/*! 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} Q60E Crude oil biodegradation. Refer ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 7: Q60E (page 418)

Crude oil biodegradation. Refer to the Journal of Petroleum Geology (April 2010) study of the environmental factors associated with biodegradation in crude oil reservoirs, Exercise 6.38 (p. 350). Recall that 16 water specimens were randomly selected from various locations in a reservoir on the floor of a mine and that the amount of dioxide (milligrams/liter)鈥攁 measure of biodegradation鈥攁s well as presence of oil were determined for each specimen. These data are reproduced in the accompanying table.

a. Conduct a test to determine if the true mean amount of dioxide present in water specimens that contained oil was less than 3 milligrams/liter. Use\(\alpha = .10\)

Short Answer

Expert verified
  1. The test statistic is -14.14
    We reject the null hypothesis

Step by step solution

01

Specifying the hypothesis

The null and alternative hypothesis are given by

\(\begin{aligned}{H_0}:\mu = 3\\{H_a}:\mu < 3\end{aligned}\)

02

calculating the mean and standard deviation

a) The mean and standard deviation is calculated as

\(\begin{aligned}\bar x &= \frac{{0.5 + 1.3 + 0.4 + 0.2 + 0.5 + 0.2}}{6}\\ &= \frac{{3.1}}{6}\\ &= 0.51\end{aligned}\)

\(\begin{aligned}sd &= \sqrt {\frac{{{{\left( { - 0.01} \right)}^2} + {{\left( {0.79} \right)}^2} + {{\left( { - 0.11} \right)}^2} + {{\left( { - 0.31} \right)}^2} + {{\left( { - 0.01} \right)}^2} + {{\left( { - 0.31} \right)}^2}}}{5}} \\ &= \sqrt {\frac{{0.0001 + .6241 + 0.0121 + 0.0961 + 0.0001 + 0.0961}}{5}} \\ &= \sqrt {\frac{{0.9375}}{5}} \\ &= 0.433\end{aligned}\)

Therefore, the mean and standard deviation are 0.51 and 0.433.

03

Test statistic

The test statistic is calculated as

\(\begin{aligned}t &= \frac{{\bar x - \mu }}{{\frac{s}{{\sqrt n }}}}\\ &= \frac{{0.51 - 3}}{{\frac{{0.433}}{{\sqrt 6 }}}}\\ &= \frac{{ - 2.49}}{{0.176}}\\ &= - 14.14\end{aligned}\)

Therefore, the test statistic is -14.14.

Degrees of freedom are

\(\begin{aligned}df &= n - 1\\ &= 6 - 1\\ &= 5\end{aligned}\)

For \(\alpha = .10\,and\,df = 5\)

The tabulated value is -1.47.

The calculated value is less than the tabulated value.

Therefore, we reject the null hypothesis.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Play Golf America program. The Professional Golf Association (PGA) and Golf Digest have developed the Play Golf America program, in which teaching professionals at participating golf clubs provide a free 10-minute lesson to new customers. According to Golf Digest, golf facilities that participate in the program gain, on average, \(2,400 in greens fees, lessons, or equipment expenditures. A teaching professional at a golf club believes that the average gain in greens fees, lessons, or equipment expenditures for participating golf facilities exceeds \)2,400.

a. In order to support the claim made by the teaching professional, what null and alternative hypotheses should you test?

b. Suppose you select伪 = 0.05. Interpret this value in the words of the problem.

c. For 伪 = 0.05, specify the rejection region of a large sample test.

For each of the following situations, determine the p-value and make the appropriate conclusion.

a.\({H_0}:\mu \le 25\),\({H_a}:\mu > 25\),\(\alpha = 0.01\),\(z = 2.02\)

b.\({H_0}:\mu \ge 6\),\({H_a}:\mu < 6\),\(\alpha = 0.05\),\(z = - 1.78\)

c.\({H_0}:\mu = 110\),\({H_a}:\mu \ne 110\),\(\alpha = 0.1\),\(z = - 1.93\)

d. \({H_0}:\mu = 10\), \({H_a}:\mu \ne 10\), \(\alpha = 0.05\), \(z = 1.96\)

A border protection avatar. The National Center for Border Security and Protection has developed the "Embodied Avatar"鈥攁 kiosk with a computer-animated border guard that uses artificial intelligence to scan passports, check fingerprints, read eye pupils, and asks questions of travellers crossing the U.S. border. (National Defense Magazine, February 2014.) Based on field tests, the avatar's developer claims that the avatar can detect deceitful speech correctly 75% of the time.

a. Identify the parameter of interest.

b. Give the null and alternative hypotheses for testing the claim made by the avatar's developer.

c. Describe a Type I error in the words of the problem.

d. Describe a Type II error in the words of the problem

A random sample of 64 observations produced the following summary statistics: \(\bar x = 0.323\) and \({s^2} = 0.034\).

a. Test the null hypothesis that\(\mu = 0.36\)against the alternative hypothesis that\(\mu < 0.36\)using\(\alpha = 0.10\).

b. Test the null hypothesis that \(\mu = 0.36\) against the alternative hypothesis that \(\mu \ne 0.36\) using \(\alpha = 0.10\). Interpret the result.

If the rejection of the null hypothesis of a particular test would cause your firm to go out of business, would you want ato be small or large? Explain
See all solutions

Recommended explanations on Math Textbooks

Decision Maths

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Pure Maths

Read Explanation

Geometry

Read Explanation

Discrete Mathematics

Read Explanation

Applied Mathematics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.