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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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Most popular questions from this chapter

The time that people have to wait for an elevator in an office building has a uniform distribution over the interval from 0 to 1 minute. For this distribution, \(\mu=0.5\) and \(\sigma=0.289\) a. If \(\bar{x}\) is the average waiting time for a random sample of \(n=16\) waiting times, what are the values of the mean and standard deviation of the sampling distribution of \(\bar{x} ?\) b. Answer Part (a) for a random sample of 50 waiting times. Draw a picture of the approximate sampling distribution of \(\bar{x}\) when \(n=50\).

The report "Majoring in Money: How American College Students Manage Their Finances" (SallieMae, \(2016,\) www.news/salliemae.com, retrieved December 24,2106\()\) includes data from a survey of college students. Each person in a representative sample of 793 college students was asked if they had one or more credit cards and if so, whether they paid their balance in full each month. There were 500 who paid in full each month. For this sample of 500 students, the sample mean credit card balance was reported to be \(\$ 825 .\) The sample standard deviation of the credit card balances for these 500 students was not reported, but for purposes of this exercises, suppose that it was \(\$ 200\). Is there convincing evidence that college students who pay their credit card balance in full each month have mean balance that is lower than \(\$ 906\), the value reported for all college students with credit cards? Carry out a hypothesis test using a significance level of 0.01 .

Suppose that the population mean value of interpupillary distance (the distance between the pupils of the left and right eyes) for adult males is \(65 \mathrm{~mm}\) and that the population standard deviation is \(5 \mathrm{~mm}\). a. If the distribution of interpupillary distance is normal and a random sample of \(n=25\) adult males is to be selected, what is the probability that the sample mean distance \(\bar{x}\) for these 25 will be between 64 and \(67 \mathrm{~mm}\) ? At least \(68 \mathrm{~mm}\) ? b. Suppose that a random sample of 100 adult males is to be selected. Without assuming that interpupillary distance is normally distributed, what is the approximate probability that the sample mean distance will be between 64 and 67 \(\mathrm{mm}\) ? At least \(68 \mathrm{~mm} ?\)

The dodo was a species of flightless bird that lived on the island of Mauritius in the Indian Ocean. The first record of human interaction with the dodo occurred in 1598 , and within 100 years the dodo was extinct due to hunting by humans and other newly introduced invasive species. After the extinction, the word "dodo" became synonymous with stupidity, implying that the birds lacked the intelligence to avoid or escape extinction. The closest existing relatives of the dodo are pigeons and doves. Researchers at the American Museum of Natural History used computed tomography (CT) scans to measure the brain size ("endocranial capacity") of one of the few existing preserved dodo birds, and to measure the brain sizes in samples of eight birds that are close relatives of dodos ("The First Endocast of the Extinct Dodo and an Anatomical Comparison Amongst Close Relatives," Zoological Journal of the Linnean Society [2016]: 950-953) The brain size for the dodo was \(4.17 \log \mathrm{mm}^{3}\). The following table contains the brain sizes for the sample of birds from related species (approximate values from a graph in the paper). a. Use the output at the bottom of the page from the Shiny App "Randomization Test for One Mean" to help you to carry out a randomization test of the hypothesis that the population mean brain size for birds that are relatives of the dodo differs from the established dodo brain size of 4.17 . b. What does the result of your test indicate about the brain size of the dodo?

The authors of the paper "Driving Performance While Using a Mobile Phone: A Simulation Study of Greek Professional Drivers" (Transportation Research Part \(F\) [2016]: 164-170) describe a study to evaluate the effect of mobile phone use by taxi drivers in Greece. Fifty taxi drivers drove in a driving simulator where they were following a lead car. The drivers were asked to carry on a conversation on a mobile phone while driving, and the following distance (the distance between the taxi and the lead car) was recorded. The sample mean following distance was 3.20 meters and the sample standard deviation was 1.11 meters. a. Construct and interpret a \(95 \%\) confidence interval for \(\mu,\) the population mean following distance while talking on a mobile phone for the population of taxi drivers. b. What assumption must be made in order to generalize this confidence interval to the population of all taxi drivers in Greece?
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