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Chapter 12: Problem 16

Suppose that a random sample of size 100 is to be drawn from a population with standard deviation 10 . a. What is the probability that the sample mean will be within 20 of the value of \(\mu\) ? b. For this example \((n=100, \sigma=10)\), complete each of the following statements by calculating the appropriate value: i. Approximately \(95 \%\) of the time, \(\bar{x}\) will be within of \(\mu\). ii. Approximately \(0.3 \%\) of the time, \(\bar{x}\) will be farther than from \(\mu\).

Short Answer

Expert verified
The short answer to the problem is: a. The probability that the sample mean will be within 20 of the value of \(\mu\) is extremely close to 1. b. i. Approximately \(95 \%\) of the time, \(\bar{x}\) will be within \(\mu \pm 1.96\) of \(\mu\). ii. Approximately \(0.3 \%\) of the time, \(\bar{x}\) will be farther than \(\mu \pm 2.75\) from \(\mu\).

Step by step solution

01

Problem (a)

First, let's find the probability that the sample mean will be within 20 of the value of \(\mu\). The Standard Error (SE) of the sample mean can be calculated using the formula: SE = \(\frac{\sigma}{\sqrt{n}}\) where \(\sigma\) = Population standard deviation \(n\) = Sample size Plugging in the given values: SE = \(\frac{10}{\sqrt{100}}\) SE = \(\frac{10}{10}\) SE = 1 To find the probability that the sample mean will be within 20 of the value of \(\mu\), we can calculate the Z-scores associated with this range. Lower Z-score = \(\frac{(\bar{x}-\mu_{\bar{x}})}{SE} = \frac{(\mu-20-\mu)}{1} = -20\) Upper Z-score = \(\frac{(\bar{x}-\mu_{\bar{x}})}{SE} = \frac{(\mu+20-\mu)}{1} = 20\) Now we need to find the probability associated with these Z-scores. We can use a Z-score table or an online calculator. The probability that the sample mean will be within 20 of the value of \(\mu\) = P(-20 < Z < 20) Since the Z-scores are very large, this probability is extremely close to 1.
02

Problem (b) - Part (i)

Next, we need to find the value that will make approximately \(95 \%\) of the time, \(\bar{x}\) will be within this value of \(\mu\). Since we know that about \(95 \%\) of the data is within 1.96 standard deviations of the mean in a normal distribution, we can use this Z-score to find the value we need. Using the formula: \(Value = \mu \pm Z * SE\) We have: \(Value = \mu \pm 1.96 * 1\) Since \(\bar{x}\) is within this value of \(\mu\), we can write: \(\bar{x} \approx \mu \pm 1.96\)
03

Problem (b) - Part (ii)

Finally, we need to find the value which will make approximately \(0.3 \%\) of the time, \(\bar{x}\) will be farther than this value from \(\mu\). Since \(0.3 \%\) of the time corresponds to \(0.003\) in decimal form, we can find the Z-score corresponding to 1 - 0.003 = 0.997 cumulative probability. Using a Z-score table or online calculator, we find that the Z-score corresponding to \(0.997\) is approximately 2.75. Now, using the formula: \(Value = \mu \pm Z * SE\) We have: \(Value = \mu \pm 2.75 * 1\) The sample mean \(\bar{x}\) will be farther than \(\mu \pm 2.75\) from \(\mu\) approximately \(0.3 \%\) of the time.

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Key Concepts

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

Standard Error
The standard error (SE) helps us understand how much the sample mean is expected to vary from the true population mean. It essentially measures the precision of the sample mean as an estimate of the population mean.
Here鈥檚 how you can calculate it:
  • Use the formula: SE = \( \frac{\sigma}{\sqrt{n}} \)
  • Where \( \sigma \) represents the population standard deviation.
  • \( n \) is the sample size.
This formula calculates how much the sample mean might deviate from the expected value. In our exercise, with a population standard deviation of 10 and a sample size of 100, the SE was calculated as:
  • SE = \( \frac{10}{\sqrt{100}} = 1 \)
A smaller SE indicates a more reliable sample mean. This is because a smaller SE suggests that the sample mean is closer to the true population mean.
Z-score
The Z-score is a statistical measurement that describes a value's relationship to the mean of a group of values. It tells us how many standard deviations away a value is from the mean. To understand this better:
  • The Z-score can be calculated using the formula: \( Z = \frac{(X - \mu)}{SE} \)
  • Where \( X \) is the value, \( \mu \) is the mean, and SE is the standard error.
In the given exercise, the sample mean needs to be within a certain range of \( \mu \). For example:
  • Lower Z-score = \( \frac{(\mu - 20 - \mu)}{1} = -20 \)
  • Upper Z-score = \( \frac{(\mu + 20 - \mu)}{1} = 20 \)
These Z-scores help us determine the probability that the sample mean falls within a specified range around the true population mean. Large absolute values of Z indicate that the value lies far from the mean. A Z-score close to zero indicates the value is near the mean.
Probability Calculation
Probability calculation in this context is about finding out how likely it is that the sample mean will fall within a particular range of the population mean. Using Z-scores, you can determine these probabilities:
  • Calculate the Z-scores for the boundaries of the range.
  • Use a Z-score table or calculator to find the probability for each Z-score.
For example, to find the probability that the sample mean will be within 20 units of \( \mu \):
  • Find the probability that Z is greater than -20 and less than 20.
  • Since these Z-scores are large, the corresponding probabilities are extremely close to 1.
This essentially means almost all sample means will fall within this range. Understanding probabilities helps predict the likelihood of outcomes and make informed statistical inferences.
Normal Distribution
Normal distribution, also known as the bell curve, is one of the fundamental concepts in statistics. It describes how data, such as test scores or heights, tend to be distributed in nature:
  • Features a symmetrical, bell-shaped curve.
  • The mean, median, and mode of the distribution are all equal.
In our exercise, the sampling distribution of the sample mean is assumed to be normal. This assumption is valid for large samples due to the Central Limit Theorem. It states that regardless of the distribution of the population, the distribution of the sample means will approximate a normal distribution as the sample size becomes larger.
In normal distributions:
  • Approximately 68% of data falls within 1 standard deviation of the mean.
  • Approximately 95% falls within 2 standard deviations.
  • About 99.7% falls within 3 standard deviations.
This information is useful for calculating probabilities, as seen in the exercise, and is crucial for making statistical inferences.

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