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

A Sampling Distribution for Statistics Graduate Programs Exercise 3.31 introduced the dataset StatisticsPhD, which gives enrollment for all 82 graduate statistics programs in the US in \(2009 .\) Use StatKey or other technology to generate a sampling distribution of sample means using a sample size of \(n=10\) from the values in this dataset. What shape does the distribution have? Approximately where is it centered? What is the standard error (in other words, what is the standard deviation of the sample means)?

Short Answer

Expert verified
The distribution shape, center, and standard error will depend on the specific dataset values. The shape would likely be approximately normal (due to large repetition), the center is the mean of the sample means, and the standard error is the standard deviation of the sample means.

Step by step solution

01

Generate the Sampling Distribution

The first step is to generate a sampling distribution of sample means, using a sample size of \(n = 10\). You need to randomly select 10 data points from the dataset and calculate their mean. Repeat this process a large number of times (at least 1000 times) to obtain the sampling distribution.
02

Identify the Shape of the Distribution

After generating the sampling distribution, the next step is to identify its shape. You need to plot a histogram or a similar visual representation. Observe the shape of the distribution. It can be normal, skewed, bimodal, etc. Given the Central Limit Theorem, it's expected to have an approximate normal distribution for large enough repetitions.
03

Determine the Center of the Distribution

The center of a distribution is typically given by its mean or median. In this case, you need to calculate the average of the sample means obtained in step 1.
04

Calculate the Standard Error

The standard error is the standard deviation of the sample means. After having computed all the sample means in Step 1, calculate the standard deviation among them to get the standard 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 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

Mean number of cell phone calls made or received per day by cell phone users. In a survey of 1917 cell phone users, the mean was 13.10 phone calls a day.

A sample is given. Indicate whether each option is a possible bootstrap sample from this original sample. Original sample: 85,72,79,97,88 . Do the values given constitute a possible bootstrap sample from the original sample? (a) 79,79,97,85,88 (b) 72,79,85,88,97 (c) 85,88,97,72 (d) 88,97,81,78,85 (e) 97,85,79,85,97 (f) 72,72,79,72,79

Standard Deviation of NHL Penalty Minutes Exercise 3.102 describes data on the number of penalty minutes for Ottawa Senators NHL players. The sample has a fairly large standard deviation, \(s=27.3\) minutes. Use StatKey or other technology to create a bootstrap distribution, estimate the standard error, and give a \(95 \%\) confidence interval for the standard deviation of penalty minutes for NHL players. Assume that the data in OttawaSenators can be viewed as a reasonable sample of all NHL players.

In Exercises 3.51 to 3.56 , information about a sample is given. Assuming that the sampling distribution is symmetric and bell-shaped, use the information to give a \(95 \%\) confidence interval, and indicate the parameter being estimated. \(\hat{p}_{1}-\hat{p}_{2}=0.08\) and the margin of error for \(95 \%\) confidence is \(\pm 3 \%\).

Bisphenol A in Your Soup Cans Bisphenol A (BPA) is in the lining of most canned goods, and recent studies have shown a positive association between BPA exposure and behavior and health problems. How much does canned soup consumption increase urinary BPA concentration? That was the question addressed in a recent study \(^{34}\) in which consumption of canned soup over five days was associated with a more than \(1000 \%\) increase in urinary BPA. In the study, 75 participants ate either canned soup or fresh soup for lunch for five days. On the fifth day, urinary BPA levels were measured. After a two-day break, the participants switched groups and repeated the process. The difference in BPA levels between the two treatments was measured for each participant. The study reports that a \(95 \%\) confidence interval for the difference in means (canned minus fresh) is 19.6 to \(25.5 \mu \mathrm{g} / \mathrm{L}\). (a) Is this a randomized comparative experiment or a matched pairs experiment? Why might this type of experiment have been used? (b) What parameter are we estimating? (c) Interpret the confidence interval in terms of BPA concentrations. (d) If the study had included 500 participants instead of \(75,\) would you expect the confidence interval to be wider or narrower?

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