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Chapter 9: Problem 20

Tree Rings Tree-ring dating from archaeological excavation sites is used in conjunction with other chronologic evidence to estimate occupation dates of prehistoric Indian ruins in the southwestern United States. It is thought that Burnt Mesa Pueblo was occupied around 1300 A.D. (based on evidence from potsherds and stone tools). The following data give tree-ring dates (A.D.) from adjacent archaeological sites (Bandelier Archaeological Excavation Project: Summer 1990 Excavations at Burnt Mesa Pueblo, edited by \(\mathrm{T}\). Kohler, Washington State University Department of Anthropology, 1992): \(\begin{array}{lllll} 1189 & 1267 & 1268 & 1275 & 1275 \\ 1271 & 1272 & 1316 & 1317 & 1230 \end{array}\) i. Use a calculator with mean and standard deviation keys to verify that \(\bar{x}=\) 1268 and \(s \approx 37.29\) years. ii. Assuming the tree-ring dates in this excavation area follow a distribution that is approximately normal, does this information indicate that the population mean of tree-ring dates in the area is different from (either higher on lower than) that in 1300 A.D.? Use a \(1 \%\) level of significance.

Short Answer

Expert verified
There is not enough evidence to conclude that the mean is different from 1300.

Step by step solution

01

Calculate the Mean

First, verify the mean of the tree-ring dates provided. Add up all the given dates: \(1189 + 1267 + 1268 + 1275 + 1275 + 1271 + 1272 + 1316 + 1317 + 1230 = 12680\). Divide by the number of entries (10) to find the mean: \(\bar{x} = \frac{12680}{10} = 1268\).
02

Calculate the Standard Deviation

Next, find the standard deviation of the data. First, calculate each date's deviation from the mean, square each deviation, sum those squares, and then divide by the number of entries minus one. Finally, take the square root to find the standard deviation.
03

Formulate Hypotheses

Establish the null hypothesis \(H_0: \mu = 1300\) and the alternative hypothesis \(H_a: \mu eq 1300\). This is a two-tailed test since we are checking if the mean is different, either higher or lower, than 1300.
04

Calculate the Test Statistic

Use the formula for the t-statistic: \(t = \frac{\bar{x} - \mu}{s/\sqrt{n}}\). Substitute \(\bar{x} = 1268\), \(\mu = 1300\), \(s = 37.29\), and \(n = 10\) into the formula to get \(t = \frac{1268 - 1300}{37.29/\sqrt{10}} \approx -2.728\).
05

Determine Critical Value and Decision

At a \(1\%\) significance level, using a t-distribution table and 9 degrees of freedom, find the critical t-value for a two-tailed test, which is approximately \(\pm 3.25\). Since our calculated t-statistic \(-2.728\) does not exceed the critical value, we do not reject the null hypothesis.

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Key Concepts

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

standard deviation
The standard deviation is a crucial statistic that tells us how spread out the values in a dataset are. It's like a measure of consistency, showing how much variation exists from the average. When we calculated the standard deviation of the tree-ring dates, we found it to be approximately 37.29 years. This indicates the average distance of a tree-ring date from the mean year of 1268 is about 37.29 years.
Calculating the standard deviation involves several steps:
  • First, find the deviation of each data point from the mean.
  • Square each of these deviations to remove negative signs and emphasize larger differences.
  • Sum all these squared deviations.
  • Finally, divide by the number of data points minus one, which gives us the sample variance, and take the square root of that result to find the standard deviation.
Understanding the standard deviation helps us gauge the reliability of our data. A smaller standard deviation implies the data points are closely clustered around the mean, while a larger one suggests more variability.
t-statistic
The t-statistic is a value that helps you determine how far your sample mean is from the population mean, taking into account the variability (standard deviation) and sample size. This is especially useful when dealing with small sample sizes or when the population standard deviation is unknown.
The formula for the t-statistic is:\[t = \frac{\bar{x} - \mu}{s/\sqrt{n}}\]where:
  • \(\bar{x}\) is the sample mean, which is 1268 in our case.
  • \(\mu\) is the population mean, hypothesized to be 1300.
  • \(s\) is the standard deviation, about 37.29 years here.
  • \(n\) is the sample size, which is 10.
Plugging these into the formula gives us a t-statistic of approximately -2.728.
The t-statistic forms the basis for hypothesis testing, allowing us to compare the observed differences to what we might expect purely by chance. In hypothesis testing, this helps us decide whether to reject the null hypothesis.
normal distribution
Normal distribution is a fundamental concept in statistics characterized by the symmetric "bell curve" shape. Many natural phenomena, including tree-ring dates, are approximately normally distributed. This means that most values are centered around the mean, with fewer values appearing as you move away from the mean.
In our exercise, we assume tree-ring dates follow this normal distribution, which allows us to use the t-statistic for hypothesis testing. The properties of a normal distribution鈥攕uch as being symmetrical around the mean and having a peak at the mean鈥攁re essential in calculating probabilities and making predictions about our data set. For small samples, like the one in this exercise with 10 tree-ring dates, using a normal distribution assumption can help model the data's variability. This assumption underpins the steps we follow in hypothesis testing, leading to conclusions about whether the data reflects a different population mean from what we expect.
null hypothesis
The null hypothesis (H_0) is a statement we make about our population, suggesting no effect or no difference exists. It's the default or "starting" assumption in hypothesis testing. In our tree-ring example, the null hypothesis is that the mean tree-ring date is 1300 A.D., i.e., \(H_0: \mu = 1300\). This presumes that the average tree-ring date is not significantly different from 1300.
Hypothesis testing involves:
  • Formulating the null hypothesis.
  • Proposing an alternative hypothesis (H_a), which suggests the opposite, such as "the mean is not 1300".
  • Using a statistical test to see if there's enough evidence to reject the null hypothesis in favor of the alternative.
In our analysis, the calculated t-statistic was compared to the critical value from the t-distribution table. Since the t-statistic (-2.728) was not beyond the critical value of 卤3.25, we could not reject the null hypothesis. This means we didn't find enough evidence to say the mean date differs from 1300 A.D.

