/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 13 Assume that the population of \(... [FREE SOLUTION] | 91Ó°ÊÓ

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Assume that the population of \(x\) values has an approximately normal distribution. Archaeology: Tree Rings At Burnt Mesa Pueblo, the method of tree-ring dating gave the following years A.D. for an archacological excavation site (Bandelier Archaeological Excavation Project: Summer 1990 Excavations at Burnt Mesa Pueblo, edited by Kohler, Washington State University): \(1268 \quad 1316\) \(\begin{array}{cccc}1189 & 1271 & 1267 & 1272\end{array}\) 1275 (a) Use a calculator with mean and standard deviation keys to verify that the sample mean year is \(\bar{x}=1272,\) with sample standard deviation \(s \approx 37\) years. (b) Find a \(90 \%\) confidence interval for the mean of all tree-ring dates from this archaeological site. (c) Interpretation What does the confidence interval mean in the context of this problem?

Short Answer

Expert verified
The 90% confidence interval for the mean tree-ring date is [1245, 1299] AD, indicating the interval within which the true mean likely falls.

Step by step solution

01

Calculate the Sample Mean

Add all the tree-ring dates and divide by the number of dates. The tree-ring dates given are: 1268, 1316, 1189, 1271, 1267, 1272, and 1275. Compute the sum: \ 1268 + 1316 + 1189 + 1271 + 1267 + 1272 + 1275 = 9688. The number of dates is 7. Hence, the sample mean (\(\bar{x}\)) is: \ \(\bar{x} = \frac{9688}{7} = 1272\). This verifies the given mean of 1272.
02

Calculate the Sample Standard Deviation

To calculate the sample standard deviation, use the formula: \ \[ s = \sqrt{\frac{1}{n-1} \sum_{i=1}^{n}(x_i - \bar{x})^2} \] \ Here \(n = 7\), \(\bar{x} = 1272\). Calculate each \((x_i - \bar{x})^2\) and sum them: \ - For 1268: \((1268 - 1272)^2 = 16\) \ - For 1316: \((1316 - 1272)^2 = 1936\) \ - For 1189: \((1189 - 1272)^2 = 6889\) \ - For 1271: \((1271 - 1272)^2 = 1\) \ - For 1267: \((1267 - 1272)^2 = 25\) \ - For 1272: \((1272 - 1272)^2 = 0\) \ - For 1275: \((1275 - 1272)^2 = 9\) \ Sum = 8876 \ Now, compute \(s\): \ \[ s = \sqrt{\frac{8876}{6}} \approx 37 \] \ This confirms the given standard deviation of approximately 37.
03

Calculate the 90% Confidence Interval

Use the formula for the confidence interval: \ \[ \text{CI} = \bar{x} \pm t \frac{s}{\sqrt{n}} \] \ where \(t\) is the t-score from the t-distribution table for 90% confidence and 6 degrees of freedom (\(n-1 = 6\)), approximately 1.943. Substitute the values: \ - \(\bar{x} = 1272\) \ - \(s = 37\) \ - \(n = 7\) \ \[ \text{Margin of Error} = 1.943 \times \frac{37}{\sqrt{7}} \approx 27 \] \ Therefore, the confidence interval is: \ \[ 1272 \pm 27 \] \ which gives \([1245, 1299]\).
04

Interpret the Confidence Interval

The confidence interval \([1245, 1299]\) means that we can be 90% confident that the true mean year for all tree-ring dates from this archaeological site falls within this interval.

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Key Concepts

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

Sample Mean
The sample mean is a way to find the "average" of a set of numbers. Here's how it works:

1. **Add Up the Numbers**: First, you sum all the numbers in your dataset. In our case, the tree-ring dates are added together to make a total of 9688.

2. **Count the Numbers**: Count how many numbers are in your dataset. Here, we have seven dates.

3. **Divide the Sum by the Count**: Take the total you calculated and divide it by the number of numbers. For our tree-ring dates: \[ \bar{x} = \frac{9688}{7} = 1272 \] This calculation gives us the sample mean of 1272 for the tree-ring data.

When you calculate a sample mean, you're finding the central point of your data. It helps us understand the "middle" or typical value of our data points.
Sample Standard Deviation
Standard deviation is important because it tells us how much the values in our dataset vary from the sample mean. To compute it:1. **Subtract the Mean from Each Value**: Take each number in your dataset, subtract the sample mean, and square the result. Squaring ensures that you only get positive values.

