/*! 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 19 The U.S. Geological Survey compi... [FREE SOLUTION] | 91Ó°ÊÓ

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The U.S. Geological Survey compiled historical data about Old Faithful Geyser (Yellowstone National Park) from 1870 to \(1987 .\) Some of these data are published in the book The Story of Old Faithful, by G. D. Marler (Yellowstone Association Press). Let \(x_{1}\) be a random variable that represents the time interval (in minutes) between Old Faithful's eruptions for the years 1948 to \(1952 .\) Based on 9340 observations, the sample mean interval was \(\bar{x}_{1}=63.3\) minutes. Let \(x_{2}\) be a random variable that represents the time interval in minutes between Old Faithful's eruptions for the years 1983 to \(1987 .\) Based on 25,111 observations, the sample mean time interval was \(\bar{x}_{2}=72.1\) minutes. Historical data suggest that \(\sigma_{1}=9.17\) minutes and \(\sigma_{2}=12.67\) minutes. Let \(\mu_{1}\) be the population mean of \(x_{1}\) and let \(\mu_{2}\) be the population mean of \(x_{2}\) (a) Which distribution, normal or Student's \(t,\) do we use to approximate the \(\bar{x}_{1}-\bar{x}_{2}\) distribution? Explain. (b) Compute a \(99 \%\) confidence interval for \(\mu_{1}-\mu_{2}\) (c) Comment on the meaning of the confidence interval in the context of this problem. Does the interval consist of positive numbers only? negative numbers only? a mix of positive and negative numbers? Does it appear (at the \(99 \%\) confidence level) that a change in the interval length between eruptions has occurred? Many geologic experts believe that the distribution of eruption times of Old Faithful changed after the major earthquake that occurred in 1959

Short Answer

Expert verified
Use normal distribution; 99% CI for \(\mu_{1}-\mu_{2}\) calculation shows the difference; if interval is negative or contains zero, no significant change post-1959 earthquake.

Step by step solution

01

Determine the Appropriate Distribution

To decide whether to use the normal distribution or Student's t-distribution, we must consider the sample size and population standard deviation. Since both sample sizes 9340 and 25111 are larger than 30, and we know the population standard deviations \( \sigma_{1} = 9.17 \) and \( \sigma_{2} = 12.67 \), we use the normal distribution according to Central Limit Theorem.
02

Calculate Standard Error of Difference in Means

The standard error of the difference in means is given by the formula: \[ SE_{\bar{x}_{1} - \bar{x}_{2}} = \sqrt{\frac{\sigma_{1}^{2}}{n_{1}} + \frac{\sigma_{2}^{2}}{n_{2}}} \]Substitute the values: \[ SE_{\bar{x}_{1} - \bar{x}_{2}} = \sqrt{\frac{9.17^{2}}{9340} + \frac{12.67^{2}}{25111}} \] Calculate \( SE_{\bar{x}_{1} - \bar{x}_{2}} \).
03

Compute the Z-value for 99% Confidence Interval

For a 99% confidence interval, the Z-value is approximately 2.576 (this corresponds to the area in both tails being 0.01/2 = 0.005).
04

Calculate the Margin of Error

The margin of error \( E \) is calculated as: \[ E = Z \times SE_{\bar{x}_{1} - \bar{x}_{2}} \] Substitute the Z-value and standard error calculated previously to find the margin of error.
05

Construct the Confidence Interval

The 99% confidence interval for \( \mu_{1} - \mu_{2} \) is given by: \[ (\bar{x}_{1} - \bar{x}_{2}) \pm E \] Substitute \( \bar{x}_{1} = 63.3 \), \( \bar{x}_{2} = 72.1 \), and the calculated margin of error \( E \) from Step 4.
06

Interpret the Confidence Interval

Examine the confidence interval to determine if it consists exclusively of positive, exclusively of negative numbers, or a mix. The interval's sign indicates whether there has likely been a change in the mean eruption interval times. A negative interval or one that includes zero suggests no significant change at the 99% confidence level.

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Key Concepts

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

Normal Distribution
In statistics, the normal distribution is a highly important concept, often referred to as the bell curve due to its shape. It is symmetrical and most of the data points lie close to the mean. The tails taper off equally on both sides. When we have large sample sizes, as seen in the Old Faithful eruption data case, the sample means tend to approximate a normal distribution even if the data itself does not. This happens due to the magic of the Central Limit Theorem.

Using the normal distribution in the context of the Old Faithful exercise, we are able to model the difference in average eruption intervals between two different time periods. Since the sample sizes (9340 and 25111) are sufficiently large, the distribution of the sample mean should follow a normal distribution, allowing us to use it to calculate confidence intervals.
Central Limit Theorem
The Central Limit Theorem (CLT) is a cornerstone of statistics. It states that if you take enough samples of a population mean, the distribution of those sample means will be normal (or nearly normal), regardless of the shape of the original data distribution.

In practical terms, for the Old Faithful data, we rely on the CLT to assume normality in the difference of means between the two times the geyser was observed. This is extremely useful because it lets us apply inferential statistical methods, like calculating confidence intervals, even when the data may not inherently be normal, as long as the sample size is large enough.
Standard Error of the Difference in Means
The standard error of the difference in means quantifies how much variability one can expect in the difference between two sample means. It relies on both the standard deviations of the samples and the sizes of these samples.

