Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Assume the samples are random samples from distributions that are reasonably normally distributed, and that a t-statistic will be used for inference about the difference in sample means. State the degrees of freedom used. Find the endpoints of the t-distribution with \(2.5 \%\) beyond them in each tail if the samples have sizes \(n_{1}=15\) and \(n_{2}=25\).

Short Answer

Expert verified
The endpoints of the t-distribution with 2.5% beyond them in each tail when the sample sizes are \(n_{1}=15\) and \(n_{2}=25\) are approximately ±2.024.

Step by step solution

01

Degrees of Freedom

The formula for degrees of freedom for a two-sample t-test when sample sizes are different and variance is assumed unequal is given as: \[ df = {(s_{1}^2/n_{1} + s_{2}^2/n_{2})^2 \over {(s_{1}^2/n_{1})^2/(n_{1}-1) + (s_{2}^2/n_{2})^2/(n_{2}-1)} } \] Where: \n \(s_{1}^2\) and \(s_{2}^2\) are the sample variances \n \(n_{1}\) and \(n_{2}\) are the sample sizes Since we don't have standard deviations here, we will use the simpler formula to calculate degrees of freedom which is: \[ df = n_{1} + n_{2} - 2 \] Substituting with given values, we get : \[ df = 15 + 25 - 2 = 38 \]
02

Calculating the t-distribution endpoints

The next step is to use a t-table or a t-distribution calculator to obtain the t-distribution values, accounting for the 2.5% in each tail. The t-distribution value that leaves 2.5% in each tail for 38 degrees of freedom is approximately ±2.024.

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 Vaia!

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

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

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 \(n=300\).

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, with \(95 \%\) confidence, a margin of error within \(\pm 3 \%\) when estimating a proportion. First, find the sample size needed if we have no prior knowledge about the population proportion \(p\). Then find the sample size needed if we have reason to believe that \(p \approx 0.7\). Finally, find the sample size needed if we assume \(p \approx 0.9 .\) Comment on the relationship between the sample size and estimates of \(p\).

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:. $$ \begin{array}{lccccc} \hline \text { Case } & \mathbf{1} & \mathbf{2} & \mathbf{3} & \mathbf{4} & \mathbf{5} \\ \hline \text { Treatment 1 } & 22 & 28 & 31 & 25 & 28 \\ \text { Treatment 2 } & 18 & 30 & 25 & 21 & 21 \\ \hline \end{array} $$

Dark Chocolate for Good Health A study \(^{47}\) examines chocolate's effects on blood vessel function in healthy people. In the randomized, doubleblind, placebo-controlled study, 11 people received 46 grams (1.6 ounces) of dark chocolate (which is naturally flavonoid-rich) every day for two weeks, while a control group of 10 people received a placebo consisting of dark chocolate with low flavonoid content. Participants had their vascular health measured (by means of flow-mediated dilation) before and after the two-week study. The increase over the two-week period was measured, with larger numbers indicating greater vascular health. For the group getting the good dark chocolate, the mean increase was 1.3 with a standard deviation of \(2.32,\) while the control group had a mean change of -0.96 with a standard deviation of 1.58 . (a) Explain what "randomized, double-blind, placebo-controlled study" means. (b) Find and interpret a \(95 \%\) confidence interval for the difference in means between the two groups. Be sure to clearly define the parameters you are estimating. You may assume that neither sample shows significant departures from normality. (c) Is it plausible that there is "no difference" between the two kinds of chocolate? Justify your answer using the confidence interval found in \(\operatorname{part}(\mathrm{b})\)

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.

Sign-up for free