2. **Sum the Squared Differences**: Add all these squared values together. For our dataset, the sum is 8876.

3. **Divide by One Less than the Count**: This step accounts for using a sample rather than a whole population: \[ s = \sqrt{\frac{8876}{6}} \approx 37 \] This is called the sample standard deviation, which is about 37 here. It measures the amount of variation, indicating how spread out the dates are around the mean.

The standard deviation can help in understanding if the data points are close together or spread out widely.
t-Distribution
The t-distribution is a fundamental concept when constructing confidence intervals, particularly with small sample sizes. Here’s why we use it:

- **Normal-Like Shape**: The t-distribution looks similar to the normal distribution but has heavier tails, meaning there is more probability in the extremes. This characteristic is useful when dealing with small sample sizes, as it provides a more accurate interval.

- **Degrees of Freedom**: The shape of the t-distribution depends on the degrees of freedom (df), calculated as the sample size minus one (n-1). For our tree-ring dataset, the degrees of freedom are 6.

- **t-Score**: For constructing a confidence interval, we use a t-score from the t-distribution. This score replaces the z-score (used in normal distribution) to accommodate the added variability found in smaller samples. In our example, the t-score for a 90% confidence level and 6 degrees of freedom is approximately 1.943.When we use the t-distribution, we can estimate a range (confidence interval) that we believe contains the true mean. In the tree-ring problem, the interval calculated was \[ 1272 \pm 27 \] This range reflects added variability and gives a more reliable estimate of the true mean tree-ring age.

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Most popular questions from this chapter

Sam computed a \(95 \%\) confidence interval for \(\mu\) from a specific random sample. His confidence interval was \(10.1<\mu<12.2 .\) He claims that the probability that \(\mu\) is in this interval is \(0.95 .\) What is wrong with his claim?

Air Temperature How hot is the air in the top (crown) of a hot air balloon? Information from Ballooning: The Complete Guide to Riding the Winds by Wirth and Young (Random House) claims that the air in the crown should be an average of \(100^{\circ} \mathrm{C}\) for a balloon to be in a state of equilibrium. However, the temperature does not need to be exactly \(100^{\circ} \mathrm{C}\). What is a reasonable and safe range of temperatures? This range may vary with the size and (decorative) shape of the balloon. All balloons have a temperature gauge in the crown. Suppose that 56 readings (for a balloon in equilibrium) gave a mean temperature of \(\bar{x}=97^{\circ} \mathrm{C} .\) For this balloon, \(\sigma \approx 17^{\circ} \mathrm{C}\) (a) Compute a \(95 \%\) confidence interval for the average temperature at which this balloon will be in a steady-state equilibrium. (b) Interpretation If the average temperature in the crown of the balloon goes above the high end of your confidence interval, do you expect that the balloon will go up or down? Explain.

If a \(90 \%\) confidence interval for the difference of means \(\mu_{1}-\mu_{2}\) contains all negative values, what can we conclude about the relationship between \(\mu_{1}\) and \(\mu_{2}\) at the \(90 \%\) confidence level?

In a combined study of northern pike, cutthroat trout, rainbow trout, and lake trout, it was found that 26 out of 855 fish died when caught and released using barbless hooks on flies or lures. All hooks were removed from the fish (Source: \(A\) National Symposium on Catch and Release Fishing, Humboldt State University Press). (a) Let \(p\) represent the proportion of all pike and trout that die (i.e., \(p\) is the mortality rate) when caught and released using barbless hooks. Find a point estimate for \(p\). (b) Find a \(99 \%\) confidence interval for \(p,\) and give a brief explanation of the meaning of the interval. (c) Is the normal approximation to the binomial justified in this problem? Explain.

Basic Computation: Confidence Interval A random sample of size 81 has sample mean 20 and sample standard deviation 3 (a) Check Requirements Is it appropriate to use a Student's \(t\) distribution to compute a confidence interval for the population mean \(\mu ?\) Explain. (b) Find a \(95 \%\) confidence interval for \(\mu\) (c) Interpretation Explain the meaning of the confidence interval you computed.

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