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91Ó°ÊÓ

You take a random sample of size \(n=49\) from a distribution with mean \(\mu=53\) and \(\sigma=21\). The sampling distribution of \(\bar{x}\) will be approximately__ with a mean of ___ and a standard deviation (or standard error) of ____

Short Answer

Expert verified
Answer: The sampling distribution of the sample mean will be approximately normal with a mean of 53 and a standard deviation (or standard error) of 3.

Step by step solution

01

Population mean (μ) and population standard deviation (σ)

The problem states that the population mean (μ) is 53, and the population standard deviation (σ) is 21.
02

Sample size (n)

Given that the sample size (n) is 49.
03

Sampling distribution shape

According to the central limit theorem, when the sample size (n) is at least 30, the sampling distribution of the sample mean (x̄) will be approximately normal.
04

Mean of the sampling distribution (μx̄)

The mean of the sampling distribution (μx̄) is equal to the population mean (μ). So, μx̄ = 53.
05

Standard deviation of the sampling distribution (σx̄)

The standard deviation of the sampling distribution of the sample mean, also called the standard error (SE), is calculated using the formula: σx̄ = σ / √n Where σ is the population standard deviation, and n is the sample size. Plugging in the given values: σx̄ = 21 / √49 σx̄ = 21 / 7 = 3 The standard deviation of the sampling distribution (σx̄) is 3. So, the sampling distribution of the sample mean (x̄) will be approximately normal with a mean of 53 and a standard deviation (or standard error) of 3.

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Key Concepts

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

Sampling Distribution
A sampling distribution is a crucial concept in statistics. It refers to the probability distribution of a given statistic based on a random sample. In simpler terms, if you were to take many samples from a population and calculate a statistic like the mean for each sample, the distribution of those means forms the sampling distribution.

In the provided exercise, we take a sample of size 49 from a larger population. The aim is to understand the behavior of the sample mean \(\bar{x}\). The sample size is significant. When it's large enough—in this case, greater than 30—the Central Limit Theorem assures us that the distribution of the sample mean tends to be normal, even if the original population distribution is not.

Key Points to Remember about Sampling Distribution:
  • It describes how the sample mean varies from sample to sample.
  • Its mean is the same as the population mean (in this exercise, 53).
  • Its spread or dispersion is less than that of the population distribution, depending on the sample size.
Standard Error
The standard error is a term you'll often encounter in statistics, which measures the precision of a sample mean estimate. Specifically, it's the standard deviation of the sampling distribution of the sample mean. It helps us understand how much variability can be expected and is crucial for constructing confidence intervals and hypothesis tests.

In this exercise, the standard error \(\sigma_{\bar{x}}\) is calculated using the formula \(\sigma / \sqrt{n}\), where \(\sigma\) is the population standard deviation and \(n\) is the sample size. Applying this formula to our numbers:

\[\sigma_{\bar{x}} = \frac{21}{\sqrt{49}} = \frac{21}{7} = 3\]

Thus, the standard error is 3. This indicates that for different samples of this size (49), the sample mean will typically bounce around 3 points above or below the true population mean of 53.
  • The smaller the standard error, the more reliable your sample mean is as an estimate of the population mean.
  • Increasing the sample size reduces the standard error, resulting in a more precise estimate.
Normal Distribution
A normal distribution is a fundamental concept in statistics and is often called a "bell curve" due to its shape. This type of distribution is symmetric, with most observations clustering around the central peak and probabilities tapering off equally on both sides.

In the context of this exercise, the Central Limit Theorem plays a key role. This theorem states that, given a sufficiently large sample size, the sampling distribution of the mean \(\bar{x}\) will approximate a normal distribution, regardless of the shape of the population distribution. Our sample size of 49 fits this criterion.

Some Key Characteristics of a Normal Distribution are:
  • The mean, median, and mode are all equal and located at the center of the distribution.
  • The curve is symmetric about the mean.
  • About 68% of data falls within one standard deviation of the mean, 95% within two, and 99.7% within three.

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Most popular questions from this chapter

The sample means were calculated for 30 samples of size \(n=10\) for a process that was judged to be in control. The means of the \(30 \bar{x}\) -values and the standard deviation of the combined 300 measurements were \(\overline{\bar{x}}=20.74\) and \(s=.87,\) respectively. a. Use the data to determine the upper and lower control limits for an \(\bar{x}\) chart. b. What is the purpose of an \(\bar{x}\) chart? c. Construct an \(\bar{x}\) chart for the process and explain how it can be used.

The number of wiring packages that can be assembled by a company's employees has a normal distribution, with a mean equal to 16.4 per hour and a standard deviation of 1.3 per hour. a. What are the mean and standard deviation of the number \(x\) of packages produced per worker in an 8-hour day? b. Do you expect the probability distribution for \(x\) to be mound-shaped and approximately normal? Explain. c. What is the probability that a worker will produce at least 135 packages per 8 -hour day?

The sample means were calculated for 40 samples of size \(n=5\) for a process that was judged to be in control. The means of the 40 values and the standard deviation of the combined 200 measurements were \(\overline{\bar{x}}=155.9\) and \(s=4.3,\) respectively. a. Use the data to determine the upper and lower control limits for an \(\bar{x}\) chart. b. Construct an \(\bar{x}\) chart for the process and explain how it can be used.

In Exercise 1.67 Allen Shoemaker derived a distribution of human body temperatures with a distinct mound shape. \(^{8}\) Suppose we assume that the temperatures of healthy humans is approximately normal with a mean of 98.6 degrees and a standard deviation of 0.8 degrees. a. If 130 healthy people are selected at random, what is the probability that the average temperature for these people is 98.25 degrees or lower? b. Would you consider an average temperature of 98.25 degrees to be an unlikely occurrence, given that the true average temperature of healthy people is 98.6 degrees? Explain.

The normal daily human potassium requirement is in the range of 2000 to 6000 milligrams (mg), with larger amounts required during hot summer weather. The amount of potassium in food varies, depending on the food. For example, there are approximately \(7 \mathrm{mg}\) in a cola drink, \(46 \mathrm{mg}\) in a beer, \(630 \mathrm{mg}\) in a banana, \(300 \mathrm{mg}\) in a carrot, and \(440 \mathrm{mg}\) in a glass of orange juice. Suppose the distribution of potassium in a banana is normally distributed, with mean equal to \(630 \mathrm{mg}\) and standard deviation equal to \(40 \mathrm{mg}\) per banana. You eat \(n=3\) bananas per day, and \(T\) is the total number of milligrams of potassium you receive from them. a. Find the mean and standard deviation of \(T\). b. Find the probability that your total daily intake of potassium from the three bananas will exceed \(2000 \mathrm{mg}\). (HINT: Note that \(T\) is the sum of three random variables, \(x_{1}, x_{2}\), and \(x_{3}\), where \(x_{1}\) is the amount of potassium in banana number 1 , etc.)

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