\[SE = \frac{s_{d}}{\sqrt{n_{d}}}\]Substitute \(s_{d}=143.6\) and \(n_{d}=100\) to get the standard error. Next, calculate the margin of error (E) using a T-distribution (as we do not have the population standard deviation) which is given by the formula:\[E = SE * T\]where T is the t-value from the t-distribution table corresponding to desired confidence level (90% in this case) and df= \(n_{d} - 1\) degrees of freedom.
02

Calculate the Confidence Interval

Once the margin of error is found, the confidence interval can be calculated. The confidence interval is found using the formula:\[CI = \bar{x}_{d} \pm E\]Substitute \(\bar{x}_{d}=556.9\) and the margin of error E from Step 1 into the equation.

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.

Paired Difference
In statistical analyses involving two related or paired samples, a paired difference approach is often utilized. This approach focuses on the differences in observations between two sets of data, such as before and after measurements on the same subjects. It helps in addressing the changes or effects due to a specific treatment or condition.
The main advantage of using a paired difference method is that it reduces variability. Instead of directly comparing the differences between two distinct groups, you zero in on how each subject's score has changed. This often results in more precise statistical results.
In the original exercise, the paired difference approach was used to estimate the confidence interval for the mean difference (\[\mu_{1}-\mu_{2}\]). This is performed by calculating the average of the differences (\[\bar{x}_{d} = 556.9\]) and using the standard deviation of those differences (\[s_{d} = 143.6\]). It’s important to ensure that the data is paired properly to maintain the validity of the statistical analysis.
Margin of Error
The concept of the margin of error is central to statistical confidence intervals. This represents the amount of uncertainty around the sample statistic, providing a range within which we can expect the true population parameter to lie.
To calculate the margin of error, you need the standard error and the appropriate critical value. The standard error is derived from the sample standard deviation, adjusted by the square root of the sample size. Hence, it is calculated using:\[SE = \frac{s_{d}}{\sqrt{n_{d}}}\]where \(s_{d} = 143.6\) is the sample standard deviation, and \(n_{d} = 100\) is the sample size.
Once the standard error is obtained, it's multiplied by the critical t-value, which corresponds to the desired confidence level (in this case, 90%) and degrees of freedom (df = \(n_{d} - 1\)). The product is your margin of error:\[E = SE \times T\]
The margin of error helps in constructing the confidence interval by adding and subtracting it to/from the sample mean difference.
T-distribution
The T-distribution is crucial in statistical analyses involving small sample sizes or unknown population standard deviations. It allows for estimating population parameters when the exact variance is unknown, serving as an excellent tool when working with sample data.
In a paired difference scenario, where sample sizes are often limited, the T-distribution simulates the spread of all possible sample means around a true population mean, accounting for uncertainty better than the normal distribution when working with smaller samples.
For calculating our confidence interval, the T-distribution is employed to identify the t-value, which aligns with a 90% confidence level and corresponds to \(df = n_{d} - 1\). This t-value is then used to compute the margin of error.
Using the T-distribution is particularly beneficial because as the sample size grows larger, it resembles the normal distribution. When the population standard deviation is unknown, as is often the case, the T-distribution offers a more realistic approximation, especially for smaller sample sizes. This precise approximation leads to more accurate estimation of confidence intervals.

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

Who Is More Trusting: Internet Users or Non-users? In a randomly selected sample of 2237 US adults, 1754 identified themselves as people who use the Internet regularly while the other 483 indicated that they do not. In addition to Internet use, participants were asked if they agree with the statement "most people can be trusted." The results show that 807 of the Internet users agree with this statement, while 130 of the non-users agree. \(^{54}\) (a) Which group is more trusting in the sample (in the sense of having a larger percentage who agree): Internet users or people who don't use the Internet? (b) Can we generalize the result from the sample? In other words, does the sample provide evidence that the level of trust is different between the two groups in the broader population? (c) Can we conclude that Internet use causes people to be more trusting? (d) Studies show that formal education makes people more trusting and also more likely to use the Internet. Might this be a confounding factor in this case?

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 (and using the sample proportions to estimate the population proportions). Sample A has a count of 90 successes with \(n=120\) and Sample \(\mathrm{B}\) has a count of 180 successes with \(n=300\).

We see in the AllCountries dataset that the percent of the population that is elderly (over 65 years old) is 17.0 in Austria and 15.9 in Denmark. Suppose we take random samples of size 200 from each of these countries and compute the difference in sample proportions \(\hat{p}_{A}-\hat{p}_{D}\), where \(\hat{p}_{A}\) represents the proportion of the sample that is elderly in Austria and \(\hat{p}_{D}\) represents the proportion of the sample that is elderly in Denmark. Find the mean and standard deviation of the differences in sample proportions.

If random samples of the given sizes are drawn from populations with the given proportions: (a) Find the mean and standard error of the distribution of differences in sample proportions, \(\hat{p}_{A}-\hat{p}_{B}\) (b) If the sample sizes are large enough for the Central Limit Theorem to apply, draw a curve showing the shape of the sampling distribution. Include at least three values on the horizontal axis. Samples of size 100 from population \(A\) with proportion 0.20 and samples of size 50 from population \(B\) with proportion 0.30

In Exercises 6.1 to \(6.6,\) if random samples of the given size are drawn from a population with the given proportion: (a) Find the mean and standard error of the distribution of sample proportions. (b) If the sample size is large enough for the Central Limit Theorem to apply, draw a curve showing the shape of the sampling distribution. Include at least three values on the horizontal axis. Samples of size 30 from a population with proportion 0.27

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.