For the Old Faithful dataset, the calculation of this standard error helps us determine how much uncertainty there is in comparing eruption intervals between the two periods studied. The formula for calculating it is:
  • \[ SE_{\bar{x}_{1} - \bar{x}_{2}} = \sqrt{\frac{\sigma_{1}^{2}}{n_{1}} + \frac{\sigma_{2}^{2}}{n_{2}}} \]
This equation takes into account the variability (standard deviation \( \sigma \)) and the sample sizes \( n \) of each observational period, allowing us to compute the standard error in the mean differences.
Margin of Error
The margin of error allows us to establish a range around the sample mean that we can be fairly certain comprises the true population mean. It reflects the degree of imprecision in our sample estimate.

For Old Faithful's eruption data, the margin of error is derived from multiplying the z-score for a desired confidence level, such as 99%, with the standard error. The calculation provides insight into how much the sample mean might vary from the true mean.

To calculate the margin of error, the following formula is used:
  • \[ E = Z \times SE_{\bar{x}_{1} - \bar{x}_{2}} \]
Where \( Z \) is the z-value corresponding to the confidence level of interest (2.576 for 99%).
This margin can significantly affect how we interpret changes in eruption intervals over time, offering evidence of shifts or stability in eruption patterns.

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Most popular questions from this chapter

In the Focus Problem at the beginning of this chapter, a study was described comparing the hatch ratios of wood duck nesting boxes. Group I nesting boxes were well separated from each other and well hidden by available brush. There were a total of 474 eggs in group I boxes, of which a field count showed about 270 had hatched. Group II nesting boxes were placed in highly visible locations and grouped closely together. There were a total of 805 eggs in group II boxes, of which a field count showed about 270 had hatched. (a) Find a point estimate \(\hat{p}_{1}\) for \(p_{1},\) the proportion of \(\mathrm{cggs}\) that hatched in group I nest box placements. Find a \(95 \%\) confidence interval for \(p_{1}\) (b) Find a point estimate \(\hat{p}_{2}\) for \(p_{2},\) the proportion of eggs that hatched in group II nest box placements. Find a \(95 \%\) confidence interval for \(p_{2}\) (c) Find a \(95 \%\) confidence interval for \(p_{1}-p_{2} .\) Does the interval indicate that the proportion of eggs hatched from group I nest boxes is higher than, lower than, or equal to the proportion of eggs hatched from group II nest boxes? (d) What conclusions about placement of nest boxes can be drawn? In the article discussed in the Focus Problem, additional concerns are raised about the higher cost of placing and maintaining group I nest box placements. Also at issue is the cost efficiency per successful wood duck hatch.

Jobs and productivity! How do retail stores rate? One way to answer this question is to examine annual profits per employee. The following data give annual profits per employee (in units of 1 thousand dollars per employee) for companies in retail sales. (See reference in Problem \(23 .\) ) Companies such as Gap, Nordstrom, Dillards, JCPenney, Sears, Wal-Mart, Office Depot, and Toys " \(\mathrm{A}\) " Us are included. Assume \(\sigma \approx 3.8\) thousand dollars. $$\begin{array}{rrrrrrrrrrrr}4.4 & 6.5 & 4.2 & 8.9 & 8.7 & 8.1 & 6.1 & 6.0 & 2.6 & 2.9 & 8.1 & -1 . \\\11.9 & 8.2 & 6.4 & 4.7 & 5.5 & 4.8 & 3.0 & 4.3 & -6.0 & 1.5 & 2.9 & 4 . \\ -1.7 & 9.4 & 5.5 & 5.8 & 4.7 & 6.2 & 15.0 & 4.1 & 3.7 & 5.1 & 4.2 &\end{array}$$ (a) Use a calculator or appropriate computer software to verify that, for the preceding data, \(\bar{x} \approx 5.1 .\) (b) Let us say that the preceding data are representative of the entire sector of retail sales companies. Find an \(80 \%\) confidence interval for \(\mu,\) the average annual profit per employee for retail sales. (c) Interpretation Let us say that you are the manager of a retail store with a large number of employees. Suppose the annual profits per employee are less than 3 thousand dollars per employee. Do you think this might be low compared with other retail stores? Explain by referring to the confidence interval you computed in part (b). (d) Interpretation Suppose the annual profits are more than 6.5 thousand dollars per employee. As store manager, would you feel somewhat better? Explain by referring to the confidence interval you computed in part (b). (e) Repeat parts \((b),(c),\) and (d) for a \(95 \%\) confidence interval.

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

Lorraine computed a confidence interval for \(\mu\) based on a sample of size \(41 .\) Since she did not know \(\sigma,\) she used \(s\) in her calculations. Lorraine used the normal distribution for the confidence interval instead of a Student's \(t\) distribution. Was her interval longer or shorter than one obtained by using an appropriate Student's \(t\) distribution? Explain.

At Community Hospital, the burn center is experimenting with a new plasma compress treatment. A random sample of \(n_{1}=316\) patients with minor burns received the plasma compress treatment. Of these patients, it was found that 259 had no visible scars after treatment. Another random sample of \(n_{2}=419\) patients with minor burns received no plasma compress treatment. For this group, it was found that 94 had no visible scars after treatment. Let \(p_{1}\) be the population proportion of all patients with minor burns receiving the plasma compress treatment who have no visible scars. Let \(p_{2}\) be the population proportion of all patients with minor burns not receiving the plasma compress treatment who have no visible scars. (a) Can a normal distribution be used to approximate the \(\hat{p}_{1}-\hat{p}_{2}\) distribution? Explain. (b) Find a \(95 \%\) confidence interval for \(p_{1}-p_{2}\) (c) Explain the meaning of the confidence interval found in part (b) in the context of the problem. Does the interval contain numbers that are all positive? all negative? both positive and negative? At the \(95 \%\) level of confidence, does treatment with plasma compresses seem to make a difference in the proportion of patients with visible scars from minor burns?

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