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Cellphone Calls Answers.com claims that the mean length of all cell phone conversations in the United States is \(3.25\) minutes (3 minutes and 15 seconds). Assume that this is correct, and also assume that the standard deviation is \(4.2\) minutes. (Source: wiki .answers.com, accessed January 16, 2011) * a. Describe the shape of the distribution of the length of cell phone conversations in this population. Do you expect it to be approximately Normally distributed, right-skewed, or left-skewed? Explain your reasoning. b. Suppose that, using a phone company's records, we randomly sample 100 phone calls. We calculate the mean length from this sample and record the value. We repeat this thousands of times. What will be the (approximate) mean value of the distribution of these thousands of sample means? c. Refer to part b. What will be the standard deviation of this distribution of thousands of sample means?

Short Answer

Expert verified
a. The distribution is assumed to be approximately Normally distributed due to the Central Limit Theorem. b. The mean value of the distribution of these thousands of sample means will be \(3.25\) minutes. c. The standard deviation of the distribution of these thousands of sample means will be \(0.42\) minutes.

Step by step solution

01

Understand and Describe the Distribution

We assume the distribution to be approximately Normally distributed as per the Central Limit Theorem, which states that if we have a large number of independent, identically distributed variables, then the distribution of the sum of these variables tends to a normal distribution. This applies here, as all individual call lengths are independent of each other.
02

Mean Value of Sample Means

The Central Limit Theorem states that the mean of the sampling distribution of the mean (the mean of all sample means) would be approximately equal to the mean of the population. Hence, the mean value of the distribution of these thousands of sample means will be the same as the mean of the population, which is \(3.25\) minutes.
03

Standard Deviation of Sample Means

The standard deviation of the distribution of sample means, also called the standard error, can be computed using the formula: \( \sigma_{\overline{x}} = \frac{\sigma}{\sqrt{n}} \) where \(\sigma_{\overline{x}}\) is the standard deviation of the sampling distribution, \(\sigma\) is the standard deviation of the population, and \(n\) is the size of the samples. Therefore, the standard deviation of the distribution of these thousands of sample means will be \( \frac{4.2}{\sqrt{100}} = 0.42 \) minutes.

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Key Concepts

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

Sampling Distribution
In statistical terms, a sampling distribution is the probability distribution of a given random-sample-based statistic. By nature, it represents what you get when you take numerous samples from a population and calculate a statistic, like a mean, for each one. In the context of the Central Limit Theorem, which is fundamental here, a sampling distribution specifically considers the means of these samples.
It is important to remember that the mean of the sampling distribution is typically equal to the mean of the population, provided that the sample size is sufficiently large and that samples are random and independent. This is what the Central Limit Theorem ensures — even when the population distribution is not normal or is unknown, the distribution of the sample means tends toward normality as the number of samples increases.
In our exercise, with thousands of samples taken, the approximation to normality becomes reliable, making the sampling distribution normally distributed around the mean of 3.25 minutes.
Standard Error
The standard error plays a crucial role in understanding the variation within a sampling distribution. It is defined as the standard deviation of the sampling distribution of a statistic, most commonly the sample mean. In simpler terms, it quantifies the precision of a sample mean estimate of a population mean.
The formula to compute standard error involves the population standard deviation divided by the square root of the sample size: \[ \sigma_{\overline{x}} = \frac{\sigma}{\sqrt{n}} \] where \( \sigma \) represents the population standard deviation, and \( n \) is the sample size.
Our exercise provides that the standard error of the mean phone call length for a sample size of 100 is \( \frac{4.2}{\sqrt{100}} = 0.42 \) minutes. This small standard error suggests that the sample mean is quite a precise estimate of the population mean of 3.25 minutes.
Normal Distribution
A normal distribution is a bell-shaped curve where most of the observations cluster around the mean. It is symmetrical and perfectly even on both sides of a central peak, which is the mean, median, and mode. The normal distribution becomes central in inferential statistics, since many statistical tests and methods assume data to be normally distributed.
According to the Central Limit Theorem, even if the underlying distribution of data points (such as cell phone call lengths) is not normal, the distribution of the sample means can be assumed to be approximately normal if the sample size is sufficiently large. This makes it a vital concept for many statistical analyses.
In the exercise, despite the individual call lengths possibly not forming a perfect normal distribution, the Central Limit Theorem enables us to assume normality for the sampling distribution of means because of the large number of samples taken.

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Most popular questions from this chapter

Suppose that 500 employees each took a random sample (with replacement) of 100 employees at their office and recorded the salaries of the employees in their sample. Then each employee used his or her data to calculate an \(80 \%\) confidence interval for the mean salary of all employees at the office. How many of the 500 intervals would you expect not to capture the true population mean? Explain by showing your calculation.

GPAs (Example 11) In finding a confidence interval for a random sample of 30 students GPAs, one interval was \((2.60,3.20)\) and the other was \((2.65,3.15)\). a. One of them is a \(95 \%\) interval and one is a \(90 \%\) interval. Which is which, and how do you know? b. If we used a larger sample size \((n=120\) instead of \(n=30\) ). would the \(95 \%\) interval be wider or narrower than the one reported here?

Potatoes Use the data from Exercise \(9.35\). a. If you use the four-step procedure with a two-sided alternative hypothesis, should you be able to reject the hypothesis that the population mean is 20 pounds using a significance level of \(0.05 ?\) Why or why not? The confidence interval is reported here: I am \(95 \%\) confident that the population mean is between \(20.4\) and \(21.7\) pounds. b. Now test the hypothesis that the population mean is not 20 pounds using the four-step procedure. Use a significance level of \(0.05\). c. Choose one of the following conclusions: i. We cannot reject a population mean of 20 pounds. ii. We can reject a population mean of 20 pounds. iii. The population mean is \(21.05\) pounds.

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