/*! 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 6 According to a 2017 report by Co... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

According to a 2017 report by ComScore .com, the mean time spent on smartphones daily by the American adults is \(2.85\) hours. Assume this is correct and assume the standard deviation is \(1.4\) hours. a. Suppose 150 American adults are randomly surveyed and asked how long they spend on their smartphones daily. The mean of the sample is recorded. Then we repeat this process, taking 1000 surveys of 150 American adults and recording the sample means. What will be the shape of the distribution of these sample means? b. Refer to part (a). What will be the mean and the standard deviation of the distribution of these sample means?

Short Answer

Expert verified
The distribution of these sample means will be approximately normally distributed. The mean will be \(2.85\) hours, and the standard deviation (standard error) will be \(1.4 / \sqrt{150}\).

Step by step solution

01

Identify the distribution shape

According to the Central Limit Theorem, if a large number of small samples is drawn from any population, the sampling distribution of the mean approximates a normal distribution. Therefore, with 1000 surveys of 150 people, the distribution will be approximately normal.
02

Calculate the mean

The mean of the distribution of sample means, also known as the expected value, will be the same as the mean of the population. Therefore, it will be \(2.85\) hours.
03

Calculate the standard deviation

The standard deviation of the sampling distribution (also known as the standard error) can be found by dividing the standard deviation of the population by the square root of the sample size. Since the population standard deviation is \(1.4\) hours and the sample size is 150, the standard deviation of the distribution of the sample means is \(1.4 / \sqrt{150}\).

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.

Central Limit Theorem
Understanding the Central Limit Theorem (CLT) is crucial in statistics, especially when dealing with sample means.
Imagine you're a researcher analyzing the time adults spend on their smartphones. Even if the time spent is not normally distributed in the entire population, the CLT states that if you take many samples and calculate their means, the distribution of these sample means will tend to be normally shaped as the sample size grows large enough.

This theorem is fundamental because it allows statisticians to make inferences about population parameters even when the population itself is not normally distributed. In practice, this means that for our exercise, when we gather 1000 surveys of 150 American adults, we can predict the behavior of our sample means even if our original data is skewed or has outliers. The CLT transforms the complex, often non-normal data into something much more manageable and universally understood: a normal distribution.
Standard Error
The term 'standard error' might sound complex, but it's actually just a measure of how much we expect sample means to vary from the true population mean. It's essentially the standard deviation for the sampling distribution of the mean.

To calculate the standard error, as we did in the exercise, we divide the population's standard deviation by the square root of the sample size. The formula is given by: \[ SE = \frac{\sigma}{\sqrt{n}} \]
where \(\sigma\) is the population standard deviation and \(n\) is the sample size. In our smartphone usage example, with a population standard deviation of 1.4 hours and groups of 150 adults, the standard error tells us how much the sample means will spread around the true population mean of 2.85 hours. This concept is vital because it directly impacts confidence intervals and hypothesis tests, which are cornerstones of statistical analysis.
Normal Distribution
The normal distribution, often represented by a bell curve, is a continuous probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean.

In the context of our example, the normal distribution describes the behavior of sample means when we conduct multiple surveys. Key characteristics of the normal distribution include being determined entirely by its mean and standard deviation and having the so-called 68-95-99.7 (empirical) rule. This rule states that approximately 68% of the data falls within one standard deviation of the mean, 95% within two standard deviations, and 99.7% within three.

The normal distribution applies to many real-world phenomena, and because of the Central Limit Theorem, it allows researchers to use normal probability to make predictions about sample means even when the population from which the sample is drawn isn't normally distributed. This is precisely what makes it possible to predict the behavior of our smartphone usage sample means.

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

Choose a test for each situation: one-sample \(t\) -test, two-sample \(t\) -test, paired \(t\) -test, and no \(t\) -test. a. A random sample of students who transfered to a 4 -year university from community colleges are asked their GPAs. Our goal is to determine whether the mean GPA for transfer students is significantly different from the population mean GPA for all students at the university. b. Students observe the number of office hours posted for a random sample of tenured and a random sample of untenured professors. c. A researcher goes to the parking lot at a large grocery chain and observes whether each person is male or female and whether they return the cart to the correct spot before leaving (yes or no).

State whether each situation has independent or paired (dependent) samples. a. A researcher wants to know whether pulse rates of people go down after brief meditation. She collects the pulse rates of a random sample of people before meditation and then collects their pulse rates after meditation. b. A researcher wants to know whether professors with tenure have fewer posted office hours than professors without tenure do. She observes the number of office hours posted on the doors of tenured and untenured professors.

The distribution of the scores on a certain exam is \(N(80,5)\) which means that the exam scores are Normally distributed with a mean of 80 and a standard deviation of 5 . a. Sketch or use technology to create the curve and label on the \(x\) -axis the position of the mean, the mean plus or minus one standard deviation, the mean plus or minus two standard deviations, and the mean plus or minus three standard deviations. b. Find the probability that a randomly selected score will be greater than 90\. Shade the region under the Normal curve whose area corresponds to this probability.

The undergraduate grade point average (GPA) for students accepted at a random sample of 10 medical schools in the United States was taken. The mean GPA for these accepted students was \(3.75\) with a standard error of \(0.06\). The distribution of undergraduate GPAs is Normal. (Source: Accepted.com) a. Decide whether each of the following statements is worded correctly for the confidence interval. Fill in the blanks for the correctly worded one(s). Explain the error for the ones that are incorrectly worded. i. We are \(95 \%\) confident that the sample mean is between ____ \(-\) and ____. ii. We are \(95 \%\) confident that the population mean is between ____. iii. There is a \(95 \%\) probability that the population mean is between ____ and ____. b. Based on your confidence interval, would you believe that the population mean GPA is \(3.80\) ? Why or why not?

A random sample of 25 baseball play ers from the 2017 Major League Baseball season was taken and the sample data was used to construct two confidence intervals for the population mean. One interval was \((22.0,42.8)\). The other interval was \((19.9,44.0)\). (Source: mlb.com) a. One interval is a \(95 \%\) interval, and one is a \(90 \%\) interval. Which is which, and how do you know? b. If a larger sample size was used, for example, 40 instead of 25 , how would this affect the width of the intervals? Explain.

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.