/*! 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 43 How much oil will ultimately be ... [FREE SOLUTION] | 91Ó°ÊÓ

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How much oil will ultimately be produced by wells in a given field is key information in deciding whether to drill more wells. Here are the estimated total amounts of oil recovered from 64 wells in the Devonian Richmond Dolomite area of the Michigan basin, in thousands of barrels: \({ }^{29}\) $$ \begin{array}{llllllll} 21.7 & 53.2 & 46.4 & 42.7 & 50.4 & 97.7 & 103.1 & 51.9 43.4 & 69.5 & 156.5 & 34.6 & 37.9 & 12.9 & 2.5 & 31.4 \\ 79.5 & 26.9 & 18.5 & 14.7 & 32.9 & 196 & 24.9 & 118.2 \\ 82.2 & 35.1 & 47.6 & 54.2 & 63.1 & 69.8 & 57.4 & 65.6 \\ 56.4 & 49.4 & 44.9 & 34.6 & 92.2 & 37.0 & 58.8 & 21.3 \\ 36.6 & 64.9 & 14.8 & 17.6 & 29.1 & 61.4 & 38.6 & 32.5 \\ 12.0 & 28.3 & 204.9 & 44.5 & 10.3 & 37.7 & 33.7 & 81.1 \\ 12.1 & 20.1 & 30.5 & 7.1 & 10.1 & 18.0 & 3.0 & 2.0 \end{array} $$ Take these wells to be an SRS of wells in this area. (a) Give a \(95 \%\) confidence interval for the mean amount of oil recovered from all wells in this area. (b) Make a graph of the data. The distribution is very skewed, with several high outliers. A computer-intensive method that gives accurate confidence intervals without assuming any specific shape for the distribution gives a \(95 \%\) confidence interval of \(40.28\) to \(60.32\). How does the \(t\) interval compare with this? Should the \(t\) procedures be used with these data?

Short Answer

Expert verified
The t-interval is roughly similar, but data skewness and outliers suggest using alternative methods may be better.

Step by step solution

01

Organize the Data

We begin by listing the 64 oil production volumes and calculating the necessary statistics: the sample mean \( \bar{x} \) and the sample standard deviation \( s \).
02

Calculate Sample Mean

The sample mean \( \bar{x} \) is calculated as follows: \( \bar{x} = \frac{\sum_{i=1}^{n} x_i}{n} \). Substitute the values provided and compute the mean.
03

Calculate Sample Standard Deviation

The sample standard deviation \( s \) is given by \( s = \sqrt{\frac{1}{n-1} \sum_{i=1}^{n} (x_i - \bar{x})^2} \). Substitute the values, compute \( \bar{x} \) and determine \( s \).
04

Determine Degrees of Freedom

In this case, the degrees of freedom (df) is \( n - 1 \) where \( n = 64 \). Thus, \( df = 63 \).
05

Calculate Standard Error

The standard error of the mean (SE) is given by \( \text{SE} = \frac{s}{\sqrt{n}} \). Calculate this value using the computed \( s \) and \( n = 64 \).
06

Find Critical Value for t-Distribution

For a 95% confidence interval and 63 degrees of freedom, find the critical value \( t^* \) using a t-table. This value \( t^* \) is typically around 2.00 for large sample sizes.
07

Construct Confidence Interval

The 95% confidence interval is given by \( \bar{x} \pm t^* \times \text{SE} \). Substitute the values of \( \bar{x} \), \( t^* \), and \( \text{SE} \) to find the interval.
08

Compare with Alternative Method

The alternative 95% confidence interval from the bootstrapped method is \( [40.28, 60.32] \). Compare this with the \( t \)-based interval and assess the appropriateness given the data's skewness and presence of outliers.

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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 an average number that represents the central value of a set of numbers. It's particularly useful in identifying a trend or average case in a pool of data. In a set of numbers like the amount of oil produced by various wells as given in the exercise, the sample mean helps us understand the typical amount of oil a well produces. To calculate the sample mean, you add up all the observed values and then divide by the number of values. If you're given 64 wells, their production numbers would be added together and then divided by 64. The formula is:\[ \bar{x} = \frac{\sum_{i=1}^{n} x_i}{n} \]Where:
  • \( \sum_{i=1}^{n} x_i \) is the sum of all production numbers from the wells.
  • \( n \) is the number of wells, which is 64 in this case.
  • \( \bar{x} \) represents the sample mean.
This calculation provides a significant baseline for further statistical analysis.
Standard Deviation
Standard deviation is a measure that helps us understand the amount of variation or dispersion in a set of values. It shows how much the individual values typically differ from the mean. In practical terms, for the oil wells' production, it tells you whether all wells produce a similar amount of oil or if some wells produce much more or much less. The formula for standard deviation in a sample is:\[ s = \sqrt{\frac{1}{n-1} \sum_{i=1}^{n} (x_i - \bar{x})^2} \]Where:
  • \( s \) is the sample standard deviation.
  • \( x_i \) represents each production value.
  • \( \bar{x} \) is the sample mean.
  • \( n \) is the sample size.
When values are spread out over a wider range, the standard deviation becomes larger. In assessing oil production, a high standard deviation might mean inconsistency among the wells’ production levels.
t Distribution
The t distribution is a probability distribution that is symmetric and bell-shaped, similar to the normal distribution but with heavier tails. This makes it useful when dealing with smaller sample sizes or when the population standard deviation is unknown.In the context of confidence intervals, the t distribution helps determine the likelihood of a sample mean being close to the true population mean. When constructing a confidence interval for the mean, especially with a small sample, the t distribution is preferred because it compensates for the potential underestimation of variability:- As sample sizes increase, the t distribution approaches the normal distribution.The critical value from the t distribution, denoted \( t^* \), is chosen based on the desired level of confidence (e.g., 95%) and the degrees of freedom. This critical value is used in forming the confidence interval for a given sample mean.
Degrees of Freedom
Degrees of freedom (df) is a concept that reflects the number of values in a calculation that are free to vary. In statistics, it is crucial for understanding how spread out data is in trials or samples and is often used in conjunction with variance estimates.For the sample variance or standard deviation:- The formula changes from using \( n \) to \( n-1 \) to allow for this variation, making the calculation an unbiased estimator.In the oil wells example, with 64 samples, the degrees of freedom would be 64 minus 1, which equals 63. This concept feeds into the computation of the t statistic, where:
  • The larger the degrees of freedom, the closer the t distribution is to the normal distribution, which provides more reliable results with larger samples.
  • For determining the t critical value, you use the degrees of freedom to reference a "t-table" to find the required values for your confidence interval calculations.
Understanding degrees of freedom helps ensure that statistical tests are fair and results are accurate, providing confidence in the conclusions drawn from the data analyses.