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Most popular questions from this chapter

Message mania! A professional employee in a large corporation receives an average of \(\mu=41.7\) e-mails per day. Most of these e-mails are from other employees in the company. Because of the large number of e-mails, employees find themselves distracted and are unable to concentrate when they return to their tasks (Reference: The Wall Street Journal). In an effort to reduce distraction caused by such interruptions, one company established a priority list that all employees were to use before sending an e-mail. One month after the new priority list was put into place, a random sample of 45 employees showed that they were receiving an average of \(\bar{x}=36.2 \mathrm{e}\) -mails per day. The computer server through which the e-mails are routed showed that \(\sigma=18.5\). Has the new policy had any effect? Use a \(5 \%\) level of significance to test the claim that there has been a change (either way) in the average number of e-mails received per day per employee.

The U.S. Department of Transportation, National Highway Traffic Safety Administration, reported that \(77 \%\) of all fatally injured automobile drivers were intoxicated. A random sample of 27 records of automobile driver fatalities in Kit Carson County, Colorado, showed that 15 involved an intoxicated driver. Do these data indicate that the population proportion of driver fatalities related to alcohol is less than \(77 \%\) in Kit Carson County? Use \(\alpha=0.01\).

What is your favorite color? A large survey of countries, including the United States, China, Russia, France, Turkey, Kenya, and others. indicated that most people prefer the color blue. In fact, about \(24 \%\) of the population claim blue as their favorite color. (Reference: Study by \(J .\) Bunge and \(A\). Freeman-Gallant, Statistics Center, Cornell University.) Suppose a random sample of \(n=56\) college students were surveyed and \(r=12\) of them said that blue is their favorite color. Does this information imply that the color preference of all college students is different (either way) from that of the general population? Use \(\alpha=0.05\).

Is there a relationship between confidence intervals and two-tailed hypothesis tests? Let \(c\) be the level of confidence used to construct a confidence interval from sample data. Let \(\alpha\) be the level of significance for a two-tailed hypothesis test. The following statement applies to hypothesis tests of the mean. For a two-tailed hypothesis test with level of significance \(\alpha\) and null hypothesis \(H_{0}: \mu=k\), we reject \(H_{0}\) whenever \(k\) falls outside the \(c=1-\alpha\) confidence interval for \(\mu\) based on the sample data. When \(k\) falls within the \(c=1-\alpha\) confidence interval, we do not reject \(H_{0}\). (A corresponding relationship between confidence intervals and two-tailed hypothesis tests also is valid for other parameters, such as \(p, \mu_{1}-\mu_{2}\), or \(p_{1}-p_{2}\) which we will study in Sections \(9.3\) and \(9.5 .\) ) Whenever the value of \(k\) given in the null hypothesis falls outside the \(c=1-\alpha\) confidence interval for the parameter, we reject \(H_{0} .\) For example, consider a two-tailed hypothesis test with \(\alpha=0.01\) and \(H_{0}: \mu=20 \quad H_{1}: \mu \neq 20\) A random sample of size 36 has a sample mean \(\bar{x}=22\) from a population with standard deviation \(\sigma=4\). (a) What is the value of \(c=1-\alpha\) ? Using the methods of Chapter 8 , construct a \(1-\alpha\) confidence interval for \(\mu\) from the sample data. What is the value of \(\mu\) given in the null hypothesis (i.e., what is \(k)\) ? Is this value in the confidence interval? Do we reject or fail to reject \(H_{0}\) based on this information? (b) Using methods of Chapter 9 , find the \(P\) -value for the hypothesis test. Do we reject or fail to reject \(H_{0}\) ? Compare your result to that of part (a).

If we fail to reject (i.e., "accept") the null hypothesis, does this mean that we have proved it to be true beyond all doubt? Explain your answer.
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