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One of the most famous large fractures (cracks) in the earth's crust is the San Andreas fault in California. A geologist attempting to study the movement of the earth's crust at a particular location found many fractures in the local rock structure. In an attempt to determine the mean angle of the breaks, she sampled \(n=50\) fractures and found the sample mean and standard deviation to be \(39.8^{\circ}\) and \(17.2^{\circ},\) respectively. Estimate the mean angular direction of the fractures and find the margin of error for your estimate.

Short Answer

Expert verified
Question: Estimate the mean angular direction of the fractures and the margin of error using the provided sample statistics and a 95% confidence interval. Answer: The estimated mean angular direction of the fractures lies between approximately 34.92° and 44.68° with a margin of error of approximately 4.88° at a 95% confidence level.

Step by step solution

01

Determine the Degrees of Freedom

We need to determine the degrees of freedom for the t-distribution, which is given by \(df = n - 1\), where \(n\) is the number of samples. In this case, \(n=50\). \(df = n - 1 = 50 - 1 = 49\)
02

Choose the Confidence Level

We are not specifically given a confidence level to use, so we will use a commonly accepted value: \(95\%\). This corresponds to an alpha level (\(\alpha\)) of \(0.05\).
03

Find the t-score

To determine the t-score, first, look up the critical value corresponding to the chosen confidence level and degrees of freedom (in this case, \(df=49\)). The critical value is the t-score that corresponds to the middle \(95\%\) of the t-distribution. For \(df=49\) and a \(95\%\) confidence interval, the critical value (\(t^*\)) is approximately \(2.0096\).
04

Calculate the Margin of Error

To calculate the margin of error, we use the following formula: \(E = t^* \frac{s}{\sqrt{n}}\) Where \(E\) is the margin of error, \(t^*\) is the critical value, \(s\) is the sample standard deviation, and \(n\) is the number of samples. In this case, \(t^*=2.0096\), \(s=17.2^\circ\), and \(n=50\). \(E = 2.0096\frac{17.2}{\sqrt{50}} \approx 4.88\)
05

Calculate the Confidence Interval for the Mean

Now that we have the margin of error, we can calculate the lower and upper bounds of our \(95\%\) confidence interval: Lower Bound: \(\bar{x}-E = 39.8 - 4.88 \approx 34.92^\circ\) Upper Bound: \(\bar{x}+E = 39.8 + 4.88 \approx 44.68^\circ\) Thus, we can estimate at \(95\%\) confidence level that the true mean angular direction of the fractures lies between approximately \(34.92^\circ\) and \(44.68^\circ\). And the margin of error is approximately \(4.88^\circ\).

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Key Concepts

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

Degrees of Freedom
Degrees of freedom in statistics play a crucial role when it comes to analyzing a sample. You can think of degrees of freedom as the number of values in a calculation that are free to vary. It’s a concept used to ensure that we get an accurate representation of the data we are studying.
In the context of a t-distribution—a type of distribution used when the sample size is small or the population standard deviation is unknown—the degrees of freedom are essential for determining the shape of the distribution. This is calculated by the formula:
  • \( df = n - 1 \)
Here, \( n \) represents the number of observations in our sample. In our example, with 50 fractures sampled, the degrees of freedom would be 49. A crucial factor since it directly influences the critical value we choose from the t-distribution table.In most cases, as the sample size increases, the t-distribution approaches a normal distribution, and the degrees of freedom become a key factor in narrowing down the range in which our true population parameter lies.
Confidence Interval
A confidence interval gives us a range of values which we expect the true population parameter to fall within. It's like saying, "We’re pretty sure the average angle of the fractures is between these two numbers." This interval is bracketed by an upper and a lower limit that specifies with a certain level of confidence how close our sample mean is to the actual population mean.
The confidence level is an indicator of how certain we are about this. Typically, a 95% confidence level is used, which means that if we were to take 100 different samples and compute a confidence interval for each, we would expect about 95 of them to contain the true population mean.
In our scenario, the 95% confidence interval calculated (from a mean of \(39.8^\circ\) with a margin of error of \(4.88^\circ\)) indicates that we are confident that the true mean angle of the fractures lies somewhere between \(34.92^\circ\) and \(44.68^\circ\). This shows how our measurement sits within the broader context of our study.
Margin of Error
The margin of error offers insight into the precision of our sample estimate. It is the range in which the true population parameter is expected to differ from the sample estimate—it's like a buffer zone on either side. The smaller the margin of error, the more accurate our estimate is expected to be.
To calculate the margin of error, we utilize the formula:
  • \( E = t^* \frac{s}{\sqrt{n}} \)
In this formula:
  • \( E \) is the margin of error.
  • \( t^* \) is the critical value from the t-distribution.
  • \( s \) is the sample standard deviation.
  • \( n \) is the sample size.
Using the numbers from our exercise (sample standard deviation \(s = 17.2^\circ\), \( t^* \approx 2.0096 \), \( n = 50 \)), we find the margin of error to be approximately \(4.88^\circ\). This means that the true mean angle could reasonably vary by up to \(4.88^\circ\) from our sample mean of \(39.8^\circ\). Effectively, the margin of error gives us a clearer picture of how reliable and close our sample statistic is to the actual population parameter.
Sample Standard Deviation
Sample standard deviation is a measure of the amount of variability or dispersion in a sample data set. It tells us how much the data points in the sample deviate from the mean of the sample.
In our scenario, the sample standard deviation is \(17.2^\circ\). A higher standard deviation indicates that the data points are spread out over a wider range of values, whereas a lower standard deviation would indicate that they are closer to the mean.
The sample standard deviation is crucial because it’s used in the calculation of both the margin of error and the confidence interval. This variability directly impacts the width of the confidence interval; the more variability in the data, the wider the interval.
  • It’s important to note that the sample standard deviation is an estimate of the population standard deviation.
  • This is particularly useful when the standard deviation of the entire population is unknown.
Understanding this concept is key in assessing the reliability of a study's findings and how broadly we can apply the results beyond the sample.

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Most popular questions from this chapter

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