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Chapter 7: Q57E (page 418)

Oxygen bubble velocity in a purification process. Refer to the Chemical Engineering Research and Design (March 2013) study of a method of purifying nuclear fuel waste, Exercise 6.35 (p. 349). Recall that the process involves oxidation in molten salt and tends to produce oxygen bubbles with a rising velocity. To monitor the process, the researchers collected data on bubble velocity (measured in meters per second) for a random sample of 25 photographic bubble images. These data (simulated) are reproduced in the accompanying table. When oxygen is inserted into the molten salt at a rate (called the sparging rate) of \(3.33 \times {10^{ - 6}}\) , the researchers discovered that the true mean bubble rising velocity \(\mu = .338\)

a. Conduct a test of hypothesis to determine if the true mean bubble rising velocity for the population from which the sample is selected is\(\mu = .338\)Use\(\alpha = .10\).

0.275 0.261 0.209 0.266 0.265 0.312 0.285 0.317 0.229 0.251 0.256 0.339 0.213 0.178 0.217 0.307 0.264 0.319 0.298 0.169 0.342 0.270 0.262 0.228 0.22

Short Answer

Expert verified

a) The test statistic is -9.60.

Step by step solution

01

Constructing the hypothesis

The null and alternative hypothesis are given by

\(\begin{aligned}{H_0}:\mu = .338\\{H_a}:\mu \ne .338\end{aligned}\)

02

Computing the mean and standard deviation

a) The mean and standard deviation is calculated as

\(\begin{aligned}\bar x &= \frac{\begin{aligned}{l}0.275 + 0.261 + 0.209 + 0.266 + 0.265 + 0.312 + 0.285 + 0.317 + 0.229\\ + 0.251 + 0.256 + 0.399 + 0.213 + 0.178 + 0.217 + 0.307 + 0.264 + 0.319\\ + 0.298 + 0.169 + 0.342 + 0.270 + 0.262 + 0.228 + 0.22\end{aligned}}{{25}}\\ &= \frac{{6.016}}{{25}}\\ &= 0.24\end{aligned}\)

\(\begin{aligned}sd &= \sqrt {\frac{\begin{aligned}{l}{\left( {0.035} \right)^2} + {\left( {0.021} \right)^2} + {\left( { - 0.031} \right)^2} + {\left( {0.026} \right)^2} + {\left( {0.025} \right)^2} + {\left( {0.072} \right)^2} + \\{\left( {0.045} \right)^2} + {\left( {0.077} \right)^2} + {\left( { - 0.011} \right)^2} + {\left( {0.011} \right)^2} + {\left( {0.016} \right)^2} + {\left( {0.099} \right)^2} + \\{\left( { - 0.027} \right)^2} + {\left( { - 0.062} \right)^2} + {\left( { - 0.023} \right)^2} + {\left( {0.067} \right)^2} + {\left( {0.024} \right)^2} + {\left( {0.079} \right)^2} + \\{\left( {0.058} \right)^2} + {\left( { - 0.071} \right)^2} + {\left( {0.102} \right)^2} + {\left( {0.03} \right)^2} + {\left( {0.022} \right)^2} + {\left( { - 0.012} \right)^2} + {\left( {0.02} \right)^2}\end{aligned}}{{24}}} \\ &= \sqrt {\frac{\begin{aligned}{l}0.001225 + 0.000441 + 0.000961 + 0.000676 + 0.000625 + 0.005184 + \\0.002025 + 0.005929 + 0.000121 + 0.000121 + 0.000256 + 0.009801 + \\0.000729 + 0.003844 + 0.000529 + 0.004489 + 0.000576 + 0.006241 + \\0.003364 + 0.005041 + 0.010404 + 0.0009 + 0.000484 + 0.000144 + 0.0004\end{aligned}}{{24}}} \\ &= \sqrt {\frac{{0.06451}}{{24}}} \\ &= 0.051\end{aligned}\)

Therefore, the mean and standard deviation are 0.24 and 0.051.

03

Test statistic

The test statistic is calculated as

\(\begin{aligned}t &= \frac{{\bar x - \mu }}{{\frac{s}{{\sqrt n }}}}\\ &= \frac{{0.24 - .338}}{{\frac{{0.051}}{{\sqrt {25} }}}}\\ &= \frac{{ - 0.098}}{{0.0102}}\\ &= - 9.60\end{aligned}\)

Therefore, the test statistic is -9.60.

Degrees of freedom are

\(\begin{aligned}df &= n - 1\\ &= 25 - 1\\ &= 24\end{aligned}\)

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.

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Most popular questions from this chapter

In a test of the hypothesis \({H_0}:\mu = 10\) versus \({H_a}:\mu \ne 10\), a sample of n = 50 observations possessed mean \(\bar x = 10.7\) and standard deviation s = 3.1. Find and interpret the p-value for this test.

A two-tailed test was conducted with the null and alternative hypotheses stated being \({H_0}:p = .69\) against \({H_a}:p \ne .69\), respectively, with a sample size of 150. The test results were z = -.98 and two-tailed p-value = .327

a. Determine the conditions required for a valid large sample test

Americans鈥� favorite sport. The Harris Poll (December 2013) conducted an online survey of American adults to determine their favorite sport. Your friend believes professional (National Football League [NFL]) football鈥攚ith revenue of about $13 billion per year鈥攊s the favorite sport for 40% of American adults. Specify the null and alternative hypotheses for testing this belief. Be sure to identify the parameter of interest.

Authorizing computer users with palm prints. Access to computers, email, and Facebook accounts is achieved via a password鈥攁 collection of symbols (usually letters and numbers) selected by the user. One problem with passwords is that persistent hackers can create programs that enter millions of combinations of symbols into a target system until the correct password is found. An article in IEEE Pervasive Computing (October-December 2007) investigated the effectiveness of using palm prints to identify authorized users. For example, a system developed by Palmguard, Inc. tests the hypothesis

\({H_0}\): The proposed user is authorized

\({H_a}\): The proposed user is unauthorized

by checking characteristics of the proposed user鈥檚 palm print against those stored in the authorized users鈥� data bank.

a. Define a Type I error and Type II error for this test. Which is the more serious error? Why?

Beta value of a stock. The 鈥渂eta coefficient鈥� of a stock is a measure of the stock鈥檚 volatility (or risk) relative to the market as a whole. Stocks with beta coefficients greater than 1 generally bear the greater risk (more volatility) than the market, whereas stocks with beta coefficients less than 1 are less risky (less volatile) than the overall market (Alexander, Sharpe, and Bailey, Fundamentals of Investments, 2000). A random sample of 15 high-technology stocks was selected at the end of 2009, and the mean and standard deviation of the beta coefficients were calculated: x=1.23s = .37.

a. Set up the appropriate null and alternative hypotheses to test whether the average high-technology stock is riskier than the market as a whole.

b. Establish the appropriate test statistic and rejection region for the test. Use a=0.10.

c. What assumptions are necessary to ensure the validity of the test?

d. Calculate the test statistic and state your conclusion.

e. What is the approximate p-value associated with this test? Interpret it.

f. Conduct a test to determine if the variance of the stock beta values differs from .15. Use a=0.05.
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