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Most popular questions from this chapter

Twenty-nine college students, identified as having a positive attitude about Mitt Romney as compared to Barack Obama in the 2012 presidential election, were asked to rate how trustworthy the face of Mitt Romney appeared, as represented in their mental image of Mitt Romney's face. Ratings were on a scale of 0 to 7 , with 0 being "not at all trustworthy" and 7 being "extremely trustworthy." Here are the 29 ratings: \({ }^{19}\) \(\begin{array}{llllllllll}2.6 & 3.2 & 3.7 & 3.3 & 3.4 & 3.6 & 3.7 & 3.8 & 3.9 & 4.1\end{array}\) \(\begin{array}{lllllllllll}4.2 & 4.9 & 5.7 & 4.2 & 3.9 & 3.2 & 4.5 & 5.0 & 5.0 & 4.6\end{array}\) \(\begin{array}{lllllllll}4.6 & 3.9 & 3.9 & 5.3 & 2.8 & 2.6 & 3.0 & 3.3 & 3.7\end{array}\) (a) Suppose we can consider this an SRS of all U.S. college students. Make a stemplot. Is there any sign of major deviation from Normality? (b) Give a \(95 \%\) confidence interval for the mean rating. (c) Is there significant evidence at the \(5 \%\) level that the mean rating is greater than \(3.5\) (a neutral rating)?

What critical value \(t^{*}\) from Table \(C\) would you use for a confidence interval for the mean of the population in each of the following situations? (If you have access to software, you can use software to determine the critical values.) (a) A \(90 \%\) confidence interval based on \(n=2\) observations (b) A \(95 \%\) confidence interval from an SRS of 20 observations (c) A 99\% confidence interval from a sample of size 1001

We prefer the \(t\) procedures to the z procedures for inference about a population mean because (a) \(z\) requires that you know the observations are from a Normal population, while \(t\) does not. (b) \(z\) requires that you know the population standard deviation \(\sigma\), while \(t\) does not. (c) \(z\) requires that you can regard your data as an SRS from the population, while \(t\) does not.

In a study of exhaust emissions from school buses, the pollution intake by passengers was determined for a sample of nine school buses used in the Southern California Air Basin. The pollution intake is the amount of exhaust emissions, in grams per person, that would be inhaled while traveling on the bus during its usual 18-mile trip on congested freeways from South Central LA to a magnet school in West LA. (As a reference, the average intake of motor emissions of carbon monoxide in the LA area is estimated to be about \(0.000046\) gram per person.) Here are the amounts for the nine buses when driven with the windows open: 20 \(\begin{array}{lllllllll}1.15 & 0.33 & 0.40 & 0.33 & 1.35 & 0.38 & 0.25 & 0.40 & 0.35\end{array}\) (a) Make a stemplot. Are there outliers or strong skewness that would preclude use of the \(t\) procedures? (b) A good way to judge the effect of outliers is to do your analysis twice, once with the outliers and a second time without them. Give two \(90 \%\) confidence intervals, one with all the data and one with the outliers removed, for the mean pollution intake among all school buses used in the Southern California Air Basin that travel the route investigated in the study. (c) Compare the two intervals in part (b). What is the most important effect of removing the outliers?

Velvetleaf is a particularly annoying weed in corn fields. It produces lots of seeds, and the seeds wait in the soil for years until conditions are right. How many seeds do velvetleaf plants produce? Here are counts from 28 plants that came up in a corn field when no herbicide was used: 28 \(\begin{array}{ll}245025042114111021378015 & 1623\end{array}\) \(\begin{array}{lllllllllll}721 & 863 & 1136 & 2819 & 1911 & 2101 & 1051 & 218 & 1711 & 164\end{array}\) \(22283635973 \quad 105019611809 \quad 130 \quad 880\) We would like to give a confidence interval for the mean number of seeds produced by velvetleaf plants. Alas, the \(t\) interval can't be safely used for these data. Why not?

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