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Smartphones 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
a) The shape of the distribution of these sample means will be approximately normal due to the Central Limit Theorem. b) The mean of the distribution of these sample means will be \( 2.85 \) hours and the standard deviation will be approximately \( 0.1144 \) hours.

Step by step solution

01

Determining the Shape of the Sample Mean Distribution

According to the Central Limit Theorem, if the sample size is large enough (normally greater than 30), then the distribution of the sample means will be approximately normal regardless of the shape of the population distribution. Here, since our sample size of 150 is larger than 30, the distribution of these sample means will be approximately normal.
02

Calculating the Mean of the Sample Means

The mean of the sample means (also known as the expected value of the sample means) is equal to the population mean. The population mean, in this case, is the mean time spent on smartphones daily by American adults, which is given as \( 2.85 \) hours. Therefore, the mean of the sample means will also be \( 2.85 \) hours.
03

Calculating the Standard Deviation of the Sample Means

The standard deviation of the sample means (also known as the standard error) can be found using the formula: \[\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}}\]where:\(\sigma_{\bar{x}}\) is the standard deviation of the sample means,\(\sigma\) is the population standard deviation,\(n\) is the size of the sample.In this case, \( \sigma \) is \( 1.4 \) hours, and \( n \) is 150. Substituting these values into the formula, we get:\[\sigma_{\bar{x}} = \frac{1.4}{\sqrt{150}} = 0.1144 \, hours\]

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Key Concepts

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

Central Limit Theorem
The Central Limit Theorem (CLT) is a fundamental statistical principle that helps in understanding the behavior of sample means. It states that, regardless of the population's distribution, the distribution of the sample means tends to be normal, or bell-shaped, as the sample size becomes large. A 'large' sample size typically means greater than or equal to 30.

This theorem is crucial when dealing with samples from a population because it allows us to make inferences about the population using the sample data. For example, in our exercise about smartphone usage, the sample size of 150 American adults is well above the threshold of 30. Hence, according to the CLT, the distribution of the sample mean of the time they spend on their smartphones will be approximately normal, even if the actual time spent by all American adults is not normally distributed.
Standard Deviation
Standard deviation is a measure that is used to quantify the amount of variation or dispersion of a set of values. A low standard deviation indicates that the values tend to be close to the mean (also called the expected value), whereas a high standard deviation indicates that the values are spread out over a wider range.

In the context of our smartphone usage exercise, the standard deviation of 1.4 hours tells us that the amount of time spent on smartphones varies around the mean by this amount on average. When considering a distribution of sample means, the standard deviation of this distribution is known as the standard error. It's important for students to understand that the 'standard error' decreases as the sample size increases, which means with larger samples, our estimate of the population mean becomes more precise.
Normal Distribution
A normal distribution, also known as a Gaussian distribution, is a 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 graph form, this distribution will appear as a bell curve. The normal distribution is widely used in statistics because it models many natural phenomena.

Returning to our smartphone example, because of the CLT, we can say that the distribution of the sample means of time spent on smartphones by 150 American adults will take the shape of a normal distribution. This is powerful as it allows the use of probability theory to make predictions about our sample mean, helping us understand how close we can expect this sample mean to be to the true population mean.

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Most popular questions from this chapter

Babies Weights (Example 2) Some sources report that the weights of full-term newborn babies have a mean of 7 pounds and a standard deviation of \(0.6\) pound and are Normally distributed. a. What is the probability that one newborn baby will have a weight within \(0.6\) pound of the mean - that is, between \(6.4\) and \(7.6\) pounds, or within one standard deviation of the mean? b. What is the probability the average of four babies' weights will be within \(0.6\) pound of the mean - that is, between \(6.4\) and \(7.6\) pounds? c. Explain the difference between a and b.

Travel Time to School A random sample of \(50 \mathrm{1} 2\) th-grade students was asked how long it took to get to school. The sample mean was \(16.2\) minutes, and the sample standard deviation was \(12.3\) minutes. (Source: AMSTAT Census at School) a. Find a \(95 \%\) confidence interval for the population mean time it takes 12 th-grade students to get to school. b. Would a \(90 \%\) confidence interval based on this sample data be wider or narrower than the \(95 \%\) confidence interval? Explain. Check your answer by constructing a \(90 \%\) confidence interval and comparing this width of the interval with the width of the \(95 \%\) confidence interval you found in part a.

Ages A study of all the students at a small college showed a mean age of \(20.7\) and a standard deviation of \(2.5\) years. Are these numbers statistics or parameters? Explain. b. Label both numbers with their appropriate symbol (such as \(\bar{x}, \mu, s\), or \(\sigma)\).

Again In exercise \(9.31\), two intervals were given for the same data, one for \(95 \%\) confidence and one for \(90 \%\) confidence. a. How would a \(99 \%\) interval compare? Would it be narrower than both, wider than both, or between the two in width. Explain. b. If we wanted to use a \(99 \%\) confidence level and get a narrower width, how could we change our data collection?

Medical School Acceptance Rates The acceptance rate for a random sample of 15 medical schools in the United States was taken. The mean acceptance rate for this sample was \(5.77\) with a standard error of \(0.56 .\) Assume the distribution of acceptance rates is Normal. (Source: Accepted.com) a. Decide whether cach 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 \(\quad\) and iii. There is a \(95 \%\) probability that the population mean is between and b. Based on your confidence interval, would you believe the average acceptance rate for medical schools is \(6.5\) ? Explain.

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